Selina Concise Solutions for ICSE Class 6 Mathematics Chapter 22 Simple Linear Equations

Important Points

1. Simple Equations: A mathematical statement that shows two expressions are equal is called a simple equation.

2. Properties of Simple Equation:

(i) If we add the same number to both sides of a simple equation, both sides still stay equal.
For Example:
\( x = 6 \)
\( \implies x + a = 6 + a \) [Adding a to both sides]

(ii) If we subtract the same number from both sides of a simple equation, both sides still stay equal.
For Example:
\( x = 6 \)
\( \implies x - a = 6 - a \) [Subtracting a from both sides]

(iii) If we multiply both sides of an equation by the same number, the products remain equal.
For Example:
\( x = 6 \)
\( \implies a \times x = a \times 6 \) i.e. \( ax = 6a \) [Multiplying both sides by a]

(iv) If we divide both sides of a simple equation by the same non-zero number, the quotients remain equal.
For Example:
\( x = 6 \)
\( \implies \frac{x}{a} = \frac{6}{a} \) [Dividing both sides by a]

Exercise 22(A)

Question 1. (i) Solve: \( x + 2 = 6 \)
Answer:
Given equation:
\( x + 2 = 6 \)
Subtracting 2 from both sides:
\( \implies x = 6 - 2 \)
\( \implies x = 4 \)
In simple words: To find the value of x, we move the 2 to the other side of the equals sign. When +2 moves over, it becomes -2, so we do 6 minus 2 to get 4.

Exam Tip: Remember to change a plus sign to a minus sign when you move a number to the other side of the equals mark.

Question 1. (ii) Solve: \( x + 6 = 2 \)
Answer:
Given equation:
\( x + 6 = 2 \)
Subtracting 6 from both sides:
\( \implies x = 2 - 6 \)
\( \implies x = -4 \)
In simple words: We move 6 to the other side where it becomes minus 6. Since 2 minus 6 is less than zero, our answer is minus 4.

Exam Tip: Subtracting a larger positive number from a smaller one always gives a negative result.

Question 1. (iii) Solve: \( y + 8 = 5 \)
Answer:
Given equation:
\( y + 8 = 5 \)
Subtracting 8 from both sides:
\( \implies y = 5 - 8 \)
\( \implies y = -3 \)
In simple words: Move 8 to the other side, which makes it minus 8. Subtracting 8 from 5 gives us minus 3.

Exam Tip: Double check your signs when subtracting a larger number from a smaller one to avoid careless mistakes.

Question 1. (iv) Solve: \( x + 4 = -3 \)
Answer:
Given equation:
\( x + 4 = -3 \)
Subtracting 4 from both sides:
\( \implies x = -3 - 4 \)
\( \implies x = -7 \)
In simple words: We move 4 to the other side, turning it into minus 4. Since both numbers are now negative, we add them together to get minus 7.

Exam Tip: When you add two negative numbers, the final answer remains negative.

Question 1. (v) Solve: \( y + 2 = -8 \)
Answer:
Given equation:
\( y + 2 = -8 \)
Subtracting 2 from both sides:
\( \implies y = -8 - 2 \)
\( \implies y = -10 \)
In simple words: Moving 2 to the right side makes it minus 2. Two negative numbers together add up to a bigger negative number, so we get minus 10.

Exam Tip: Treat negative numbers like debt; if you owe 8 and then owe 2 more, you owe 10 in total.

Question 1. (vi) Solve: \( b + 2.5 = 4.2 \)
Answer:
Given equation:
\( b + 2.5 = 4.2 \)
Subtracting 2.5 from both sides:
\( \implies b = 4.2 - 2.5 \)
\( \implies b = 1.7 \)
In simple words: We move 2.5 to the other side and subtract it from 4.2 to get 1.7.

Exam Tip: Align the decimal points carefully when you subtract decimal numbers.

Question 1. (vii) Solve: \( p + 4.6 = 8.5 \)
Answer:
Given equation:
\( p + 4.6 = 8.5 \)
Subtracting 4.6 from both sides:
\( \implies p = 8.5 - 4.6 \)
\( \implies p = 3.9 \)
In simple words: Move 4.6 to the other side and subtract it from 8.5 to find that p is 3.9.

Exam Tip: Make sure to borrow correctly across the decimal point when doing subtraction.

Question 1. (viii) Solve: \( y + 3.2 = -6.5 \)
Answer:
Given equation:
\( y + 3.2 = -6.5 \)
Subtracting 3.2 from both sides:
\( \implies y = -6.5 - 3.2 \)
\( \implies y = -9.7 \)
In simple words: Moving 3.2 to the right side turns it into minus 3.2. We add these two negative decimals together to get minus 9.7.

Exam Tip: Keep the negative sign in front of the final decimal sum when combining two negative numbers.

Question 1. (ix) Solve: \( a + 8.9 = -12.6 \)
Answer:
Given equation:
\( a + 8.9 = -12.6 \)
Subtracting 8.9 from both sides:
\( \implies a = -12.6 - 8.9 \)
\( \implies a = -21.5 \)
In simple words: Move 8.9 to the right side where it becomes minus 8.9. Adding these two negative values gives us minus 21.5.

Exam Tip: Be careful with carry-overs when adding negative decimal values together.

Question 1. (x) Solve: \( x + 2\frac{1}{3} = 5 \)
Answer:
Given equation:
\( x + 2\frac{1}{3} = 5 \)
Converting the mixed fraction to an improper fraction:
\( \implies x + \frac{7}{3} = 5 \)
Subtracting \( \frac{7}{3} \) from both sides:
\( \implies x = 5 - \frac{7}{3} \)
Taking 3 as the common denominator:
\( \implies x = \frac{15 - 7}{3} \)
\( \implies x = \frac{8}{3} \)
Converting back to a mixed fraction:
\( \implies x = 2\frac{2}{3} \)
In simple words: Change the mixed fraction to 7/3 first. Then move it to the right and subtract it from 5 by using a common denominator of 3. This gives 8/3, which is 2 and 2/3.

Exam Tip: Converting mixed fractions to improper fractions first makes it much easier to solve equations with fractions.

Question 1. (xi) Solve: \( z + 2 = 4\frac{1}{5} \)
Answer:
Given equation:
\( z + 2 = 4\frac{1}{5} \)
Converting the mixed fraction:
\( \implies z + 2 = \frac{21}{5} \)
Subtracting 2 from both sides:
\( \implies z = \frac{21}{5} - 2 \)
Taking 5 as the common denominator:
\( \implies z = \frac{21 - 10}{5} \)
\( \implies z = \frac{11}{5} \)
Converting back to a mixed fraction:
\( \implies z = 2\frac{1}{5} \)
In simple words: Write the mixed fraction as 21/5. Move the 2 over and subtract it to get 11/5, which is the same as 2 and 1/5.

Exam Tip: You can also solve this quickly by just subtracting 2 from the whole number part of \( 4\frac{1}{5} \).

Question 1. (xii) Solve: \( m + 3\frac{1}{2} = 4\frac{1}{4} \)
Answer:
Given equation:
\( m + 3\frac{1}{2} = 4\frac{1}{4} \)
Converting the mixed fractions:
\( \implies m + \frac{7}{2} = \frac{17}{4} \)
Subtracting \( \frac{7}{2} \) from both sides:
\( \implies m = \frac{17}{4} - \frac{7}{2} \)
Taking 4 as the common denominator:
\( \implies m = \frac{17 - 14}{4} \)
\( \implies m = \frac{3}{4} \)
In simple words: Convert both mixed fractions to standard fractions first. Then move 7/2 over, make the bottom numbers equal, and subtract.

Exam Tip: Always find the lowest common multiple (LCM) of the denominators before subtracting fractions.

Question 1. (xiii) Solve: \( x + 2 = 1\frac{1}{4} \)
Answer:
Given equation:
\( x + 2 = 1\frac{1}{4} \)
Converting the mixed fraction:
\( \implies x + 2 = \frac{5}{4} \)
Subtracting 2 from both sides:
\( \implies x = \frac{5}{4} - 2 \)
Taking 4 as the common denominator:
\( \implies x = \frac{5 - 8}{4} \)
\( \implies x = -\frac{3}{4} \)
In simple words: Change 1 and 1/4 to 5/4. Shift the 2 over and subtract it using a common denominator of 4, leaving us with -3/4.

Exam Tip: Remember that fraction subtraction can result in a negative fraction. Keep the negative sign clear.

Question 1. (xiv) Solve: \( y + 5\frac{1}{3} = 4 \)
Answer:
Given equation:
\( y + 5\frac{1}{3} = 4 \)
Converting the mixed fraction:
\( \implies y + \frac{16}{3} = 4 \)
Subtracting \( \frac{16}{3} \) from both sides:
\( \implies y = 4 - \frac{16}{3} \)
Taking 3 as the common denominator:
\( \implies y = \frac{12 - 16}{3} \)
\( \implies y = -\frac{4}{3} \)
Converting to a mixed fraction:
\( \implies y = -1\frac{1}{3} \)
In simple words: Change the mixed fraction to 16/3. Move it over and subtract it from 4 to get -4/3, which simplifies to -1 and 1/3.

Exam Tip: Do not forget to keep the negative sign when converting an improper negative fraction back to a mixed fraction.

Question 1. (xv) Solve: \( a + 3\frac{1}{5} = 1\frac{1}{2} \)
Answer:
Given equation:
\( a + 3\frac{1}{5} = 1\frac{1}{2} \)
Converting mixed fractions to improper fractions:
\( \implies a + \frac{16}{5} = \frac{3}{2} \)
Subtracting \( \frac{16}{5} \) from both sides:
\( \implies a = \frac{3}{2} - \frac{16}{5} \)
Taking 10 as the common denominator:
\( \implies a = \frac{15 - 32}{10} \)
\( \implies a = -\frac{17}{10} \)
Converting back to a mixed fraction:
\( \implies a = -1\frac{7}{10} \)
In simple words: Convert the mixed fractions, then subtract 16/5 from 3/2 by making the bottom numbers 10. The result is -17/10, or -1 and 7/10.

Exam Tip: When subtracting, double check that you do the first numerator minus the second numerator to get the correct negative sign.

Question 2. (i) Solve: \( x - 3 = 2 \)
Answer:
Given equation:
\( x - 3 = 2 \)
Adding 3 to both sides:
\( \implies x = 2 + 3 \)
\( \implies x = 5 \)
In simple words: Move -3 to the other side where it becomes +3. Adding 2 and 3 gives 5.

Exam Tip: Remember that a minus sign changes to a plus sign when moved to the other side of the equals sign.

Question 2. (ii) Solve: \( m - 2 = -5 \)
Answer:
Given equation:
\( m - 2 = -5 \)
Adding 2 to both sides:
\( \implies m = -5 + 2 \)
\( \implies m = -3 \)
In simple words: Shifting -2 to the right side makes it +2. Adding 2 to -5 gives -3.

Exam Tip: When adding a positive and a negative number, subtract the smaller value from the larger one and use the sign of the larger value.

Question 2. (iii) Solve: \( b - 5 = 7 \)
Answer:
Given equation:
\( b - 5 = 7 \)
Adding 5 to both sides:
\( \implies b = 7 + 5 \)
\( \implies b = 12 \)
In simple words: We move the minus 5 over to make it plus 5, and 7 plus 5 is 12.

Exam Tip: Double check simple addition to ensure you do not make careless errors.

Question 2. (iv) Solve: \( a - 2.5 = -4 \)
Answer:
Given equation:
\( a - 2.5 = -4 \)
Adding 2.5 to both sides:
\( \implies a = -4 + 2.5 \)
\( \implies a = -1.5 \)
In simple words: Moving -2.5 to the other side makes it +2.5. Adding 2.5 to -4 leaves us with -1.5.

Exam Tip: Work with decimals carefully by keeping track of the negative sign when the larger value is negative.

Question 2. (v) Solve: \( y - 3\frac{1}{2} = 6 \)
Answer:
Given equation:
\( y - 3\frac{1}{2} = 6 \)
Converting the mixed fraction:
\( \implies y - \frac{7}{2} = 6 \)
Adding \( \frac{7}{2} \) to both sides:
\( \implies y = 6 + \frac{7}{2} \)
Taking 2 as the common denominator:
\( \implies y = \frac{12 + 7}{2} \)
\( \implies y = \frac{19}{2} \)
Converting back to a mixed fraction:
\( \implies y = 9\frac{1}{2} \)
In simple words: Change 3 and 1/2 to 7/2. Move it to the right as plus 7/2 and add it to 6 to get 19/2, which is 9 and 1/2.

Exam Tip: When converting to a mixed fraction, divide the numerator by the denominator to find the whole number and remainder.

