ML Aggarwal Class 7 Maths Solutions Chapter 08 Algebraic Expressions

Question 1. Form the algebraic expressions using variables, constants and arithmetic operations:
(i) 6 more than thrice a number x
(ii) 5 times x is subtracted from 13
(iii) The numbers x and y both squared and added.
(iv) Number 7 is added to 3 times the product of p and q
(v) Three times of x is subtracted from the product of x with itself.
(vi) Sum of the numbers m and n is subtracted from their product.
Answer:
(i) When you multiply a number x by 3, you get 3x. Adding 6 to this gives you the expression 3x + 6.
(ii) When you multiply 5 by x, you get 5x. Taking this away from 13 gives the expression 13 - 5x.
(iii) When x is squared, you write x². When y is squared, you write y². Adding both together gives you the expression x² + y².
(iv) Multiplying p and q gives you pq. Multiplying this by 3 gives 3pq. Adding 7 to this gives you the expression 3pq + 7.
(v) Multiplying x by itself gives x². Multiplying 3 by x gives 3x. Subtracting 3x from x² gives you the expression x² - 3x.
(vi) Multiplying m and n gives mn. Adding m and n gives m + n. Subtracting this sum from the product gives you the expression mn - (m + n).
In simple words: Translate each phrase step by step using variables like x, y, p, q, and combine them with addition, subtraction, and multiplication to form the final expression.

Exam Tip: Break down each phrase into smaller parts and translate each action (more than, subtracted from, squared, added) into the matching math operation. Always check that your variables and signs match the words in the question.

Question 2. A taxi charges Rs. 9 per km and a fixed charge of Rs. 50. If the taxi is hired for x km, write an algebraic expression of the invoice.
Answer: The taxi costs Rs. 9 for every kilometer driven. For x kilometers, the charge is Rs. 9x. On top of this, there is a fixed charge of Rs. 50 that is added no matter what. So the total bill is Rs. 9x + Rs. 50. The algebraic expression for the invoice is Rs. (9x + 50).
In simple words: Multiply the cost per km by the number of km, then add the fixed charge to get the total cost.

Exam Tip: Identify the variable charge (depends on usage) and the fixed charge (does not change). Add them together to form the final expression. Always write Rs. or the currency symbol clearly in your answer.

Question 3. Write the algebraic expressions whose terms are:
(i) 5a, -3b, c
(ii) x², -5x, 6
(iii) x²y, xy, -xy²
Answer:
(i) You need to combine the three given terms by putting them together with their signs: 5a - 3b + c.
(ii) Putting together the three terms x², -5x, and 6 creates the expression x² - 5x + 6.
(iii) Joining the three terms x²y, xy, and -xy² together gives you the expression x²y + xy - xy².
In simple words: Take each term as it is given and join them with addition or subtraction signs to form a complete expression.

Exam Tip: Write each term exactly as it appears, including its sign. Make sure all terms are combined in the order given in the question for clarity.

Question 4. Write all the terms of each of the following algebraic expressions:
(i) 3 - 7x
(ii) 2 - 5a + \( \frac{2}{3}b \)
(iii) 3x⁵ + 4y³ - 7xy² + 3
Answer:
(i) The expression 3 - 7x has two separate parts: the term 3 and the term -7x.
(ii) The expression 2 - 5a + \( \frac{2}{3}b \) breaks down into three pieces: the term 2, the term -5a, and the term \( \frac{2}{3}b \).
(iii) The expression 3x⁵ + 4y³ - 7xy² + 3 consists of four distinct terms: 3x⁵, 4y³, -7xy², and 3.
In simple words: A term is each separate piece in an expression. Look for numbers and variables connected by + or - signs, and list them one by one.

Exam Tip: Remember that constants (plain numbers) are also terms. Count all pieces separated by + or - signs, including any constant at the beginning or end of the expression.