Question 2. (vi) Solve: \( z - 2\frac{1}{3} = -6 \)
Answer:
Given equation:
\( z - 2\frac{1}{3} = -6 \)
Converting the mixed fraction:
\( \implies z - \frac{7}{3} = -6 \)
Adding \( \frac{7}{3} \) to both sides:
\( \implies z = -6 + \frac{7}{3} \)
Taking 3 as the common denominator:
\( \implies z = \frac{-18 + 7}{3} \)
\( \implies z = -\frac{11}{3} \)
Converting back to a mixed fraction:
\( \implies z = -3\frac{2}{3} \)
In simple words: Change the mixed fraction to 7/3. Move it over and add it to -6 using a common denominator, which gives -11/3 or -3 and 2/3.

Exam Tip: Always carry the negative sign carefully through all intermediate fraction steps.

Question 2. (vii) Solve: \( p - 5.4 = 2.7 \)
Answer:
Given equation:
\( p - 5.4 = 2.7 \)
Adding 5.4 to both sides:
\( \implies p = 2.7 + 5.4 \)
\( \implies p = 8.1 \)
In simple words: Move -5.4 to the other side to make it plus 5.4. Then add 2.7 and 5.4 to get 8.1.

Exam Tip: Line up the decimal point when adding decimal numbers to prevent alignment mistakes.

Question 2. (viii) Solve: \( x - 1.5 = -4.9 \)
Answer:
Given equation:
\( x - 1.5 = -4.9 \)
Adding 1.5 to both sides:
\( \implies x = -4.9 + 1.5 \)
\( \implies x = -3.4 \)
In simple words: Move -1.5 over as +1.5. Adding 1.5 to -4.9 results in -3.4.

Exam Tip: Since the negative number has a larger absolute value, the final sum must be negative.

Question 2. (ix) Solve: \( n - 4 = -4\frac{1}{5} \)
Answer:
Given equation:
\( n - 4 = -4\frac{1}{5} \)
Converting the mixed fraction:
\( \implies n - 4 = -\frac{21}{5} \)
Adding 4 to both sides:
\( \implies n = -\frac{21}{5} + 4 \)
Taking 5 as the common denominator:
\( \implies n = \frac{-21 + 20}{5} \)
\( \implies n = -\frac{1}{5} \)
In simple words: Convert -4 and 1/5 to -21/5. Move the -4 over to become +4, and combine them to get -1/5.

Exam Tip: Adding a whole number to a fraction is easiest when you convert the whole number to a fraction with the same denominator.

Question 3. (i) Solve: \( 3x = 12 \)
Answer:
Given equation:
\( 3x = 12 \)
Dividing both sides by 3:
\( \implies x = \frac{12}{3} \)
\( \implies x = 4 \)
In simple words: To isolate x, divide 12 by 3, which equals 4.

Exam Tip: When a number is multiplied by a variable, divide both sides by that number to solve it.

Question 3. (ii) Solve: \( 2y = 9 \)
Answer:
Given equation:
\( 2y = 9 \)
Dividing both sides by 2:
\( \implies y = \frac{9}{2} \)
Converting to a mixed fraction:
\( \implies y = 4\frac{1}{2} \)
In simple words: Divide 9 by 2. This gives us 9/2 or 4 and 1/2.

Exam Tip: Express improper fractions as mixed fractions for a more complete final answer.

Question 3. (iii) Solve: \( 5z = 8.5 \)
Answer:
Given equation:
\( 5z = 8.5 \)
Dividing both sides by 5:
\( \implies z = \frac{8.5}{5} \)
\( \implies z = 1.7 \)
In simple words: Divide 8.5 by 5 to find that z is 1.7.

Exam Tip: Decimal division can be checked by multiplying your answer back by the divisor.

Question 3. (iv) Solve: \( 2.5m = 7.5 \)
Answer:
Given equation:
\( 2.5m = 7.5 \)
Dividing both sides by 2.5:
\( \implies m = \frac{7.5}{2.5} \)
Multiplying numerator and denominator by 10:
\( \implies m = \frac{75}{25} \)
\( \implies m = 3 \)
In simple words: Dividing 7.5 by 2.5 is the same as dividing 75 by 25, which gives 3.

Exam Tip: Multiply both numerator and denominator by 10 to clear decimals before dividing.

Question 3. (v) Solve: \( 3.2p = 16 \)
Answer:
Given equation:
\( 3.2p = 16 \)
Dividing both sides by 3.2:
\( \implies p = \frac{16}{3.2} \)
Multiplying numerator and denominator by 10:
\( \implies p = \frac{16 \times 10}{32} \)
\( \implies p = \frac{160}{32} \)
\( \implies p = 5 \)
In simple words: Divide 16 by 3.2. If we multiply both by 10, it becomes 160 divided by 32, which is 5.

Exam Tip: Eliminating decimals in the denominator makes long division much simpler.

Question 3. (vi) Solve: \( 2a = 4.6 \)
Answer:
Given equation:
\( 2a = 4.6 \)
Dividing both sides by 2:
\( \implies a = \frac{4.6}{2} \)
\( \implies a = 2.3 \)
In simple words: Divide 4.6 by 2 to get 2.3.

Exam Tip: When dividing a decimal by a whole number, place the decimal point in the quotient directly above the decimal point in the dividend.

Question 4. (i) Solve: \( \frac{x}{2} = 5 \)
Answer:
Given equation:
\( \frac{x}{2} = 5 \)
Multiplying both sides by 2:
\( \implies x = 5 \times 2 \)
\( \implies x = 10 \)
In simple words: Since x is divided by 2, we multiply the other side by 2 to solve it. 5 times 2 is 10.

Exam Tip: The opposite of division is multiplication; always use multiplication to isolate a variable in the numerator.

Question 4. (ii) Solve: \( \frac{y}{3} = -2 \)
Answer:
Given equation:
\( \frac{y}{3} = -2 \)
Multiplying both sides by 3:
\( \implies y = -2 \times 3 \)
\( \implies y = -6 \)
In simple words: Multiply -2 by 3 to get -6.

Exam Tip: Remember that a negative number multiplied by a positive number always yields a negative product.

Question 4. (iii) Solve: \( \frac{a}{5} = -15 \)
Answer:
Given equation:
\( \frac{a}{5} = -15 \)
Multiplying both sides by 5:
\( \implies a = -15 \times 5 \)
\( \implies a = -75 \)
In simple words: Multiply -15 by 5 to find that a is -75.

Exam Tip: Keep careful track of the negative sign in your final calculation.

Question 4. (iv) Solve: \( \frac{z}{4} = 3\frac{1}{4} \)
Answer:
Given equation:
\( \frac{z}{4} = 3\frac{1}{4} \)
Converting the mixed fraction:
\( \implies \frac{z}{4} = \frac{13}{4} \)
Multiplying both sides by 4:
\( \implies z = \frac{13}{4} \times 4 \)
\( \implies z = 13 \)
In simple words: Change 3 and 1/4 to 13/4. Since both sides are divided by 4, they cancel out, leaving z equal to 13.

Exam Tip: When denominators on both sides of an equation are equal, their numerators are also equal.

Question 4. (v) Solve: \( \frac{m}{6} = 2\frac{1}{2} \)
Answer:
Given equation:
\( \frac{m}{6} = 2\frac{1}{2} \)
Converting the mixed fraction:
\( \implies \frac{m}{6} = \frac{5}{2} \)
Multiplying both sides by 6:
\( \implies m = \frac{5}{2} \times 6 \)
\( \implies m = 15 \)
In simple words: Change 2 and 1/2 to 5/2. Multiply by 6 to solve, which gives 15.

Exam Tip: Simplify the multiplication by dividing the multiplier by the denominator first.

Question 4. (vi) Solve: \( \frac{n}{7} = -2.8 \)
Answer:
Given equation:
\( \frac{n}{7} = -2.8 \)
Multiplying both sides by 7:
\( \implies n = -2.8 \times 7 \)
\( \implies n = -19.6 \)
In simple words: Multiply -2.8 by 7 to find that n is -19.6.

Exam Tip: Pay close attention to the position of the decimal point during multiplication.

Question 5. (i) Solve: \( -2x = 8 \)
Answer:
Given equation:
\( -2x = 8 \)
Dividing both sides by -2:
\( \implies x = \frac{8}{-2} \)
\( \implies x = -4 \)
In simple words: Divide 8 by -2. A positive number divided by a negative number gives a negative result, so the answer is -4.

Exam Tip: Remember that dividing a positive number by a negative number results in a negative quotient.

Question 5. (ii) Solve: \( -3.5y = 14 \)
Answer:
Given equation:
\( -3.5y = 14 \)
Dividing both sides by -3.5:
\( \implies y = \frac{14}{-3.5} \)
Multiplying numerator and denominator by 10:
\( \implies y = \frac{-140}{35} \)
\( \implies y = -4 \)
In simple words: Divide 14 by -3.5. Multiplying by 10 gives -140 divided by 35, which equals -4.

Exam Tip: Simplify fraction division by canceling out common factors first.

Question 5. (iii) Solve: \( -5z = 4 \)
Answer:
Given equation:
\( -5z = 4 \)
Dividing both sides by -5:
\( \implies z = -\frac{4}{5} \)
\( \implies z = -0.8 \)
In simple words: Divide 4 by -5 to get -4/5, which is -0.8.

Exam Tip: Be prepared to write your final answer as either a fraction or a decimal.

Question 5. (iv) Solve: \( -5 = a + 3 \)
Answer:
Given equation:
\( -5 = a + 3 \)
Rearranging the equation:
\( \implies a + 3 = -5 \)
Subtracting 3 from both sides:
\( \implies a = -5 - 3 \)
\( \implies a = -8 \)
In simple words: Swap the sides to make it a + 3 = -5. Move 3 to the other side where it becomes minus 3, giving us -8.

Exam Tip: Swapping the entire left and right sides of an equation does not change any signs.

Question 5. (v) Solve: \( 2 = p + 5 \)
Answer:
Given equation:
\( 2 = p + 5 \)
Rearranging the equation:
\( \implies p + 5 = 2 \)
Subtracting 5 from both sides:
\( \implies p = 2 - 5 \)
\( \implies p = -3 \)
In simple words: Rewrite it as p + 5 = 2. Move 5 over to get 2 minus 5, which is -3.

Exam Tip: Make sure to subtract the larger number from the smaller one correctly to get a negative value.

Question 5. (vi) Solve: \( 4.5 = m - 2.7 \)
Answer:
Given equation:
\( 4.5 = m - 2.7 \)
Rearranging the equation:
\( \implies m - 2.7 = 4.5 \)
Adding 2.7 to both sides:
\( \implies m = 4.5 + 2.7 \)
\( \implies m = 7.2 \)
In simple words: Rearrange to m - 2.7 = 4.5. Move -2.7 to the other side as plus 2.7 and add them to get 7.2.

Exam Tip: Adding a decimal to both sides is used to isolate a variable with a subtracted decimal.

Question 5. (vii) Solve: \( 3\frac{2}{5} = x - 2\frac{1}{3} \)
Answer:
Given equation:
\( 3\frac{2}{5} = x - 2\frac{1}{3} \)
Converting to improper fractions:
\( \implies \frac{17}{5} = x - \frac{7}{3} \)
Rearranging the equation:
\( \implies x - \frac{7}{3} = \frac{17}{5} \)
Adding \( \frac{7}{3} \) to both sides:
\( \implies x = \frac{17}{5} + \frac{7}{3} \)
Taking 15 as the common denominator:
\( \implies x = \frac{51 + 35}{15} \)
\( \implies x = \frac{86}{15} \)
Converting back to a mixed fraction:
\( \implies x = 5\frac{11}{15} \)
In simple words: Change mixed fractions to improper ones. Rearrange to put x on the left. Move -7/3 to the right side as a plus, then add them using 15 as the common denominator to get 86/15, which is 5 and 11/15.

Exam Tip: Be very precise when finding common denominators for large fractions to avoid simple addition errors.

Question 5. (viii) Solve: \( 5 = m + 3\frac{4}{7} \)
Answer:
Given equation:
\( 5 = m + 3\frac{4}{7} \)
Converting the mixed fraction:
\( \implies 5 = m + \frac{25}{7} \)
Rearranging the equation:
\( \implies m + \frac{25}{7} = 5 \)
Subtracting \( \frac{25}{7} \) from both sides:
\( \implies m = 5 - \frac{25}{7} \)
Taking 7 as the common denominator:
\( \implies m = \frac{35 - 25}{7} \)
\( \implies m = \frac{10}{7} \)
Converting to a mixed fraction:
\( \implies m = 1\frac{3}{7} \)
In simple words: Change 3 and 4/7 to 25/7. Move it over and subtract it from 5 using 7 as the bottom number. This leaves 10/7, which is 1 and 3/7.

Exam Tip: Practice converting improper fractions to mixed fractions to ensure speed and accuracy in exams.

Question 5. (ix) Solve: \( -2\frac{1}{5} = y - 4 \)
Answer:
Given equation:
\( -2\frac{1}{5} = y - 4 \)
Converting the mixed fraction:
\( \implies -\frac{11}{5} = y - 4 \)
Rearranging the equation:
\( \implies y - 4 = -\frac{11}{5} \)
Adding 4 to both sides:
\( \implies y = -\frac{11}{5} + 4 \)
Taking 5 as the common denominator:
\( \implies y = \frac{-11 + 20}{5} \)
\( \implies y = \frac{9}{5} \)
Converting to a mixed fraction:
\( \implies y = 1\frac{4}{5} \)
In simple words: Change -2 and 1/5 to -11/5. Swap sides, then move the -4 over to become +4. Add them to get 9/5, which is 1 and 4/5.

Exam Tip: Make sure the sign of the whole number changes from minus to plus when shifting it to the other side.

Exercise 22(B)

Question 1. (i) Solve: \( 2x + 5 = 17 \)
Answer:
Given equation:
\( 2x + 5 = 17 \)
Subtracting 5 from both sides:
\( \implies 2x = 17 - 5 \)
\( \implies 2x = 12 \)
Dividing both sides by 2:
\( \implies x = \frac{12}{2} \)
\( \implies x = 6 \)
In simple words: First, move the 5 over to get 17 minus 5, which is 12. Then, divide 12 by 2 to find that x is 6.

Exam Tip: In two-step equations, always perform addition or subtraction before division or multiplication.

Question 1. (ii) Solve: \( 3y - 2 = 1 \)
Answer:
Given equation:
\( 3y - 2 = 1 \)
Adding 2 to both sides:
\( \implies 3y = 1 + 2 \)
\( \implies 3y = 3 \)
Dividing both sides by 3:
\( \implies y = \frac{3}{3} \)
\( \implies y = 1 \)
In simple words: Move -2 to the other side where it becomes +2. Adding 1 and 2 gives 3. Then divide 3 by 3 to get 1.

Exam Tip: Isolating the variable term first is key to solving linear equations.

Question 1. (iii) Solve: \( 5p + 4 = 29 \)
Answer:
Given equation:
\( 5p + 4 = 29 \)
Subtracting 4 from both sides:
\( \implies 5p = 29 - 4 \)
\( \implies 5p = 25 \)
Dividing both sides by 5:
\( \implies p = \frac{25}{5} \)
\( \implies p = 5 \)
In simple words: Subtract 4 from 29 to get 25. Then divide 25 by 5 to find that p is 5.

Exam Tip: Perform operations on both sides systematically to keep the equation balanced.

Question 1. (iv) Solve: \( 4a - 3 = -27 \)
Answer:
Given equation:
\( 4a - 3 = -27 \)
Adding 3 to both sides:
\( \implies 4a = -27 + 3 \)
\( \implies 4a = -24 \)
Dividing both sides by 4:
\( \implies a = \frac{-24}{4} \)
\( \implies a = -6 \)
In simple words: Move -3 to the other side to make it +3. Adding 3 to -27 gives -24. Then divide -24 by 4 to get -6.

Exam Tip: Be careful when combining negative and positive integers; subtraction rules apply here.

Question 1. (v) Solve: \( 2z + 3 = -19 \)
Answer:
Given equation:
\( 2z + 3 = -19 \)
Subtracting 3 from both sides:
\( \implies 2z = -19 - 3 \)
\( \implies 2z = -22 \)
Dividing both sides by 2:
\( \implies z = \frac{-22}{2} \)
\( \implies z = -11 \)
In simple words: Move 3 over as -3. Two negative numbers combine to make -22. Dividing by 2 gives -11.

Exam Tip: Keep absolute values in mind when combining numbers with like negative signs.

Question 1. (vi) Solve: \( 7m - 1 = 20 \)
Answer:
Given equation:
\( 7m - 1 = 20 \)
Adding 1 to both sides:
\( \implies 7m = 20 + 1 \)
\( \implies 7m = 21 \)
Dividing both sides by 7:
\( \implies m = \frac{21}{7} \)
\( \implies m = 3 \)
In simple words: Move -1 to the other side as +1, which gives 21. Then divide 21 by 7 to get 3.

Exam Tip: Always perform a quick mental check by putting the solution back into the original equation.

Question 1. (vii) Solve: \( 2.4x - 3 = 4.2 \)
Answer:
Given equation:
\( 2.4x - 3 = 4.2 \)
Adding 3 to both sides:
\( \implies 2.4x = 4.2 + 3 \)
\( \implies 2.4x = 7.2 \)
Dividing both sides by 2.4:
\( \implies x = \frac{7.2}{2.4} \)
Multiplying numerator and denominator by 10:
\( \implies x = \frac{72}{24} \)
\( \implies x = 3 \)
In simple words: Add 3 to 4.2 to get 7.2. Then divide 7.2 by 2.4, which is the same as 72 divided by 24, giving us 3.

Exam Tip: Clearing decimals from fractions makes it easier to work with integer divisions.

Question 1. (viii) Solve: \( 4m + 9.4 = 5 \)
Answer:
Given equation:
\( 4m + 9.4 = 5 \)
Subtracting 9.4 from both sides:
\( \implies 4m = 5 - 9.4 \)
\( \implies 4m = -4.4 \)
Dividing both sides by 4:
\( \implies m = \frac{-4.4}{4} \)
\( \implies m = -1.1 \)
In simple words: Move 9.4 over to get 5 minus 9.4, which is -4.4. Divide by 4 to get -1.1.

Exam Tip: Pay attention to sign rules when subtracting a larger positive decimal from a smaller positive whole number.

Question 1. (ix) Solve: \( 6y + 4 = -4.4 \)
Answer:
Given equation:
\( 6y + 4 = -4.4 \)
Subtracting 4 from both sides:
\( \implies 6y = -4.4 - 4 \)
\( \implies 6y = -8.4 \)
Dividing both sides by 6:
\( \implies y = \frac{-8.4}{6} \)
\( \implies y = -1.4 \)
In simple words: Subtract 4 from -4.4 to get -8.4. Dividing -8.4 by 6 gives -1.4.

Exam Tip: When dividing a negative decimal by a positive integer, the result is negative. Keep the decimal place aligned.

Question 2. (i) Solve: \( \frac{x}{3} - 5 = 2 \)
Answer:
Given equation:
\( \frac{x}{3} - 5 = 2 \)
Adding 5 to both sides:
\( \implies \frac{x}{3} = 2 + 5 \)
\( \implies \frac{x}{3} = 7 \)
Multiplying both sides by 3:
\( \implies x = 7 \times 3 \)
\( \implies x = 21 \)
In simple words: First, move the -5 to the other side as +5, which gives 7. Then, multiply 7 by 3 to find that x is 21.

Exam Tip: Always add or subtract the standalone number before multiplying to clear the denominator.

Question 2. (ii) Solve: \( \frac{y}{2} - 3 = 8 \)
Answer:
Given equation:
\( \frac{y}{2} - 3 = 8 \)
Adding 3 to both sides:
\( \implies \frac{y}{2} = 8 + 3 \)
\( \implies \frac{y}{2} = 11 \)
Multiplying both sides by 2:
\( \implies y = 11 \times 2 \)
\( \implies y = 22 \)
In simple words: Move -3 to the other side to make it +3, giving 11. Multiply 11 by 2 to get 22.

Exam Tip: Clear subtraction first, then clear division by multiplying.

Question 2. (iii) Solve: \( \frac{z}{7} + 1 = 2\frac{1}{2} \)
Answer:
Given equation:
\( \frac{z}{7} + 1 = 2\frac{1}{2} \)
Converting the mixed fraction:
\( \implies \frac{z}{7} + 1 = \frac{5}{2} \)
Subtracting 1 from both sides:
\( \implies \frac{z}{7} = \frac{5}{2} - 1 \)
Taking 2 as the common denominator:
\( \implies \frac{z}{7} = \frac{5 - 2}{2} \)
\( \implies \frac{z}{7} = \frac{3}{2} \)
Multiplying both sides by 7:
\( \implies z = \frac{3}{2} \times 7 \)
\( \implies z = \frac{21}{2} \)
Converting to a mixed fraction:
\( \implies z = 10\frac{1}{2} \)
In simple words: Convert the mixed fraction to 5/2. Subtract 1 to get 3/2. Multiply by 7 to get 21/2, which equals 10 and 1/2.

Exam Tip: Be careful when multiplying a fraction by a whole number; multiply only the numerator.

Question 2. (iv) Solve: \( \frac{a}{2.4} - 5 = 2.4 \)
Answer:
Given equation:
\( \frac{a}{2.4} - 5 = 2.4 \)
Adding 5 to both sides:
\( \implies \frac{a}{2.4} = 2.4 + 5 \)
\( \implies \frac{a}{2.4} = 7.4 \)
Multiplying both sides by 2.4:
\( \implies a = 7.4 \times 2.4 \)
\( \implies a = 17.76 \)
In simple words: Add 5 to 2.4 to get 7.4. Then multiply 7.4 by 2.4 to find that a is 17.76.

Exam Tip: Use standard decimal multiplication to find the product of two decimals.

Question 2. (v) Solve: \( \frac{b}{1.6} + 3 = -2.5 \)
Answer:
Given equation:
\( \frac{b}{1.6} + 3 = -2.5 \)
Subtracting 3 from both sides:
\( \implies \frac{b}{1.6} = -2.5 - 3 \)
\( \implies \frac{b}{1.6} = -5.5 \)
Multiplying both sides by 1.6:
\( \implies b = -5.5 \times 1.6 \)
\( \implies b = -8.8 \)
In simple words: Subtract 3 from -2.5 to get -5.5. Multiply by 1.6 to get -8.8.

Exam Tip: Do not forget to attach the negative sign to the final product when multiplying a negative decimal by a positive decimal.

Question 2. (vi) Solve: \( \frac{m}{4} - 4.6 = -3.1 \)
Answer:
Given equation:
\( \frac{m}{4} - 4.6 = -3.1 \)
Adding 4.6 to both sides:
\( \implies \frac{m}{4} = -3.1 + 4.6 \)
\( \implies \frac{m}{4} = 1.5 \)
Multiplying both sides by 4:
\( \implies m = 1.5 \times 4 \)
\( \implies m = 6 \)
In simple words: Add 4.6 to -3.1, which leaves 1.5. Multiply 1.5 by 4 to get 6.

Exam Tip: Ensure correct subtraction when adding a positive decimal to a negative decimal.

Question 3. Solve:
(i) \( -8m - 2 = -10 \)
(ii) \( 4x + 2x = 3 + 5 \)
(iii) \( 4x - x + 5 = 8 \)
(iv) \( 6x + 2 = 2x + 10 \)
(v) \( 18 - (2a - 12) = 8a \)
(vi) \( 3x + 5 + 2x + 6 + x = 4x + 21 \)
(vii) \( 3.5x - 9 - 3 = x + 1 \)
(viii) \( 8x + 6 + 2x - 4 = 4x + 8 \)
(ix) \( -m + (3m - 6m) = -8 - 14 \)
(x) \( 5x - 14 = x - (24 + 4x) \)
Answer:
(i) \( -8m - 2 = -10 \)
\( \implies -8m = -10 + 2 \)
\( \implies -8m = -8 \)
\( \implies m = \frac{-8}{-8} \)
\( \implies m = 1 \)

(ii) \( 4x + 2x = 3 + 5 \)
\( \implies 6x = 8 \)
\( \implies x = \frac{8}{6} \)
\( \implies x = \frac{4}{3} \)
\( \implies x = 1\frac{1}{3} \)

(iii) \( 4x - x + 5 = 8 \)
\( \implies 3x + 5 = 8 \)
\( \implies 3x = 8 - 5 \)
\( \implies 3x = 3 \)
\( \implies x = \frac{3}{3} \)
\( \implies x = 1 \)

(iv) \( 6x + 2 = 2x + 10 \)
\( \implies 6x - 2x = 10 - 2 \)
\( \implies 4x = 8 \)
\( \implies x = \frac{8}{4} \)
\( \implies x = 2 \)

(v) \( 18 - (2a - 12) = 8a \)
\( \implies 18 - 2a + 12 = 8a \)
\( \implies -2a - 8a = -18 - 12 \)
\( \implies -10a = -30 \)
\( \implies a = \frac{-30}{-10} \)
\( \implies a = 3 \)

(vi) \( 3x + 5 + 2x + 6 + x = 4x + 21 \)
\( \implies 6x + 11 = 4x + 21 \)
\( \implies 6x - 4x = 21 - 11 \)
\( \implies 2x = 10 \)
\( \implies x = \frac{10}{2} \)
\( \implies x = 5 \)

(vii) \( 3.5x - 9 - 3 = x + 1 \)
\( \implies 3.5x - x = 1 + 9 + 3 \)
\( \implies 2.5x = 13 \)
\( \implies x = \frac{13}{2.5} \)
\( \implies x = \frac{13 \times 10}{25} \)
\( \implies x = \frac{26}{5} \)
\( \implies x = 5\frac{1}{5} \)

(viii) \( 8x + 6 + 2x - 4 = 4x + 8 \)
\( \implies 10x + 2 = 4x + 8 \)
\( \implies 10x - 4x = 8 - 2 \)
\( \implies 6x = 6 \)
\( \implies x = \frac{6}{6} \)
\( \implies x = 1 \)

(ix) \( -m + (3m - 6m) = -8 - 14 \)
\( \implies -m + 3m - 6m = -22 \)
\( \implies -7m + 3m = -22 \)
\( \implies -4m = -22 \)
\( \implies m = \frac{-22}{-4} \)
\( \implies m = \frac{11}{2} \)
\( \implies m = 5\frac{1}{2} \)

(x) \( 5x - 14 = x - (24 + 4x) \)
\( \implies 5x - 14 = x - 24 - 4x \)
\( \implies 5x - 14 = -3x - 24 \)
\( \implies 5x + 3x = -24 + 14 \)
\( \implies 8x = -10 \)
\( \implies x = \frac{-10}{8} \)
\( \implies x = -\frac{5}{4} \)
\( \implies x = -1\frac{1}{4} \)
In simple words: To solve these equations, we put the terms with letters on one side and the normal numbers on the other side. Then, we simplify them to find the value of the letter.