Question 5. Identify the terms and their factors in the algebraic expressions given below:
(i) -4x + 5y
(ii) xy + 2x²y²
(iii) 1.2ab - 2.4b + 3.6a
Answer:
(i) The expression -4x + 5y splits into two terms: -4x and 5y. The term -4x is made up of the factors -4 and x. The term 5y is made up of the factors 5 and y.
(ii) The expression xy + 2x²y² is made of two terms: xy and 2x²y². For the term xy, the factors are x and y. For the term 2x²y², the factors are 2, x, x, y, and y.
(iii) The expression 1.2ab - 2.4b + 3.6a breaks down into three terms: 1.2ab, -2.4b, and 3.6a. The factors of 1.2ab are 1.2, a, and b. The factors of -2.4b are -2.4 and b. The factors of 3.6a are 3.6 and a.
In simple words: Terms are the big pieces separated by + or -. Factors are the smaller pieces multiplied together to make each term.

Exam Tip: Write each factor separately, including coefficients and repeated variables. For example, x² counts as the factor x listed twice.

Question 6. Show the terms and their factors by tree diagrams of the following algebraic expressions:
(i) 8x + 3y²
(ii) y - y³
(iii) 5xy² + 7x²y
(iv) -ab + 2b² - 3a²
Answer:
(i) The expression 8x + 3y² breaks into two terms. The first term is 8x, which has factors 8 and x. The second term is 3y², which splits into factors 3, y, and y (since y appears twice).


(ii) The expression y - y³ contains two terms: y and -y³. The term y has only one factor, which is y itself. The term -y³ has factors -1, y, y, and y (three y's because of the cube).


(iii) The expression 5xy² + 7x²y is split into two terms. The term 5xy² has factors 5, x, y, and y. The term 7x²y has factors 7, x, x, and y.


(iv) The expression -ab + 2b² - 3a² has three terms. The term -ab breaks into the factors -1, a, and b. The term 2b² breaks into the factors 2, b, and b. The term -3a² breaks into the factors -3, a, and a.


In simple words: Start with the whole expression at the top. Draw lines down to each term. Then from each term, draw lines down to show all its factors.

Exam Tip: Make sure you list every factor, including numerical coefficients and all instances of repeated variables. The tree should show how each term breaks down completely into its basic building blocks.

Question 7. Write the numerical coefficient of each of the following:
(i) -7x
(ii) -2x³y²
(iii) 6abcd²
(iv) \( \frac{2}{3}pq² \)
Answer:
(i) In the term -7x, the numerical coefficient (the number in front) is -7.
(ii) In the term -2x³y², the numerical coefficient is -2.
(iii) In the term 6abcd², the numerical coefficient is 6.
(iv) In the term \( \frac{2}{3}pq² \), the numerical coefficient is \( \frac{2}{3} \).
In simple words: The numerical coefficient is the number at the start of a term. It shows how many times you take the variables.

Exam Tip: Identify only the number part, not the variable letters. If there is no number written, the coefficient is 1 (or -1 if there is a minus sign).

Question 8. Write the coefficient of x in the following:
(i) -4bx
(ii) 5xyz
(iii) -x
(iv) -3x²y
Answer:
(i) You can rewrite -4bx as (-4b) times x. This means the coefficient of x is -4b.
(ii) You can rewrite 5xyz as (5yz) times x. This means the coefficient of x is 5yz.
(iii) You can rewrite -x as (-1) times x. This means the coefficient of x is -1.
(iv) You can rewrite -3x²y as (-3xy) times x. This means the coefficient of x is -3xy.
In simple words: The coefficient of x is everything else in the term (all other numbers and letters) that is multiplied by x.

Exam Tip: Rewrite the term by grouping all non-x parts together. This grouped part is your coefficient of x. Do not include any x itself in the coefficient.

Question 9. In -7xy²z³, write down the coefficient of:
(i) 7x
(ii) -xy²
(iii) xyz
(iv) 7yz²
Answer:
(i) You can express -7xy²z³ as (7x) times (-y²z³). So the coefficient of 7x is -y²z³.
(ii) You can express -7xy²z³ as (-xy²) times (7z³). So the coefficient of -xy² is 7z³.
(iii) You can express -7xy²z³ as (xyz) times (-7yz²). So the coefficient of xyz is -7yz².
(iv) You can express -7xy²z³ as (7yz²) times (-xyz). So the coefficient of 7yz² is -xyz.
In simple words: To find the coefficient of any part, divide the whole term by that part. What is left over is the coefficient you are looking for.

Exam Tip: Always rewrite the term as (given part) times (everything else). The "everything else" is your answer. Check your work by multiplying the part and the coefficient back together to verify you get the original term.