Exam Tip: Be very careful with signs when moving terms across the equals sign. A minus becomes a plus, and a plus becomes a minus.

Exercise 22(C)

Question 1. 5 - x = 3
Answer:
\( \implies 5 - 3 = x \)
\( \implies x = 2 \)
In simple words: To find the value of \( x \), we can move \( x \) to the right side to make it positive, and move \( 3 \) to the left side. This leaves us with \( 5 - 3 \), which equals \( 2 \).

Exam Tip: Keeping the variable positive makes it much easier to avoid basic calculation errors.

Question 2. 2 - y = 8
Answer:
\( \implies 2 - 8 = y \)
\( \implies y = -6 \)
In simple words: Shift \( y \) to the other side to make it positive, and bring \( 8 \) over as \( -8 \). Then, subtracting \( 8 \) from \( 2 \) gives \( -6 \).

Exam Tip: Remember that when you subtract a larger number from a smaller number, the result is always negative.

Question 3. 8.4 - x = -2
Answer:
\( \implies 8.4 + 2 = x \)
\( \implies x = 10.4 \)
In simple words: Moving the minus \( x \) to the right side makes it positive, and bringing \( -2 \) to the left side changes it to \( +2 \). Adding these gives \( 10.4 \).

Exam Tip: Be careful with decimal addition. Align the place values correctly: \( 8.4 + 2.0 = 10.4 \).

Question 4. x + 2\frac{1}{5} = 3
Answer:
\( \implies x + \frac{11}{5} = 3 \)
\( \implies x = 3 - \frac{11}{5} \)
\( \implies x = \frac{15 - 11}{5} \)
\( \implies x = \frac{4}{5} \)
In simple words: First, change the mixed fraction to an improper fraction. Then, subtract it from \( 3 \) by finding a common denominator of \( 5 \) to get the final answer.

Exam Tip: When converting mixed numbers like \( 2\frac{1}{5} \), multiply the whole number by the denominator and add the numerator: \( 2 \times 5 + 1 = 11 \).

Question 5. y - 3\frac{1}{2} = 2\frac{1}{3}
Answer:
\( \implies y - \frac{7}{2} = \frac{7}{3} \)
\( \implies y = \frac{7}{3} + \frac{7}{2} \)
\( \implies y \times 6 = \left(\frac{7}{3} + \frac{7}{2}\right) \times 6 \)
\( \implies 6y = 14 + 21 \)
\( \implies 6y = 35 \)
\( \implies y = \frac{35}{6} \)
\( \implies y = 5\frac{5}{6} \)
In simple words: Turn both mixed fractions into simple fractions first. Then, add them by converting both to have a common bottom number of \( 6 \) to find \( y \).

Exam Tip: Multiplying the entire equation by the LCM of the denominators is an excellent shortcut to clear out fractions early.

Question 6. 5\frac{2}{3} - z = 2\frac{1}{2}
Answer:
\( \implies \frac{17}{3} - z = \frac{5}{2} \)
\( \implies \frac{17}{3} - \frac{5}{2} = z \)
\( \implies z \times 6 = \left(\frac{17}{3} - \frac{5}{2}\right) \times 6 \)
\( \implies 6z = 34 - 15 \)
\( \implies 6z = 19 \)
\( \implies z = \frac{19}{6} \)
\( \implies z = 3\frac{1}{6} \)
In simple words: Change both fractions to improper form. Move the variable \( z \) to the other side, find a common denominator, and solve for \( z \).

Exam Tip: Don't forget to convert your final improper fraction back into a mixed fraction if the original question used mixed numbers.

Question 7. 1.6z = 8
Answer:
\( \implies z = \frac{8}{1.6} \)
\( \implies z = \frac{8 \times 10}{16} \)
\( \implies z = \frac{80}{16} \)
\( \implies z = 5 \)
In simple words: To find \( z \), divide \( 8 \) by \( 1.6 \). You can make it easier by multiplying the top and bottom by \( 10 \) to work with whole numbers.

Exam Tip: Multiplying decimal denominators by powers of 10 to clear the decimal point is a highly reliable way to avoid division mistakes.

Question 8. 3a = -2.1
Answer:
\( \implies a = \frac{-2.1}{3} \)
\( \implies a = \frac{-21}{3 \times 10} \)
\( \implies a = -\frac{7}{10} \)
\( \implies a = -0.7 \)
In simple words: Divide \( -2.1 \) by \( 3 \) to find \( a \). The result is negative because a negative number divided by a positive number is always negative.

Exam Tip: Keep careful track of the negative sign throughout your calculation so that it doesn't get lost in the final step.

Question 9. \frac{z}{4} = -1.5
Answer:
\( \implies z = -1.5 \times 4 \)
\( \implies z = -6.0 \)
In simple words: To find \( z \), multiply both sides by \( 4 \). Multiplying \( -1.5 \) by \( 4 \) gives \( -6 \).

Exam Tip: Remember that multiplication by 4 is the same as doubling a number twice, which is a quick mental check.

Question 10. \frac{z}{6} = -1\frac{2}{3}
Answer:
\( \implies \frac{z}{6} = -\frac{5}{3} \)
\( \implies z \times 3 = -5 \times 6 \)
\( \implies z = \frac{-5 \times 6}{3} \)
\( \implies z = -10 \)
In simple words: Convert the mixed number into a fraction, then multiply both sides by \( 6 \). Simplifying the expression gives the value of \( z \).

Exam Tip: When cross-multiplying, simplify the numbers before doing the final division to keep the arithmetic simple.

Question 11. -5x = 10
Answer:
\( \implies x = \frac{10}{-5} \)
\( \implies x = -2 \)
In simple words: Divide \( 10 \) by \( -5 \). This gives a result of \( -2 \) because positive divided by negative is negative.

Exam Tip: Be sure to write the negative sign clearly in the denominator before dividing.

Question 12. 2.4z = -4.8
Answer:
\( \implies z = \frac{-4.8}{2.4} \)
\( \implies z = -\frac{48}{24} \)
\( \implies z = -2 \)
In simple words: Divide both sides by \( 2.4 \). Since \( 4.8 \) is exactly double of \( 2.4 \), the answer is \( -2 \).

Exam Tip: Recognizing simple multiples like 24 and 48 can save you a lot of time on decimal division.

Question 13. 2y - 5 = -11
Answer:
\( \implies 2y = -11 + 5 \)
\( \implies 2y = -6 \)
\( \implies y = -\frac{6}{2} \)
\( \implies y = -3 \)
In simple words: Move \( -5 \) to the right side where it becomes \( +5 \). Then divide by \( 2 \) to get the final value for \( y \).

Exam Tip: Always add or subtract constants from both sides before you divide by the coefficient of the variable.

Question 14. 2x + 4.6 = 8
Answer:
\( \implies 2x = 8 - 4.6 \)
\( \implies 2x = 3.4 \)
\( \implies x = \frac{3.4}{2} \)
\( \implies x = \frac{34}{2 \times 10} \)
\( \implies x = \frac{17}{10} \)
\( \implies x = 1.7 \)
In simple words: Subtract \( 4.6 \) from \( 8 \) to get \( 3.4 \), then divide by \( 2 \) to find that \( x \) is \( 1.7 \).

Exam Tip: Ensure your decimal subtraction is correct. Doing \( 8.0 - 4.6 = 3.4 \) carefully prevents common calculation errors.

Question 15. 5y - 3.5 = 10
Answer:
\( \implies 5y = 10 + 3.5 \)
\( \implies 5y = 13.5 \)
\( \implies y = \frac{13.5}{5} \)
\( \implies y = 2.7 \)
In simple words: Add \( 3.5 \) to \( 10 \), giving \( 13.5 \). Then division by \( 5 \) gives the final result of \( 2.7 \).

Exam Tip: Think of \( 13.5 \) as \( 10 + 3.5 \) to make division by 5 easier in your head.

Question 16. 3x + 2 = -2.2
Answer:
\( \implies 3x = -2.2 - 2 \)
\( \implies 3x = -4.2 \)
\( \implies x = -\frac{4.2}{3} \)
\( \implies x = -\frac{42}{3 \times 10} \)
\( \implies x = -\frac{14}{10} \)
\( \implies x = -1.4 \)
In simple words: Shift \( 2 \) to the right as \( -2 \), which gives \( -4.2 \). Dividing by \( 3 \) gives the result \( -1.4 \).

Exam Tip: When combining negative decimals like \( -2.2 - 2 \), add their values and keep the negative sign: \( -4.2 \).

Question 17. \frac{y}{2} - 5 = 1
Answer:
\( \implies \frac{y}{2} \times 2 - 5 \times 2 = 1 \times 2 \)
\( \implies y - 10 = 2 \)
\( \implies y = 2 + 10 \)
\( \implies y = 12 \)
In simple words: Multiply the entire equation by \( 2 \) to clear the fraction. Then, move \( -10 \) over to get \( y = 12 \).

Exam Tip: Make sure to multiply every single term of the equation by 2, not just the fraction term.

Question 18. \frac{z}{3} - 1 = -5
Answer:
\( \implies \frac{z}{3} \times 3 - 1 \times 3 = -5 \times 3 \)
\( \implies z - 3 = -15 \)
\( \implies z = -15 + 3 \)
\( \implies z = -12 \)
In simple words: Multiply all terms by \( 3 \) to remove the fraction. Then add \( 3 \) to \( -15 \) to get \( -12 \).

Exam Tip: Be careful with the sum \( -15 + 3 \). Since \( 15 \) is larger, the final answer must stay negative.

Question 19. \frac{x}{4} + 3.6 = -1.1
Answer:
\( \implies \frac{x}{4} = -1.1 - 3.6 \)
\( \implies \frac{x}{4} = -4.7 \)
\( \implies x = -4.7 \times 4 \)
\( \implies x = -18.8 \)
In simple words: Move \( 3.6 \) to the other side to get \( -4.7 \). Finally, multiply by \( 4 \) to find \( x \).

Exam Tip: Subtract first before multiplying to avoid dealing with large decimal products.

Question 20. -3y - 2 = 10
Answer:
\( \implies -3y = 10 + 2 \)
\( \implies -3y = 12 \)
\( \implies y = \frac{12}{-3} \)
\( \implies y = -4 \)
In simple words: Add \( 2 \) to \( 10 \) to get \( 12 \). Then divide \( 12 \) by \( -3 \), which results in \( -4 \).

Exam Tip: Always double check that positive divided by negative yields a negative final value.

Question 21. 4z - 5 = 3 - z
Answer:
\( \implies 4z + z = 3 + 5 \)
\( \implies 5z = 8 \)
\( \implies z = \frac{8}{5} \)
\( \implies z = 1.6 \)
In simple words: Group the terms with \( z \) on one side and constant numbers on the other side. This gives \( 5z = 8 \), so \( z = 1.6 \).

Exam Tip: Convert improper fractions like \( \frac{8}{5} \) into decimals by multiplying the top and bottom by 2 to get \( \frac{16}{10} = 1.6 \).

Question 22. 7x - 3x + 2 = 22
Answer:
\( \implies 4x + 2 = 22 \)
\( \implies 4x = 22 - 2 \)
\( \implies 4x = 20 \)
\( \implies x = \frac{20}{4} \)
\( \implies x = 5 \)
In simple words: Combine \( 7x - 3x \) into \( 4x \). Then subtract \( 2 \) from \( 22 \) to get \( 20 \), and divide by \( 4 \) to get \( 5 \).

Exam Tip: Simplify the variable terms on the left side before performing any transposition steps.

Question 23. 6y + 3 = 2y + 11
Answer:
\( \implies 6y - 2y = 11 - 3 \)
\( \implies 4y = 8 \)
\( \implies y = \frac{8}{4} \)
\( \implies y = 2 \)
In simple words: Move \( 2y \) to the left side and \( 3 \) to the right side. This leaves \( 4y = 8 \), so \( y \) is \( 2 \).

Exam Tip: Group variables on the side that keeps their coefficient positive to make calculations simpler.

Question 24. 3 (x+5) = 18
Answer:
\( \implies 3x + 15 = 18 \)
\( \implies 3x = 18 - 15 \)
\( \implies 3x = 3 \)
\( \implies x = \frac{3}{3} \)
\( \implies x = 1 \)
In simple words: Expand the brackets by multiplying both terms by \( 3 \). Then, solve the resulting simple equation.