Question 10. Identify the terms (other than constants) and write their numerical coefficients in each of the following algebraic expressions:
(i) 3 - 7x
(ii) 1 + 2x - 3x²
(iii) 1.2a + 0.8b
Answer:
(i) In the expression 3 - 7x, the only term that has variables is -7x (the 3 is a constant). The numerical coefficient of -7x is -7.
(ii) In the expression 1 + 2x - 3x², the terms with variables are 2x and -3x². The numerical coefficient of 2x is 2. The numerical coefficient of -3x² is -3.
(iii) In the expression 1.2a + 0.8b, both terms have variables. The numerical coefficient of 1.2a is 1.2. The numerical coefficient of 0.8b is 0.8.
In simple words: Skip any plain number (constant) in the expression. For each term containing a variable, write down the number in front of it.

Exam Tip: Separate out the constant terms first, then focus on the variable terms. The numerical coefficient is always the number part only, not the variable part.

Question 11. Identify the terms which contain x and write the coefficient of x in each of the following expressions:
(i) 13y² - 8xy
(ii) 7x - xy²
(iii) 5 - 7xyz + 4x²y
Answer:
(i) In the expression 13y² - 8xy, only the term -8xy contains the variable x. The coefficient of x in -8xy is -8y.
(ii) In the expression 7x - xy², both terms contain x. In the first term 7x, the coefficient of x is 7. In the second term -xy², the coefficient of x is -y².
(iii) In the expression 5 - 7xyz + 4x²y, two terms hold the variable x: -7xyz and 4x²y. In the term -7xyz, the coefficient of x is -7yz. In the term 4x²y, the coefficient of x is 4xy.
In simple words: Find every term that has an x in it. For each one, figure out what is multiplied by x to make that term.

Exam Tip: Some expressions have x in multiple terms. Find all of them and give the coefficient of x for each term separately. Do not combine or add the coefficients.

Question 12. Identify the term which contain y² and write the coefficient of y² in each of the following expressions:
(i) 8 - xy²
(ii) 5y² + 7x - 3xy²
(iii) 2x²y - 15xy² + 7y²
Answer:
(i) In the expression 8 - xy², the term carrying y² is -xy². The coefficient of y² in -xy² is -x.
(ii) In the expression 5y² + 7x - 3xy², the terms that hold y² are 5y² and -3xy². In the term 5y², the coefficient of y² is 5. In the term -3xy², the coefficient of y² is -3x.
(iii) In the expression 2x²y - 15xy² + 7y², the terms containing y² are -15xy² and 7y². In the term -15xy², the coefficient of y² is -15x. In the term 7y², the coefficient of y² is 7.
In simple words: Look for y² in the expression. Pick out each term that has y². Write what you multiply by y² to get that term.

Exam Tip: Make sure you identify y² specifically, not just y. A term like 2x²y does not contain y², so skip it. Only count terms where y is squared.

Question 13. Classify into monomials, binomials and trinomials:
(i) 4y - 7z
(ii) -5xy²
(iii) x + y - xy
(iv) ab² - 5b - 3a
(v) 4p²q - 5pq²
(vi) 2017
(vii) 1 + x + x²
(viii) 5x² - 7 + 3x + 4
Answer:
An algebraic expression with a single term is called a monomial. An expression with exactly two terms is a binomial. An expression with exactly three terms is a trinomial.

(i) The expression 4y - 7z contains 2 terms, making it a Binomial.
(ii) The expression -5xy² has only 1 term, so it is a Monomial.
(iii) The expression x + y - xy holds 3 terms, so it is a Trinomial.
(iv) The expression ab² - 5b - 3a has 3 terms, making it a Trinomial.
(v) The expression 4p²q - 5pq² contains 2 terms, so it is a Binomial.
(vi) The expression 2017 is just 1 constant term, so it is a Monomial.
(vii) The expression 1 + x + x² has 3 terms, making it a Trinomial.
(viii) The expression 5x² - 7 + 3x + 4 first appears to have four pieces, but -7 + 4 = -3. After joining these, you get 5x² + 3x - 3, which has 3 terms. This makes it a Trinomial.
In simple words: Count how many separate pieces (terms) are in the expression. One piece is a monomial, two pieces is a binomial, and three pieces is a trinomial.