Exam Tip: Alternatively, you can divide both sides by 3 first: \( x + 5 = 6 \implies x = 1 \), which is often much faster!

Question 25. 5 (x-2) -2 (x+2) = 3
Answer:
\( \implies 5x - 10 - 2x - 4 = 3 \)
\( \implies 3x - 14 = 3 \)
\( \implies 3x = 3 + 14 \)
\( \implies 3x = 17 \)
\( \implies x = \frac{17}{3} \)
\( \implies x = 5\frac{2}{3} \)
In simple words: Expand both sets of brackets by multiplying. Combine the like terms and solve for \( x \) to find the fraction value.

Exam Tip: Remember that distributing a negative sign inside the parenthesis changes the sign of all terms inside: \( -2(x + 2) = -2x - 4 \).

Question 26. \( \frac{5x-3}{4} = 3 \)
Answer:
\( \implies 5x - 3 = 3 \times 4 \)
\( \implies 5x - 3 = 12 \)
\( \implies 5x = 12 + 3 \)
\( \implies 5x = 15 \)
\( \implies x = \frac{15}{5} \)
\( \implies x = 3 \)
In simple words: Multiply both sides by \( 4 \) to clear the fraction. Then add \( 3 \) to \( 12 \) to get \( 15 \), and divide by \( 5 \) to get \( 3 \).

Exam Tip: Treat the numerator as a single grouped term when multiplying both sides to clear the denominator.

Question 27. 3(2x+1) -2(x-5) -5 (5-2x) = 16
Answer:
\( \implies 6x + 3 - 2x + 10 - 25 + 10x = 16 \)
\( \implies 6x - 2x + 10x + 3 + 10 - 25 = 16 \)
\( \implies 14x - 12 = 16 \)
\( \implies 14x = 16 + 12 \)
\( \implies 14x = 28 \)
\( \implies x = \frac{28}{14} \)
\( \implies x = 2 \)
In simple words: Expand each bracket carefully, making sure to multiply the signs correctly. Combine all the \( x \) terms and constant numbers, then solve.

Exam Tip: Be extremely careful with sign changes, especially when a minus is outside the brackets: \( -5(5 - 2x) = -25 + 10x \).

Exercise 22(D)

Question 1. A number increased by 17 becomes 54. Find the number.
Answer: Let the unknown number be \( x \).
Based on the given problem:
\( x + 17 = 54 \)
\( \implies x = 54 - 17 \)
\( \implies x = 37 \)
Thus, the number is 37.
In simple words: When we add \( 17 \) to a number, we get \( 54 \). To find that number, we subtract \( 17 \) from \( 54 \).

Exam Tip: Always clearly state your assumption at the start, like "Let the number be \( x \)", to earn full marks for steps.

Question 2. A number decreased by 8 equals 26, find the number.
Answer: Let the needed number be represented by \( A \).
According to the question:
\( A - 8 = 26 \)
\( \implies A = 26 + 8 \)
\( \implies A = 34 \)
So, the correct number is 34.
In simple words: A number minus \( 8 \) equals \( 26 \). So, the starting number must be \( 8 \) more than \( 26 \), which is \( 34 \).

Exam Tip: Use consistent variables. If you define the variable as \( A \), use \( A \) throughout your working instead of switching to \( x \).

Question 3. One-fourth of a number added to two- seventh of it gives 135; find the number.
Answer: Let us call the unknown number \( x \).
From the information provided:
\( \frac{x}{4} + \frac{2}{7}x = 135 \)
\( \implies \frac{7x + 8x}{28} = 135 \) (Taking the LCM of 4 and 7 as 28)
\( \implies \frac{15x}{28} = 135 \)
\( \implies x = \frac{135 \times 28}{15} \)
\( \implies x = 9 \times 28 \)
\( \implies x = 252 \)
So, the value of the number is 252.
In simple words: Add one-fourth of the number to two-sevenths of it to equal \( 135 \). Use a common bottom number of \( 28 \) to combine the fractions and solve.

Exam Tip: Show the LCM step clearly. Writing down "LCM of 4 and 7 is 28" helps you score full steps marks.

Question 4. Two-fifths of a number subtracted from three-fourths of it gives 56, find the number.
Answer: Let the unknown value be \( x \).
As per the problem statement:
\( \frac{3}{4}x - \frac{2}{5}x = 56 \)
\( \implies \frac{15x - 8x}{20} = 56 \) (Taking the LCM of 4 and 5 as 20)
\( \implies \frac{7x}{20} = 56 \)
\( \implies x = \frac{56 \times 20}{7} \)
\( \implies x = 8 \times 20 \)
\( \implies x = 160 \)
Therefore, the required value is 160.
In simple words: When we take two-fifths of a number away from three-fourths of it, we get \( 56 \). We find the common denominator and solve to find \( 160 \).

Exam Tip: Be sure not to swap the order of subtraction. "Subtracted from" means the second term goes first in the equation.

Question 5. A number is increased by 12 and the new number obtained is multiplied by 5. If the resulting number is 95, find the original number.
Answer: Let the starting number be \( x \).
According to the given condition:
\( (x + 12) \times 5 = 95 \)
\( \implies 5x + 60 = 95 \)
\( \implies 5x = 95 - 60 \)
\( \implies 5x = 35 \)
\( \implies x = \frac{35}{5} \)
\( \implies x = 7 \)
Consequently, the starting number is 7.
In simple words: Add \( 12 \) to our number, then multiply the whole thing by \( 5 \) to get \( 95 \). We expand the equation to find the number is \( 7 \).

Exam Tip: Brackets are essential here! Write \( (x + 12) \times 5 \) and not \( x + 12 \times 5 \), because the entire sum must be multiplied.

Question 6. A number is increased by 26 and the new number obtained is divided by 3. If the resulting number is 18; find the original number.
Answer: Let the starting value be represented by \( x \).
From the given problem:
\( \frac{x + 26}{3} = 18 \)
\( \implies x + 26 = 18 \times 3 \)
\( \implies x + 26 = 54 \)
\( \implies x = 54 - 26 \)
\( \implies x = 28 \)
So, the original number is 28.
In simple words: We add \( 26 \) to a number and divide the result by \( 3 \) to get \( 18 \). Multiplying \( 18 \) by \( 3 \) and then subtracting \( 26 \) gives us \( 28 \).

Exam Tip: To solve equations with a fraction, clear the division first by multiplying both sides by the denominator.

Question 7. The age of a man is 27 years more than the age of his son. If the sum of their ages is 47 years, find the age of the son and his father.
Answer: Let us assume the son's age is \( x \) years.
Therefore, the age of the father becomes \( x + 27 \) years.
Based on the given sum:
\( x + (x + 27) = 47 \)
\( \implies 2x + 27 = 47 \)
\( \implies 2x = 47 - 27 \)
\( \implies 2x = 20 \)
\( \implies x = \frac{20}{2} \)
\( \implies x = 10 \)
Thus, the son is 10 years old.
The age of his father is \( 10 + 27 = 37 \) years.
In simple words: Since the father is \( 27 \) years older than his son and their total age is \( 47 \), we can set up an equation to find the son is \( 10 \) and the father is \( 37 \).

Exam Tip: Don't forget to write the units (years) in your final statement, as examiners cut marks for missing units.

Question 8. The difference between the ages of Gopal and his father is 26 years. If the sum of their ages is 56 years, find the ages of Gopal and his father.
Answer: Assume Gopal's age is \( x \) years.
Therefore, his father's age is \( (x + 26) \) years.
As per the given sum of their ages:
\( x + (x + 26) = 56 \)
\( \implies 2x + 26 = 56 \)
\( \implies 2x = 56 - 26 \)
\( \implies 2x = 30 \)
\( \implies x = \frac{30}{2} \)
\( \implies x = 15 \)
Thus, Gopal's age is 15 years, and his father's age is \( 15 + 26 = 41 \) years.
In simple words: If Gopal is x years old, his father is x + 26. Adding their ages gives 56, which helps us find that Gopal is 15 and his father is 41.

Exam Tip: Always define the variable for the smaller quantity first to keep the equation simpler and avoid negative numbers.

Question 9. When two consecutive natural numbers are added, the sum is 31; find the numbers.
Answer: Let the first natural number be \( x \).
So, the next consecutive natural number is \( x + 1 \).
Based on the given condition:
\( x + (x + 1) = 31 \)
\( \implies 2x + 1 = 31 \)
\( \implies 2x = 31 - 1 \)
\( \implies 2x = 30 \)
\( \implies x = \frac{30}{2} \)
\( \implies x = 15 \)
Hence, the first number is 15, and the second number is \( 15 + 1 = 16 \).
In simple words: Two numbers that follow each other add up to 31. By setting them as x and x + 1, we find the numbers are 15 and 16.

Exam Tip: Remember that consecutive natural numbers always differ by exactly 1, so they should be represented as x and x + 1.

Question 10. When three consecutive natural numbers are added, the sum is 66, find the numbers.
Answer: Let the first natural number be represented as \( x \).
Consequently, the second consecutive number is \( x + 1 \), and the third is \( x + 2 \).
According to the problem:
\( x + (x + 1) + (x + 2) = 66 \)
\( \implies 3x + 3 = 66 \)
\( \implies 3x = 66 - 3 \)
\( \implies 3x = 63 \)
\( \implies x = \frac{63}{3} \)
\( \implies x = 21 \)
So, the first number is 21, the second is 22, and the third is 23. Thus, the required numbers are 21, 22, and 23.
In simple words: Three consecutive numbers add up to 66. We can write them as x, x + 1, and x + 2, which gives us the numbers 21, 22, and 23.

Exam Tip: Clearly show the step of adding the consecutive terms and grouping the like terms (the x's) together before solving.

Question 11. A natural number decreased by 7 is 12. Find the number.
Answer: Let the unknown natural number be \( x \).
Subtracting 7 from this number gives 12:
\( x - 7 = 12 \)
\( \implies x = 12 + 7 \)
\( \implies x = 19 \)
Therefore, the required number is 19.
In simple words: When a number is reduced by 7, we get 12. To find the starting number, we simply add 7 to 12.

Exam Tip: Be sure to write down the equation first, then isolate the variable by performing the opposite operation on both sides.

Question 12. One fourth of a number added to one- sixth of itself is 15. Find the number.
Answer: Let the required number be defined as \( x \).
Adding one-fourth of the number to one-sixth of the number equals 15:
\( \frac{x}{4} + \frac{x}{6} = 15 \)
Find the common denominator for 4 and 6, which is 12:
\( \implies \frac{3x + 2x}{12} = 15 \)
\( \implies \frac{5x}{12} = 15 \)
\( \implies 5x = 15 \times 12 \)
\( \implies x = \frac{15 \times 12}{5} \)
\( \implies x = 3 \times 12 \)
\( \implies x = 36 \)
Hence, the value of the number is 36.
In simple words: We set up a fraction sum where x divided by 4 plus x divided by 6 equals 15. Solving this fraction equation gives us 36.

Exam Tip: Find the least common multiple (LCM) of the denominators to simplify equations containing fractions.

Question 13. A whole number is increased by 7 and the new number so obtained is multiplied by 5; the result is 45. Find the number.
Answer: Let the required whole number be \( x \).
If we increase this number by 7 and multiply the sum by 5, we get 45:
\( (x + 7) \times 5 = 45 \)
Divide both sides of the equation by 5:
\( \implies x + 7 = \frac{45}{5} \)
\( \implies x + 7 = 9 \)
\( \implies x = 9 - 7 \)
\( \implies x = 2 \)
Consequently, the required whole number is 2.
In simple words: First we add 7 to our number, then multiply by 5 to get 45. Working backward, we divide 45 by 5 and then subtract 7.

Exam Tip: Remember to place brackets around the addition part of the statement before multiplying by 5, as the entire sum is multiplied.

Question 14. The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man.
Answer: Let the daughter's age be \( x \) years.
Therefore, the father's age is \( (x + 23) \) years.
The sum of their ages is 41 years:
\( x + (x + 23) = 41 \)
\( \implies 2x + 23 = 41 \)
\( \implies 2x = 41 - 23 \)
\( \implies 2x = 18 \)
\( \implies x = \frac{18}{2} \)
\( \implies x = 9 \)
So, the daughter is 9 years old. The man's age is \( 9 + 23 = 32 \) years.
In simple words: Since the father is 23 years older than his daughter, we add their ages together to solve for the daughter's age, which is 9. Thus, the father is 32.

Exam Tip: Read the final question carefully: here it asks for the man's age, not the daughter's age, so do not stop after finding x.

Question 15. The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son.
Answer: Let the son's age be \( x \) years.
Thus, the mother's age is \( (x + 19) \) years.
Their total age combined is 37 years:
\( x + (x + 19) = 37 \)
\( \implies 2x + 19 = 37 \)
\( \implies 2x = 37 - 19 \)
\( \implies 2x = 18 \)
\( \implies x = \frac{18}{2} \)
\( \implies x = 9 \)
Therefore, the age of the son is 9 years.
In simple words: We can represent the son's age as x and the mother's as x + 19. Solving the sum of their ages reveals the son is 9 years old.

Exam Tip: Verify your answer by checking if the two ages (9 and 28) differ by 19 and sum up to 37.

Question 16. Two natrual numbers differ by 6 and sum of them is 36. Find the larger number.
Answer: Let the larger natural number be \( x \).
Since the two numbers differ by 6, the smaller number is \( x - 6 \).
The sum of these numbers is 36:
\( x + (x - 6) = 36 \)
\( \implies 2x - 6 = 36 \)
\( \implies 2x = 36 + 6 \)
\( \implies 2x = 42 \)
\( \implies x = \frac{42}{2} \)
\( \implies x = 21 \)
Therefore, the larger number is 21.
In simple words: If the bigger number is x, the smaller one is x - 6. Adding them gives 36, which means the larger number is 21.

Exam Tip: If the question asks for the larger number, it is easier to define the larger number directly as x so that the value of x is your final answer.

Question 17. The difference between two numbers is 15. Taking the smaller number as x; find:
(i) the expression for larger number.
(if) the larger number, if the sum of these numbers is 71.
Answer: Let the smaller number be \( x \).
Given that the difference is 15:
(i) The expression for the larger number is \( x + 15 \).
(ii) If their sum is 71, we can write:
\( x + (x + 15) = 71 \)
\( \implies 2x + 15 = 71 \)
\( \implies 2x = 71 - 15 \)
\( \implies 2x = 56 \)
\( \implies x = \frac{56}{2} \)
\( \implies x = 28 \)
Consequently, the larger number is \( 28 + 15 = 43 \).
In simple words: Since the difference is 15, the larger number is x + 15. Setting their sum to 71 tells us the smaller number is 28, and the larger one is 43.

Exam Tip: Be sure to write the distinct answers for both parts (i) and (ii) clearly as separate steps.

Question 18. The difference between two numbers is 23. Taking the larger number as x, find:
(i) the expression for smaller number.
(ii) the smaller number, if the sum of these two numbers is 91.
Answer: Let the larger number be \( x \).
Since they differ by 23:
(i) The expression for the smaller number is \( x - 23 \).
(ii) Given that the sum of these numbers is 91:
\( x + (x - 23) = 91 \)
\( \implies 2x - 23 = 91 \)
\( \implies 2x = 91 + 23 \)
\( \implies 2x = 114 \)
\( \implies x = \frac{114}{2} \)
\( \implies x = 57 \)
Thus, the smaller number is \( 57 - 23 = 34 \).
In simple words: If the larger number is x, the smaller one is x - 23. Solving the equation with their sum of 91 shows that the smaller number is 34.

Exam Tip: Double check that your final expression in part (i) correctly uses the larger variable to describe the smaller number.

Question 19. Find three consecutive integers such that their sum is 78.
Answer: Let the first of the three consecutive integers be \( x \).
Therefore, the second is \( x + 1 \) and the third is \( x + 2 \).
Their sum is 78:
\( x + (x + 1) + (x + 2) = 78 \)
\( \implies 3x + 3 = 78 \)
\( \implies 3x = 78 - 3 \)
\( \implies 3x = 75 \)
\( \implies x = \frac{75}{3} \)
\( \implies x = 25 \)
Thus, the three consecutive integers are 25, 26, and 27.
In simple words: Three consecutive integers that add up to 78 are 25, 26, and 27.

Exam Tip: Always list all the solved numbers in your final answer statement to complete the response.

Question 20. The sum of three consecutive numbers is 54. Taking the middle number as x, find:
(i) expression for the smallest number and the largest number.
(ii) the three numbers.
Answer: Let the middle number be \( x \).
(i) The expression for the smallest number is \( x - 1 \), and the expression for the largest number is \( x + 1 \).
(ii) Given that their sum is 54:
\( (x - 1) + x + (x + 1) = 54 \)
\( \implies 3x = 54 \)
\( \implies x = \frac{54}{3} \)
\( \implies x = 18 \)
Therefore, the first number is \( 18 - 1 = 17 \), and the third number is \( 18 + 1 = 19 \).
Hence, the three consecutive numbers are 17, 18, and 19.
In simple words: By choosing the middle number as x, the numbers on either side are x - 1 and x + 1. Their sum of 54 helps us find that the numbers are 17, 18, and 19.

Exam Tip: Using the middle number as x is highly efficient because the constant terms (-1 and +1) cancel out when you sum them up.

REVISION EXERCISE

Question 1(i). 2x + 3 = 7
Answer: Subtract 3 from both sides of the equation:
\( 2x + 3 - 3 = 7 - 3 \)
\( \implies 2x = 4 \)
Now, divide both sides by 2:
\( \frac{2x}{2} = \frac{4}{2} \)
\( \implies x = 2 \).
In simple words: First subtract 3 from 7 to get 4, then divide by 2 to get 2.

Exam Tip: Remember to always perform the same mathematical operation on both sides of the equation to keep it balanced.

Question 1(ii). 2x – 3 = 7
Answer: Add 3 to both sides:
\( 2x - 3 + 3 = 7 + 3 \)
\( \implies 2x = 10 \)
Divide both sides by 2:
\( \frac{2x}{2} = \frac{10}{2} \)
\( \implies x = 5 \).
In simple words: Add 3 to 7 to get 10, then divide 10 by 2 to find that x is 5.

Exam Tip: When a number is subtracted from the variable term, add it to both sides to eliminate it.

Question 1(iii). 2x ÷ 3 = 7
Answer: First, multiply both sides by 3 to eliminate the division:
\( \frac{2x}{3} \times 3 = 7 \times 3 \)
\( \implies 2x = 21 \)
Divide both sides by 2:
\( \frac{2x}{2} = \frac{21}{2} \)
\( \implies x = 10\frac{1}{2} \).
In simple words: Multiply 7 by 3 to get 21, and then divide 21 by 2 to get 10 and a half.

Exam Tip: Leave your final answer as a mixed fraction if it does not divide into a whole number.

Question 1(iv). 3x – 8 = 13
Answer: Add 8 to both sides:
\( 3x - 8 + 8 = 13 + 8 \)
\( \implies 3x = 21 \)
Divide both sides by 3:
\( \frac{3x}{3} = \frac{21}{3} \)
\( \implies x = 7 \).
In simple words: First add 8 to 13 to get 21, then divide by 3 to find x is 7.

Exam Tip: Be careful with basic addition and division steps to prevent simple calculation errors.

Question 1(v). 3y + 8 = 13
Answer: Subtract 8 from both sides:
\( 3y + 8 - 8 = 13 - 8 \)
\( \implies 3y = 5 \)
Divide both sides by 3:
\( \frac{3y}{3} = \frac{5}{3} \)
\( \implies y = 1\frac{2}{3} \).
In simple words: Subtract 8 from 13 to get 5, then divide 5 by 3 to get 1 and two-thirds.

Exam Tip: Convert improper fractions like 5/3 into mixed fractions for cleaner final presentation.

Question 1(vi). 3y ÷ 8 = 13
Answer: Multiply both sides by 8:
\( \frac{3y}{8} \times 8 = 13 \times 8 \)
\( \implies 3y = 104 \)
Divide both sides by 3:
\( \frac{3y}{3} = \frac{104}{3} \)
\( \implies y = 34\frac{2}{3} \).
In simple words: Multiply 13 by 8 to get 104, then divide by 3 to get 34 and two-thirds.

Exam Tip: Practice long division to easily convert large improper fractions into mixed numbers.

Question 1(vii). x – 3 = 5\frac{1}{2}
Answer: Add 3 to both sides:
\( x - 3 + 3 = 5\frac{1}{2} + 3 \)
\( \implies x = 8\frac{1}{2} \).
In simple words: Adding 3 to 5 and a half gives 8 and a half.

Exam Tip: When adding a whole number to a mixed fraction, you only need to add it to the whole number part.

Question 1(viii). \frac{3}{5}x + 4 = 13
Answer: Subtract 4 from both sides:
\( \frac{3}{5}x + 4 - 4 = 13 - 4 \)
\( \implies \frac{3}{5}x = 9 \)
Multiply both sides by \( \frac{5}{3} \):
\( \frac{3}{5}x \times \frac{5}{3} = 9 \times \frac{5}{3} \)
\( \implies x = 3 \times 5 \)
\( \implies x = 15 \).
In simple words: Subtract 4 from 13 to get 9, then multiply 9 by 5 and divide by 3 to find x is 15.

Exam Tip: Multiplying by the reciprocal of a fraction coefficient is a quick way to isolate the variable.

Question 1(ix). u + 3\frac{1}{4} = 4\frac{1}{3}
Answer: Convert mixed fractions to improper fractions:
\( u + \frac{13}{4} = \frac{13}{3} \)
Subtract \( \frac{13}{4} \) from both sides:
\( u = \frac{13}{3} - \frac{13}{4} \)
Find a common denominator of 12:
\( u = \frac{52 - 39}{12} \)
\( \implies u = \frac{13}{12} \)
\( \implies u = 1\frac{1}{12} \).
In simple words: To find u, subtract 3 and a quarter from 4 and a third. The result is 1 and one-twelfth.

Exam Tip: Converting mixed fractions to improper fractions first makes subtraction much easier and less prone to errors.

Question 1(x). 5x – 2.4 = 4.9
Answer: Add 2.4 to both sides:
\( 5x - 2.4 + 2.4 = 4.9 + 2.4 \)
\( \implies 5x = 7.3 \)
Divide both sides by 5:
\( \frac{5x}{5} = \frac{7.3}{5} \)
\( \implies x = 1.46 \).
In simple words: Add 2.4 to 4.9 to get 7.3, then divide by 5 to find x is 1.46.

Exam Tip: When working with decimals, align the decimal points carefully during addition.

Question 1(xi). 5y + 4.9 = 2.4
Answer: Subtract 4.9 from both sides:
\( 5y + 4.9 - 4.9 = 2.4 - 4.9 \)
\( \implies 5y = -2.5 \)
Divide both sides by 5:
\( \frac{5y}{5} = \frac{-2.5}{5} \)
\( \implies y = -0.5 \).
In simple words: Subtract 4.9 from 2.4 to get negative 2.5, then divide by 5 to get negative 0.5.

Exam Tip: Pay close attention to negative signs when subtracting a larger number from a smaller number.

Question 1(xii). 48 z + 3.6 = 1.2
Answer: Solve the equation by subtracting 3.6 from both sides:
\( 4.8z + 3.6 - 3.6 = 1.2 - 3.6 \)
\( \implies 4.8z = -2.4 \)
Divide both sides by 4.8:
\( z = \frac{-2.4}{4.8} \)
\( \implies z = -0.5 \).
In simple words: Subtracting 3.6 from 1.2 gives negative 2.4, which we then divide by 4.8 to get negative 0.5.

Exam Tip: Watch for minor typographical discrepancies in the source equations, and resolve them by maintaining consistent equation-solving steps.

Question 1(xiii). \frac{x}{2} – 3 = 5
Answer: Add 3 to both sides:
\( \frac{x}{2} - 3 + 3 = 5 + 3 \)
\( \implies \frac{x}{2} = 8 \)
Multiply both sides by 2:
\( \frac{x}{2} \times 2 = 8 \times 2 \)
\( \implies x = 16 \).
In simple words: First add 3 to 5 to get 8, then multiply by 2 to find x is 16.

Exam Tip: Isolate the fraction term first by adding or subtracting constants before multiplying to clear the denominator.

Question 1(xiv). \frac{y}{3} + 7 = 2
Answer: Subtract 7 from both sides:
\( \frac{y}{3} + 7 - 7 = 2 - 7 \)
\( \implies \frac{y}{3} = -5 \)
Multiply both sides by 3:
\( \frac{y}{3} \times 3 = -5 \times 3 \)
\( \implies y = -15 \).
In simple words: Subtract 7 from 2 to get negative 5, then multiply by 3 to get negative 15.

Exam Tip: Keep careful track of negative signs when subtracting a larger number from a smaller one.

Question 1(xv). \frac{2m}{3} = 8\frac{2}{3}
Answer: Convert the mixed fraction to an improper fraction:
\( \frac{2m}{3} = \frac{26}{3} \)
Multiply both sides by \( \frac{3}{2} \):
\( \frac{2m}{3} \times \frac{3}{2} = \frac{26}{3} \times \frac{3}{2} \)
\( \implies m = 13 \).
In simple words: Convert 8 and two-thirds to 26 over 3, then multiply by 3 over 2 to find m is 13.

Exam Tip: If denominators are the same on both sides, they can be directly canceled out to simplify the equation.

Question 1(xvi). -3x + 4 = 10
Answer: Subtract 4 from both sides:
\( -3x + 4 - 4 = 10 - 4 \)
\( \implies -3x = 6 \)
Divide both sides by -3:
\( \frac{-3x}{-3} = \frac{6}{-3} \)
\( \implies x = -2 \).
In simple words: Subtract 4 from 10 to get 6, then divide 6 by negative 3 to get negative 2.