Exam Tip: Before classifying, always combine any like constant terms first (for example, -7 + 4). Count the terms after simplifying. Terms are separated by + or - signs.

Question 14. State whether the given pair of terms is of like or unlike terms:
(i) -7x, \( \frac{5}{2}x \)
(ii) -29x, -29y
(iii) 2xy, 2xyz
(iv) 4m²p, 4mp²
(v) 12xz, 12x²z²
(vi) -5pq, 7qp
Answer:
Two terms are called like terms when they share the same variable (or variables) with the same powers.

(i) The terms -7x and \( \frac{5}{2}x \) both have the variable x with no power (or power 1). Since they share the same variable part, they are Like terms.
(ii) The terms -29x and -29y have different variables (x and y). Since the variable parts are not the same, they are Unlike terms.
(iii) The terms 2xy and 2xyz have different variables - one has x and y, the other has x, y, and z. This makes them Unlike terms.
(iv) The terms 4m²p and 4mp² both use m and p, but with different powers: in the first term m is squared, while in the second term p is squared. This difference means they are Unlike terms.
(v) The terms 12xz and 12x²z² again share x and z, but the powers differ: in the first term they are to the first power, in the second they are squared. So they are Unlike terms.
(vi) The terms -5pq and 7qp both have variables p and q. Even though they are written in different orders (pq vs qp), multiplication is the same either way. They are Like terms.
In simple words: Compare the variable parts of two terms (ignoring numbers). If they look exactly the same, they are like. If they are different, they are unlike.

Exam Tip: Write out the variables and their powers in both terms side by side. Even small differences in variables or their powers mean the terms are unlike. Remember that pq and qp are the same thing.

Question 15. Identify like terms in the following:
(i) x²y, 3xy², -2x²y, 4x²y²
(ii) 3a²b, 2abc, -6a²b, 4abc
(iii) 10pq, 7p, 8q, -p²q², -7qp, -100q, -23, 12q²p², -5p², 41, 2405p, 78qp, 13p²q, qp², 701p²
Answer:
Like terms share exactly the same variables with the same powers.

(i) Looking at each term: x²y (has x² and y), 3xy² (has x and y²), -2x²y (has x² and y), and 4x²y² (has x² and y²). The terms that match are x²y and -2x²y - both have x squared and y to the first power. Only x²y and -2x²y are Like terms.

(ii) Checking each: 3a²b (has a² and b), 2abc (has a, b, and c), -6a²b (has a² and b), and 4abc (has a, b, and c). The first and third terms both have a² and b, so 3a²b and -6a²b are Like terms. The second and fourth terms both have a, b, and c, so 2abc and 4abc are also Like terms together.

(iii) Organizing by variable pattern: pq terms (same as qp) include 10pq, -7qp, and 78qp. These three form one group of like terms. Terms with p² only: -5p² and 701p² form another group. Terms with q only: 8q and -100q. Terms with p only: 7p and 2405p. Terms with p²q: 13p²q. Terms with p²q²: -p²q² and 12q²p² (which is the same as 12p²q²). Terms with q²p or pq²: qp². Constant terms: -23 and 41. After grouping:
- Like terms involving pq: 10pq, -7qp, 78qp
- Like terms involving p²: -5p², 701p²
- Like terms involving q: 8q, -100q
- Like terms involving p: 7p, 2405p
- Like terms involving p²q²: -p²q², 12q²p²
In simple words: Write the variables and powers for each term. Group together all terms that have the exact same variable pattern.

Exam Tip: List each term and its variable composition carefully. Remember that the order of variables does not matter (pq = qp), but powers do. Always double-check that every term in a group has the same variables with the same powers.

Question 16. Write the degree of following polynomials in x:
(i) \( x^2 - 6x^7 + x^8 \)
(ii) \( 3 - 2x \)
(iii) \( -2 \)
(iv) \( 1 - x^2 \)
Answer: The degree of a polynomial in x is the largest power of x that shows up in it.
(i) In \( x^2 - 6x^7 + x^8 \), the highest power of x is 8 - Degree 8.
(ii) In \( 3 - 2x \), the highest power of x is 1 - Degree 1.
(iii) \( -2 = -2x^0 \), so the power of x is 0 - Degree 0.
(iv) In \( 1 - x^2 \), the highest power of x is 2 - Degree 2.
In simple words: The degree of a polynomial is just the biggest power of x that appears in it.

Exam Tip: Always identify the term with the highest power of x - this determines the degree, regardless of the order the terms are written in.

Question 17. Write the degree of the following polynomials:
(i) \( 3x^2 - 5xy^2 + 7 \)
(ii) \( xy^2 - y^3 + 3y^4 - 2 \)
(iii) \( 7 - 2x^3 - 5xy^3 + 9y^5 \)
Answer: When a polynomial has two or more variables, its degree is found by taking the greatest sum of the powers of the variables in any one term.
(i) In \( 3x^2 - 5xy^2 + 7 \), the degrees of the terms are 2, \( (1 + 2) = 3 \) and 0. The greatest is 3 - Degree 3.
(ii) In \( xy^2 - y^3 + 3y^4 - 2 \), the degrees of the terms are \( (1 + 2) = 3 \), 3, 4 and 0. The greatest is 4 - Degree 4.
(iii) In \( 7 - 2x^3 - 5xy^3 + 9y^5 \), the degrees of the terms are 0, 3, \( (1 + 3) = 4 \) and 5. The greatest is 5 - Degree 5.
In simple words: Add up the powers of all variables in each term, then pick the term where this sum is biggest - that sum is the degree.

Exam Tip: For multi-variable polynomials, remember to add the powers of each variable together within a term before comparing - do not just look at the highest single power.

Question 18. State true or false:
(i) If 5 is constant and y is variable, then 5y and 5 + y are variables.
(ii) 7x has two terms, 7 and x
(iii) 5 + xy is a trinomial
(iv) 7a × bc is a binomial
(v) \( 7x^3 + 2x^2 + 3x - 5 \) is a polynomial
(vi) \( 2x^2 - \frac{3}{x} \) is a polynomial
(vii) coefficient of x in -3xy is -3
Answer:
(i) True. Both 5y and 5 + y change depending on the value of y, which makes them variables.
(ii) False. 7x is a single term (a monomial) where 7 is the numerical coefficient and x is the literal coefficient.
(iii) False. 5 + xy contains 2 terms, so it is a binomial (not a trinomial).
(iv) False. \( 7a \times bc = 7abc \) is a single term (a monomial).
(v) True. All the powers of x are whole numbers and not negative.
(vi) False. \( \frac{3}{x} \) has a negative power of x, so the expression is not a polynomial.
(vii) False. The coefficient of x in −3xy is −3y (the y is also part of the coefficient).
In simple words: A monomial has 1 term, a binomial has 2 terms, a trinomial has 3 terms. A polynomial must have only whole number powers and no division by variables.

Exam Tip: Count terms carefully - a term is a complete product of numbers and variables. The literal coefficient includes all variables, not just one specific variable.

Exercise 8.2

Question 1. Add:
(i) 7x, -3x
(ii) 6x, -11x
(iii) \( 5x^2, -9x^2 \)
(iv) \( 3ab^2, -5ab^2 \)
(v) \( \frac{1}{2}pq, -\frac{1}{3}pq \)
(vi) \( 5x^3y, -\frac{2}{3}x^3y \)
Answer: When you add like terms, you combine them by adding their coefficients and keeping the variable part the same.
(i) \( 7x + (-3x) = (7 - 3)x = 4x \)
(ii) \( 6x + (-11x) = (6 - 11)x = -5x \)
(iii) \( 5x^2 + (-9x^2) = (5 - 9)x^2 = -4x^2 \)
(iv) \( 3ab^2 + (-5ab^2) = (3 - 5)ab^2 = -2ab^2 \)
(v) \( \frac{1}{2}pq + \left(-\frac{1}{3}pq\right) = \left(\frac{1}{2} - \frac{1}{3}\right)pq = \frac{3 - 2}{6}pq = \frac{1}{6}pq \)
(vi) \( 5x^3y + \left(-\frac{2}{3}x^3y\right) = \left(5 - \frac{2}{3}\right)x^3y = \frac{15 - 2}{3}x^3y = \frac{13}{3}x^3y \)
In simple words: For like terms, keep the letters the same and add or subtract only the numbers in front.

Exam Tip: Always check that terms are "like" (same variables with same powers) before combining them - unlike terms cannot be added together.