Exam Tip: Dividing a positive number by a negative number always results in a negative value.

Question 1(xvii). 5 = x – 3
Answer: Add 3 to both sides to solve for x:
\( 5 + 3 = x - 3 + 3 \)
\( \implies 8 = x \)
\( \implies x = 8 \).
In simple words: Add 3 to 5 to find that x equals 8.

Exam Tip: The variable can be on either side of the equation; simply isolate it as usual.

Question 1(xviii). 8y = 3- 3y
Answer: Solve the equation by subtracting 3 from both sides:
\( 18 - 3 = 3 - 3y - 3 \)
\( \implies 15 = -3y \)
Divide both sides by -3:
\( \frac{15}{-3} = \frac{-3y}{-3} \)
\( \implies y = -5 \).
In simple words: Subtract 3 from 18 to get 15, then divide by negative 3 to get negative 5.

Exam Tip: Be careful with signs when dividing a positive number by a negative number.

Question 1(xix). 4x + 4.9 = 6.5
Answer: Subtract 4.9 from both sides:
\( 4x + 4.9 - 4.9 = 6.5 - 4.9 \)
\( \implies 4x = 1.6 \)
Divide both sides by 4:
\( \frac{4x}{4} = \frac{1.6}{4} \)
\( \implies x = 0.4 \).
In simple words: Subtract 4.9 from 6.5 to get 1.6, then divide by 4 to get 0.4.

Exam Tip: Convert decimal subtraction carefully by aligning the decimal points before subtracting.

Question xx. 3z + 2 = -4
Answer:
Given equation: \( 3z + 2 = -4 \)
We subtract 2 from both the left and right sides:
\( \implies 3z + 2 - 2 = -4 - 2 \)
\( \implies 3z = -6 \)
Now, we divide both sides by 3:
\( \implies \frac{3z}{3} = \frac{-6}{3} \)
\( \implies z = -2 \)
In simple words: First, take away 2 from both sides to get the term with z by itself. Then, divide both sides by 3 to find that z equals -2.

Exam Tip: Always perform the same operation on both sides of the equation to keep it balanced, and double check your signs when working with negative numbers.

Question xxi. 7y - 18 = 17
Answer:
Given equation: \( 7y - 18 = 17 \)
We add 18 to both sides to isolate the variable term:
\( \implies 7y - 18 + 18 = 17 + 18 \)
\( \implies 7y = 35 \)
Next, we divide both sides by 7:
\( \implies \frac{7y}{7} = \frac{35}{7} \)
\( \implies y = 5 \)
In simple words: Add 18 to both sides to get 7y by itself. After that, divide both sides by 7 to find that y is 5.

Exam Tip: Remember to add when you see a subtraction sign, as performing the opposite action helps cancel out numbers next to your variable.

Question xxii. \( \frac{x}{1.2} - 6 = 1 \)
Answer:
Given equation: \( \frac{x}{1.2} - 6 = 1 \)
First, we add 6 to both sides of the equation:
\( \implies \frac{x}{1.2} - 6 + 6 = 1 + 6 \)
\( \implies \frac{x}{1.2} = 7 \)
Next, we multiply both sides by 1.2:
\( \implies \frac{x}{1.2} \times 1.2 = 7 \times 1.2 \)
\( \implies x = 8.4 \)
In simple words: First, add 6 to both sides so only the fraction is left on one side. Then, multiply both sides by 1.2 to find that x is 8.4.

Exam Tip: Be extra careful with decimal multiplication; counting the decimal places correctly in your final product prevents easy mistakes.

Question xxiii. \( \frac{z}{2.4} + 3.6 = 5.1 \)
Answer:
Given equation: \( \frac{z}{2.4} + 3.6 = 5.1 \)
First, subtract 3.6 from both sides of the equation:
\( \implies \frac{z}{2.4} + 3.6 - 3.6 = 5.1 - 3.6 \)
\( \implies \frac{z}{2.4} = 1.5 \)
Now, we multiply both sides by 2.4:
\( \implies \frac{z}{2.4} \times 2.4 = 1.5 \times 2.4 \)
\( \implies z = 3.6 \)
In simple words: Subtract 3.6 from both sides to clear the decimal addition. Then, multiply both sides by 2.4 to get z by itself, which gives 3.6.

Exam Tip: When you multiply decimal values like 1.5 and 2.4, you can think of them as 15 and 24, multiply them to get 360, and then put the decimal point two places from the right.

Question xxiv. \( \frac{y}{1.8} - 2.1 = -2.8 \)
Answer:
Given equation: \( \frac{y}{1.8} - 2.1 = -2.8 \)
We add 2.1 to both sides of the equation:
\( \implies \frac{y}{1.8} - 2.1 + 2.1 = -2.8 + 2.1 \)
\( \implies \frac{y}{1.8} = -0.7 \)
Next, we multiply both sides by 1.8:
\( \implies \frac{y}{1.8} \times 1.8 = -0.7 \times 1.8 \)
\( \implies y = -1.26 \)
In simple words: Add 2.1 to both sides to get the fraction by itself. Then, multiply both sides by 1.8 to find that y equals -1.26.

Exam Tip: When working with negative numbers, remember that adding a positive number to a negative number moves you closer to zero (from -2.8 to -0.7).

Question xxv. 7x - 2 = 4x + 7
Answer:
Given equation: \( 7x - 2 = 4x + 7 \)
We add 2 to both sides of the equation:
\( \implies 7x - 2 + 2 = 4x + 7 + 2 \)
\( \implies 7x = 4x + 9 \)
Next, we subtract 4x from both sides:
\( \implies 7x - 4x = 4x + 9 - 4x \)
\( \implies 3x = 9 \)
Finally, we divide both sides by 3:
\( \implies \frac{3x}{3} = \frac{9}{3} \)
\( \implies x = 3 \)
In simple words: Add 2 to both sides to clear the -2. Then, subtract 4x from both sides to group all the x terms together. Divide by 3 to find x is 3.

Exam Tip: When variables are on both sides, always shift the variable with the smaller coefficient to the other side to keep your calculations positive and simpler.

Question xxvi. 3y -(y -+2) = 4
Answer:
Given equation: \( 3y - (y + 2) = 4 \)
We open the parentheses by distributing the negative sign:
\( \implies 3y - y - 2 = 4 \)
Simplify the left side:
\( \implies 2y - 2 = 4 \)
Next, add 2 to both sides of the equation:
\( \implies 2y - 2 + 2 = 4 + 2 \)
\( \implies 2y = 6 \)
Divide both sides by 2:
\( \implies \frac{2y}{2} = \frac{6}{2} \)
\( \implies y = 3 \)
In simple words: First, open the parentheses by changing the signs inside. Group the y terms to get 2y - 2 = 4. Add 2 to both sides, then divide by 2 to find y is 3.

Exam Tip: Be very careful when distributing a negative sign across parentheses; it changes the sign of every term inside, so +(2) becomes -2.

Question xxvii. 3z – 18 = z – (12 - 4z)
Answer:
Given equation: \( 3z - 18 = z - (12 - 4z) \)
First, remove the parentheses by distributing the negative sign:
\( \implies 3z - 18 = z - 12 + 4z \)
Combine the like terms on the right side:
\( \implies 3z - 18 = 5z - 12 \)
Add 18 to both sides of the equation:
\( \implies 3z - 18 + 18 = 5z - 12 + 18 \)
\( \implies 3z = 5z + 6 \)
Now, subtract 5z from both sides:
\( \implies 3z - 5z = 5z + 6 - 5z \)
\( \implies -2z = 6 \)
Finally, divide both sides by -2:
\( \implies \frac{-2z}{-2} = \frac{6}{-2} \)
\( \implies z = -3 \)
In simple words: Expand the brackets first by changing the sign inside. Combine the z terms on the right side to get 5z - 12. Add 18 to both sides, subtract 5z from both sides, and divide by -2 to get z = -3.

Exam Tip: Don't forget that dividing a positive number by a negative number yields a negative result.

Question xxiii. \( x - 2\frac{1}{3} = 5\frac{1}{2} \)
Answer:
Given equation: \( x - 2\frac{1}{3} = 5\frac{1}{2} \)
Add \( 2\frac{1}{3} \) to both sides of the equation:
\( \implies x - 2\frac{1}{3} + 2\frac{1}{3} = 5\frac{1}{2} + 2\frac{1}{3} \)
\( \implies x = 5\frac{1}{2} + 2\frac{1}{3} \)
Convert both mixed fractions into improper fractions:
\( \implies x = \frac{11}{2} + \frac{7}{3} \)
Find a common denominator to add the fractions:
\( \implies x = \frac{33 + 14}{6} \)
\( \implies x = \frac{47}{6} \)
Convert the improper fraction back into a mixed number:
\( \implies x = 7\frac{5}{6} \)
In simple words: Add 2 and 1/3 to both sides. Change the mixed numbers to improper fractions, find a common denominator of 6 to add them, and convert the final fraction back to a mixed number.

Exam Tip: When working with mixed numbers, converting them to improper fractions before adding or subtracting makes the calculation much easier and less prone to errors.

Question xxix. \( 3\frac{2}{5} - y = 2\frac{1}{2} \)
Answer:
Given equation: \( 3\frac{2}{5} - y = 2\frac{1}{2} \)
Subtract \( 3\frac{2}{5} \) from both sides:
\( \implies 3\frac{2}{5} - y - 3\frac{2}{5} = 2\frac{1}{2} - 3\frac{2}{5} \)
\( \implies -y = \frac{5}{2} - \frac{17}{5} \)
Find a common denominator of 10 to combine the terms on the right side:
\( \implies -y = \frac{25 - 34}{10} \)
\( \implies -y = \frac{-9}{10} \)
Multiply both sides by -1:
\( \implies y = \frac{9}{10} \)
In simple words: Subtract 3 and 2/5 from both sides. Convert the mixed numbers into fractions, make their denominators the same, subtract them, and change the negative sign.

Exam Tip: Be careful with the negative sign in front of the variable (like -y); make sure to multiply or divide by -1 at the end to solve for the positive variable.

Question xxx. \( 2z - 2\frac{1}{2} = 3\frac{1}{3} \)
Answer:
Given equation: \( 2z - 2\frac{1}{2} = 3\frac{1}{3} \)
Add \( 2\frac{1}{2} \) to both sides of the equation:
\( \implies 2z - 2\frac{1}{2} + 2\frac{1}{2} = 3\frac{1}{3} + 2\frac{1}{2} \)
\( \implies 2z = \frac{10}{3} + \frac{5}{2} \)
Combine the fractions with a common denominator of 6:
\( \implies 2z = \frac{20 + 15}{6} \)
\( \implies 2z = \frac{35}{6} \)
Now, divide both sides by 2:
\( \implies \frac{2z}{2} = \frac{35}{6 \times 2} \)
\( \implies z = \frac{35}{12} \)
Convert to a mixed number:
\( \implies z = 2\frac{11}{12} \)
In simple words: Add 2 and 1/2 to both sides. Turn the mixed fractions into improper ones, add them together using a common denominator, and then divide the result by 2.

Exam Tip: When dividing a fraction by a whole number, simply multiply the denominator by that whole number.

Question xxxi. 5x – 2x +15 = 27
Answer:
Given equation: \( 5x - 2x + 15 = 27 \)
Combine the like terms on the left side:
\( \implies 3x + 15 = 27 \)
Subtract 15 from both sides of the equation:
\( \implies 3x + 15 - 15 = 27 - 15 \)
\( \implies 3x = 12 \)
Next, divide both sides by 3:
\( \implies \frac{3x}{3} = \frac{12}{3} \)
\( \implies x = 4 \)
In simple words: First combine 5x and -2x to get 3x. Subtract 15 from both sides, then divide by 3 to find x is 4.

Exam Tip: Grouping similar terms on one side of the equation before performing inverse operations keeps your work structured and reduces errors.

Question xxxii. 5y – 15 = 27 -2y
Answer:
Given equation: \( 5y - 15 = 27 - 2y \)
We add 2y to both sides of the equation:
\( \implies 5y + 2y - 15 = 27 - 2y + 2y \)
\( \implies 7y - 15 = 27 \)
Now, we add 15 to both sides:
\( \implies 7y - 15 + 15 = 27 + 15 \)
\( \implies 7y = 42 \)
Finally, divide both sides by 7:
\( \implies \frac{7y}{7} = \frac{42}{7} \)
\( \implies y = 6 \)
In simple words: Add 2y to both sides so all y terms are on one side. Then, add 15 to both sides and divide by 7 to find that y equals 6.

Exam Tip: Try to collect the variables on the side that keeps their coefficient positive, as this minimizes errors with negative numbers.