Question 2. Add:
(i) 3x, -5x, 7x
(ii) 7xy, 2xy, -8xy
(iii) -2abc, 3abc, abc
(iv) 3mn, -5mn, 8mn, -4mn
(v) \( 2x^3, 3x^3, -4x^3, -5x^3 \)
Answer:
(i) \( 3x + (-5x) + 7x = (3 - 5 + 7)x = 5x \)
(ii) \( 7xy + 2xy + (-8xy) = (7 + 2 - 8)xy = xy \)
(iii) \( -2abc + 3abc + abc = (-2 + 3 + 1)abc = 2abc \)
(iv) \( 3mn + (-5mn) + 8mn + (-4mn) = (3 - 5 + 8 - 4)mn = 2mn \)
(v) \( 2x^3 + 3x^3 + (-4x^3) + (-5x^3) = (2 + 3 - 4 - 5)x^3 = -4x^3 \)
In simple words: Collect all the numbers, add and subtract them in order, then put the variable part at the end.

Exam Tip: Work through the addition step by step, grouping all positive coefficients first, then all negative ones, to reduce mistakes.

Question 3. Simplify the following by combining like terms:
(i) \( 21b - 32 + 7b - 20b \)
(ii) \( 12m^2 - 9m + 5m - 4m^2 - 7m + 10 \)
(iii) \( -z^2 + 13z^2 - 5z + 7z^3 - 15z \)
(iv) \( 5x^2y - 5x^2 + 3yx^2 - 3y^2 + x^2 - y^2 + 8xy^2 - 3y^2 \)
(v) \( p - (p - q) - (q - p) - q \)
(vi) \( 3a - 2b - ab - (a - b + ab) + 3ab + b - a \)
(vii) \( (3y^2 + 5y - 4) - (8y - y^2 - 4) \)
Answer:
(i) \( 21b - 32 + 7b - 20b = (21 + 7 - 20)b - 32 = 8b - 32 \)
(ii) \( 12m^2 - 9m + 5m - 4m^2 - 7m + 10 = (12 - 4)m^2 + (-9 + 5 - 7)m + 10 = 8m^2 - 11m + 10 \)
(iii) \( -z^2 + 13z^2 - 5z + 7z^3 - 15z = 7z^3 + (-1 + 13)z^2 + (-5 - 15)z = 7z^3 + 12z^2 - 20z \)
(iv) \( 5x^2y - 5x^2 + 3yx^2 - 3y^2 + x^2 - y^2 + 8xy^2 - 3y^2 = (5x^2y + 3x^2y) + (-5x^2 + x^2) + (-3y^2 - y^2 - 3y^2) + 8xy^2 = 8x^2y - 4x^2 - 7y^2 + 8xy^2 \)
(v) \( p - (p - q) - (q - p) - q = p - p + q - q + p - q = (p - p + p) + (q - q - q) = p - q \)
(vi) \( 3a - 2b - ab - (a - b + ab) + 3ab + b - a = 3a - 2b - ab - a + b - ab + 3ab + b - a = (3a - a - a) + (-2b + b + b) + (-ab - ab + 3ab) = a + ab \)
(vii) \( (3y^2 + 5y - 4) - (8y - y^2 - 4) = 3y^2 + 5y - 4 - 8y + y^2 + 4 = (3y^2 + y^2) + (5y - 8y) + (-4 + 4) = 4y^2 - 3y \)
In simple words: Group terms by their variable parts, then add or subtract the numbers belonging to each group separately.

Exam Tip: When subtracting an expression in brackets, flip the sign of every term inside before combining - this is a common source of errors.

Question 4. Find the sum of the following algebraic expressions:
(i) 5xy, -7xy, \( 3x^2 \)
(ii) \( 4x^2y, -3xy^2, -5xy^2, 5x^2y \)
(iii) \( -7mn + 5, 12mn + 2, 8mn - 8, -2mn - 3 \)
(iv) \( a + b - 3, b - a + 3, a - b + 3 \)
(v) \( 14x + 10y - 12xy - 13, 18 - 7x - 10y + 8xy, 4xy \)
(vi) \( 5m - 7n, 3n - 4m + 2, 2m - 3mn - 5 \)
(vii) \( 3x^3 - 5x^2 + 2x + 1, 3x - 2x^2 - x^3, 2x^2 - 7x + 9 \)
(viii) \( 7a^2 - 5a + 2, 3a^2 - 7, 2a + 9, 1 + 2a - 5a^2 \)
Answer:
(i) \( 5xy + (-7xy) + 3x^2 = (5 - 7)xy + 3x^2 = 3x^2 - 2xy \)
(ii) \( 4x^2y + (-3xy^2) + (-5xy^2) + 5x^2y = (4 + 5)x^2y + (-3 - 5)xy^2 = 9x^2y - 8xy^2 \)
(iii) \( (-7mn + 5) + (12mn + 2) + (8mn - 8) + (-2mn - 3) = (-7 + 12 + 8 - 2)mn + (5 + 2 - 8 - 3) = 11mn - 4 \)
(iv) \( (a + b - 3) + (b - a + 3) + (a - b + 3) = (a - a + a) + (b + b - b) + (-3 + 3 + 3) = a + b + 3 \)
(v) \( (14x + 10y - 12xy - 13) + (18 - 7x - 10y + 8xy) + 4xy = (14 - 7)x + (10 - 10)y + (-12 + 8 + 4)xy + (-13 + 18) = 7x + 5 \)
(vi) \( (5m - 7n) + (3n - 4m + 2) + (2m - 3mn - 5) = (5 - 4 + 2)m + (-7 + 3)n - 3mn + (2 - 5) = 3m - 4n - 3mn - 3 \)
(vii) \( (3x^3 - 5x^2 + 2x + 1) + (3x - 2x^2 - x^3) + (2x^2 - 7x + 9) = (3 - 1)x^3 + (-5 - 2 + 2)x^2 + (2 + 3 - 7)x + (1 + 9) = 2x^3 - 5x^2 - 2x + 10 \)
(viii) \( (7a^2 - 5a + 2) + (3a^2 - 7) + (2a + 9) + (1 + 2a - 5a^2) = (7 + 3 - 5)a^2 + (-5 + 2 + 2)a + (2 - 7 + 9 + 1) = 5a^2 - a + 5 \)
In simple words: Write each expression in order, group the like terms, add up the numbers in front of each group, and write your answer in order from highest power to lowest.

Exam Tip: Set up a clear vertical alignment when adding multiple expressions - this helps ensure no terms are missed or duplicated.

Question 5. Simplify the following:
(i) \( 2x^2 + 3y^2 - 5xy + 5x^2 - y^2 + 6xy - 3x^2 \)
(ii) \( 3xy^2 - 5x^2y + 7xy - 8xy^2 - 4xy + 6x^2y \)
(iii) \( 5x^4 - 7x^2 + 8x - 1 + 3x^3 - 9x^2 + 7 - 3x^4 + 11x - 2 + 8x^2 \)
Answer:
(i) \( 2x^2 + 3y^2 - 5xy + 5x^2 - y^2 + 6xy - 3x^2 = (2 + 5 - 3)x^2 + (3 - 1)y^2 + (-5 + 6)xy = 4x^2 + 2y^2 + xy \)
(ii) \( 3xy^2 - 5x^2y + 7xy - 8xy^2 - 4xy + 6x^2y = (3 - 8)xy^2 + (-5 + 6)x^2y + (7 - 4)xy = -5xy^2 + x^2y + 3xy \)
(iii) \( 5x^4 - 7x^2 + 8x - 1 + 3x^3 - 9x^2 + 7 - 3x^4 + 11x - 2 + 8x^2 = (5 - 3)x^4 + 3x^3 + (-7 - 9 + 8)x^2 + (8 + 11)x + (-1 + 7 - 2) = 2x^4 + 3x^3 - 8x^2 + 19x + 4 \)
In simple words: Find all terms that look the same, put them together, do the arithmetic, and arrange your final answer with the highest powers first.

Exam Tip: It helps to mark off or underline each group of like terms as you work - this prevents losing track of which terms you have already combined.