Question xxxiii. 7z + 15 = 3z – 13
Answer:
Given equation: \( 7z + 15 = 3z - 13 \)
We subtract 3z from both sides:
\( \implies 7z + 15 - 3z = 3z - 13 - 3z \)
\( \implies 4z + 15 = -13 \)
Next, subtract 15 from both sides of the equation:
\( \implies 4z + 15 - 15 = -13 - 15 \)
\( \implies 4z = -28 \)
Finally, divide both sides by 4:
\( \implies \frac{4z}{4} = \frac{-28}{4} \)
\( \implies z = -7 \)
In simple words: Subtract 3z from both sides so all z terms are together. Then, subtract 15 from both sides, and divide by 4 to get z = -7.

Exam Tip: Make sure not to lose the negative sign on the right-hand side when combining -13 and -15.

Question xxxiv. 2 (x -3) – 3 (x-4) =12
Answer:
Given equation: \( 2(x - 3) - 3(x - 4) = 12 \)
Expand the terms inside the parentheses:
\( \implies 2x - 6 - 3x + 12 = 12 \)
Combine the like terms on the left side:
\( \implies -x + 6 = 12 \)
Subtract 6 from both sides of the equation:
\( \implies -x + 6 - 6 = 12 - 6 \)
\( \implies -x = 6 \)
Multiply both sides by -1:
\( \implies x = -6 \)
In simple words: First expand the brackets, watching out for the negative signs. Combine the x terms and numbers to get -x + 6 = 12. Subtract 6 and change the sign to find x is -6.

Exam Tip: Be very careful when expanding \( -3(x-4) \). Distributing the negative three yields \( -3x + 12 \), not \( -3x - 12 \).

Question xxxv. (7y +8) ÷ 7= 8
Answer:
Given equation: \( \frac{7y + 8}{7} = 8 \)
Multiply both sides by 7:
\( \implies \frac{7y + 8}{7} \times 7 = 8 \times 7 \)
\( \implies 7y + 8 = 56 \)
Subtract 8 from both sides of the equation:
\( \implies 7y + 8 - 8 = 56 - 8 \)
\( \implies 7y = 48 \)
Divide both sides by 7:
\( \implies \frac{7y}{7} = \frac{48}{7} \)
\( \implies y = 6\frac{6}{7} \)
In simple words: First write the division as a fraction. Multiply both sides by 7 to clear the denominator, then subtract 8 from 56 to get 48. Divide by 7 to find y = 6 and 6/7.

Exam Tip: Convert your final improper fraction into a mixed number to present your answer in its most standard textbook format.

Question xxxvi. 2(z-5) +3 (z+2) -(3-5z) =10
Answer:
Given equation: \( 2(z - 5) + 3(z + 2) - (3 - 5z) = 10 \)
First, expand all brackets:
\( \implies 2z - 10 + 3z + 6 - 3 + 5z = 10 \)
Group and simplify the z terms and numbers on the left side:
\( \implies 10z - 7 = 10 \)
Add 7 to both sides of the equation:
\( \implies 10z - 7 + 7 = 10 + 7 \)
\( \implies 10z = 17 \)
Divide both sides by 10:
\( \implies \frac{10z}{10} = \frac{17}{10} \)
\( \implies z = 1\frac{7}{10} \)
In simple words: Open all brackets, then group the z terms and numbers. This gives 10z - 7 = 10. Add 7 to 10 to get 17, then divide by 10.

Exam Tip: Be very careful when expanding the negative bracket \( -(3-5z) \) as the minus sign outside changes \( -5z \) into \( +5z \).

Question 2. A natural number decreased by 7 is 12. Find the number.
Answer:
Let us assume the unknown number is \( x \).
According to the given condition:
\( \implies x - 7 = 12 \)
Now, we add 7 to both sides of the equation:
\( \implies x - 7 + 7 = 12 + 7 \)
\( \implies x = 19 \)
Thus, the required number is 19.
In simple words: Let the unknown number be x. If we take away 7 from x, we get 12. Adding 7 back to 12 gives us the original number, which is 19.

Exam Tip: Always clearly state your assumption (like "Let the number be x") at the beginning of any word problem to earn full presentation marks.

Question 3. One-fourth of a number added to one-sixth of It is 15. Find the number.
Answer:
Let the unknown number be \( x \).
Based on the problem, we set up the equation:
\( \implies \frac{x}{4} + \frac{x}{6} = 15 \)
Find a common denominator of 12 for the fractions:
\( \implies \frac{3x + 2x}{12} = 15 \)
\( \implies \frac{5x}{12} = 15 \)
Multiply both sides by 12 and divide by 5:
\( \implies x = \frac{15 \times 12}{5} \)
\( \implies x = 36 \)
Therefore, the required number is 36.
In simple words: Let the number be x. Add x/4 and x/6 by using a common denominator of 12. This gives 5x/12 = 15. Solve for x to get 36.

Exam Tip: When dealing with fractions in word problems, finding the LCM of the denominators (here, LCM of 4 and 6 is 12) is the key to simplifying the equation quickly.

Question 4. A whole number is increased by 7 and the number so obtained is multiplied by 5; the result is 45. Find the whole number.
Answer:
Let the whole number we want to find be \( x \).
According to the given scenario:
\( \implies (x + 7) \times 5 = 45 \)
Divide both sides by 5 to simplify:
\( \implies \frac{(x + 7) \times 5}{5} = \frac{45}{5} \)
\( \implies x + 7 = 9 \)
Subtract 7 from both sides:
\( \implies x = 9 - 7 \)
\( \implies x = 2 \)
Thus, the required whole number is 2.
In simple words: Let the number be x. Adding 7 and multiplying by 5 gives 45. Divide 45 by 5 to get 9, then subtract 7 to find the number is 2.

Exam Tip: Remember to use parentheses around \( x + 7 \) because the entire sum, not just \( x \), is multiplied by 5.

Question 5. The age of a man and the age of his daughter differ by 23 years and the sum of their ages is 41 years. Find the age of the man.
Answer:
Let the age of the daughter be \( x \) years.
Since the father is 23 years older, his age is \( x + 23 \) years.
The sum of their ages is 41 years:
\( \implies x + (x + 23) = 41 \)
\( \implies 2x + 23 = 41 \)
Subtract 23 from both sides of the equation:
\( \implies 2x = 41 - 23 \)
\( \implies 2x = 18 \)
Divide by 2 to find the daughter's age:
\( \implies x = \frac{18}{2} \)
\( \implies x = 9 \)
Now, find the father's age:
\( \implies \text{Age of man} = x + 23 = 9 + 23 = 32 \text{ years} \)
So, the father is 32 years old.
In simple words: Let the daughter's age be x. The father's age is x + 23. Adding them together gives 41. We find x is 9, so the father's age is 9 + 23 = 32 years.

Exam Tip: Be sure to write the final unit ("years") in your concluding sentence, as units are essential for full credit in word problems.

Question 6. The difference between the ages of a woman and her son is 19 years and the sum of their ages is 37 years; find the age of the son.
Answer:
Let the age of the son be \( x \) years.
Therefore, the age of the woman is \( x + 19 \) years.
Given that the sum of their ages is 37 years:
\( \implies x + x + 19 = 37 \)
\( \implies 2x + 19 = 37 \)
Subtract 19 from both sides of the equation:
\( \implies 2x = 37 - 19 \)
\( \implies 2x = 18 \)
Divide both sides by 2:
\( \implies x = \frac{18}{2} \)
\( \implies x = 9 \)
So, the age of the son is 9 years.
In simple words: If the son's age is x, his mother is x + 19. Together they are 37. Solving the equation gives the son's age as 9 years.

Exam Tip: Read carefully to identify whose age is asked (here, the son's age \(x\)), so you don't accidentally calculate and write down the mother's age as the final answer.

Question 7. Two natural numbers differ by 6 and their sum is 36. Find the larger number.
Answer:
Given that the difference between the two numbers is 6 and their sum is 36.
Let us assume the larger natural number is \( x \).
This means the smaller natural number is \( x - 6 \).
The sum of these two numbers is 36:
\( \implies x + (x - 6) = 36 \)
\( \implies 2x - 6 = 36 \)
Add 6 to both sides of the equation:
\( \implies 2x = 36 + 6 \)
\( \implies 2x = 42 \)
Divide both sides by 2:
\( \implies x = \frac{42}{2} \)
\( \implies x = 21 \)
So, the larger number is 21.
In simple words: Let the larger number be x. The smaller one is x - 6. Adding them gives 36. Solving the equation shows that the larger number is 21.

Exam Tip: Defining the larger number directly as \(x\) simplifies your final step, as you do not need to do any extra subtraction to find the required value.

Question 8. The difference between two numbers is 15. Taking the smaller number as x; find :
(i) the expression for the larger number.
(ii) the larger number, if the sum of these numbers is 71.
Answer:
Given that the difference between the two numbers is 15, and the smaller number is \( x \).
(i) The expression for the larger number is:
\( \implies \text{Larger number} = x + 15 \)
(ii) If the sum of the two numbers is 71:
\( \implies x + (x + 15) = 71 \)
\( \implies 2x + 15 = 71 \)
Subtract 15 from both sides:
\( \implies 2x = 71 - 15 \)
\( \implies 2x = 56 \)
Divide both sides by 2 to find the smaller number:
\( \implies x = \frac{56}{2} \)
\( \implies x = 28 \)
Now, find the larger number by substituting \( x = 28 \):
\( \implies \text{Larger number} = x + 15 = 28 + 15 = 43 \)
In simple words: (i) Since the difference is 15, the larger number is x + 15. (ii) Adding both numbers gives 71, so we find x is 28. Adding 15 to 28 gives 43 as the larger number.

Exam Tip: Be sure to answer both sub-questions (i) and (ii) clearly and label them separately to make sure you get full marks for each part.

Question 9. The difference between two numbers is 23. Taking the larger number as x, find :
(i) the expression for smaller number.
(ii) the smaller number, if the sum of these two numbers is 91.
Answer:
Given that the difference between the two numbers is 23, and the larger number is \( x \).
(i) The expression for the smaller number is:
\( \implies \text{Smaller number} = x - 23 \)
(ii) If the sum of these two numbers is 91:
\( \implies x + (x - 23) = 91 \)
\( \implies 2x - 23 = 91 \)
Add 23 to both sides of the equation:
\( \implies 2x = 91 + 23 \)
\( \implies 2x = 114 \)
Divide both sides by 2 to find the larger number:
\( \implies x = \frac{114}{2} \)
\( \implies x = 57 \)
Now, calculate the smaller number:
\( \implies \text{Smaller number} = x - 23 = 57 - 23 = 34 \)
In simple words: (i) Since the larger number is x and they differ by 23, the smaller is x - 23. (ii) Adding them together equals 91. We solve to find x is 57, and subtracting 23 gives the smaller number, which is 34.

Exam Tip: Since the problem defines \(x\) as the larger number, remember that the smaller number must be \(x - 23\), not \(x + 23\).

Question 10. Find the three consecutive integers whose sum is 78.
Answer:
Let the three consecutive integers be \( x \), \( x + 1 \), and \( x + 2 \).
The sum of these three integers is 78:
\( \implies x + (x + 1) + (x + 2) = 78 \)
\( \implies 3x + 3 = 78 \)
Subtract 3 from both sides:
\( \implies 3x = 78 - 3 \)
\( \implies 3x = 75 \)
Divide both sides by 3:
\( \implies x = \frac{75}{3} \)
\( \implies x = 25 \)
Now, find the three integers:
First integer = \( 25 \)
Second integer = \( 25 + 1 = 26 \)
Third integer = \( 25 + 2 = 27 \)
Thus, the three required consecutive integers are 25, 26, and 27.
In simple words: Let the three numbers in a row be x, x + 1, and x + 2. Their sum is 78. Solving this tells us the first number is 25, so the numbers are 25, 26, and 27.

Exam Tip: Consecutive integers always increase by 1, so setting them up as \(x\), \(x+1\), and \(x+2\) is the standard method to solve such problems.

Question 11. The sum of three consecutive numbers is 54. Taking the middle number as x, find :
(i) the expressions for the smallest number and the largest number.
(ii) the three numbers.
Answer:
Given that the middle number of the three consecutive integers is \( x \), and their sum is 54.
(i) Since the numbers are consecutive and \( x \) is the middle number:
The smallest number is \( x - 1 \)
The largest number is \( x + 1 \)
(ii) The sum of these three numbers is 54:
\( \implies (x - 1) + x + (x + 1) = 54 \)
\( \implies 3x = 54 \)
Divide both sides by 3:
\( \implies x = \frac{54}{3} \)
\( \implies x = 18 \)
Now, calculate the three numbers:
Smallest number = \( 18 - 1 = 17 \)
Middle number = \( 18 \)
Largest number = \( 18 + 1 = 19 \)
Thus, the three required numbers are 17, 18, and 19.
In simple words: (i) Since x is in the middle, the smallest is x - 1 and the largest is x + 1. (ii) Their sum is 54, which simplifies to 3x = 54. Solving gives x = 18, so the three numbers are 17, 18, and 19.

Exam Tip: Defining the middle number as \(x\) is highly efficient because the -1 and +1 cancel out when you add the terms, making the equation very quick to solve.