Question 6. Subtract:
(i) \( -5y^2 \) from \( y^2 \)
(ii) \( -7xy \) from \( -2xy \)
(iii) \( a(b - 5) \) from \( b(5 - a) \)
(iv) \( -m^2 + 5mn \) from \( 4m^2 - 3mn + 8 \)
(v) \( 5a^2 - 7ab + 5b^2 \) from \( 3ab - 2a^2 - 2b^2 \)
(vi) \( 4pq - 5q^2 - 3p^2 \) from \( 5p^2 + 3q^2 - pq \)
(vii) \( 7xy + 5x^2 - 7y^2 + 3 \) from \( 7x^2 - 8xy + 3y^2 - 5 \)
(viii) \( 2x^4 - 7x^2 + 5x + 3 \) from \( x^4 - 3x^3 - 2x^2 + 3 \)
Answer: To subtract, we change the sign of each term in the second expression and then add.
(i) \( y^2 - (-5y^2) = y^2 + 5y^2 = 6y^2 \)
(ii) \( -2xy - (-7xy) = -2xy + 7xy = 5xy \)
(iii) Here \( a(b - 5) = ab - 5a \) and \( b(5 - a) = 5b - ab \).
\( b(5 - a) - a(b - 5) = (5b - ab) - (ab - 5a) = 5b - ab - ab + 5a = 5a + 5b - 2ab \)
(iv) \( (4m^2 - 3mn + 8) - (-m^2 + 5mn) = 4m^2 - 3mn + 8 + m^2 - 5mn = 5m^2 - 8mn + 8 \)
(v) \( (3ab - 2a^2 - 2b^2) - (5a^2 - 7ab + 5b^2) = 3ab - 2a^2 - 2b^2 - 5a^2 + 7ab - 5b^2 = (-2 - 5)a^2 + (3 + 7)ab + (-2 - 5)b^2 = 10ab - 7a^2 - 7b^2 \)
(vi) \( (5p^2 + 3q^2 - pq) - (4pq - 5q^2 - 3p^2) = 5p^2 + 3q^2 - pq - 4pq + 5q^2 + 3p^2 = (5 + 3)p^2 + (3 + 5)q^2 + (-1 - 4)pq = 8p^2 + 8q^2 - 5pq \)
(vii) \( (7x^2 - 8xy + 3y^2 - 5) - (7xy + 5x^2 - 7y^2 + 3) = 7x^2 - 8xy + 3y^2 - 5 - 7xy - 5x^2 + 7y^2 - 3 = (7 - 5)x^2 + (-8 - 7)xy + (3 + 7)y^2 + (-5 - 3) = 2x^2 - 15xy + 10y^2 - 8 \)
(viii) \( (x^4 - 3x^3 - 2x^2 + 3) - (2x^4 - 7x^2 + 5x + 3) = x^4 - 3x^3 - 2x^2 + 3 - 2x^4 + 7x^2 - 5x - 3 = (1 - 2)x^4 - 3x^3 + (-2 + 7)x^2 - 5x + (3 - 3) = -x^4 - 3x^3 + 5x^2 - 5x \)
In simple words: Change every plus to minus and every minus to plus in the second expression, then add it to the first like normal addition.

Exam Tip: Always flip all the signs in the expression being subtracted before you start combining - missing even one sign flip will give you the wrong answer.

Question 7. Subtract p - 2q + r from the sum of 10p - r and 5p + 2q
Answer: First, find the sum of \( 10p - r \) and \( 5p + 2q \).
\( \text{Sum} = (10p - r) + (5p + 2q) = 15p + 2q - r \)

Now, subtract \( p - 2q + r \) from this sum.
\( (15p + 2q - r) - (p - 2q + r) = 15p + 2q - r - p + 2q - r = (15 - 1)p + (2 + 2)q + (-1 - 1)r = 14p + 4q - 2r \)
In simple words: Add the first two expressions together, then take away the third expression by flipping its signs and adding.

Exam Tip: Read the question carefully - "subtract X from the sum of Y and Z" means find (Y + Z) - X, not Y - X.

Question 8. From the sum of 4 + 3x and 5 - 4x + 2x², subtract the sum of 3x² - 5x and -x² + 2x + 5
Answer: Sum of \( 4 + 3x \) and \( 5 - 4x + 2x^2 \):
\( = (4 + 3x) + (5 - 4x + 2x^2) = 2x^2 - x + 9 \)

Sum of \( 3x^2 - 5x \) and \( -x^2 + 2x + 5 \):
\( = (3x^2 - 5x) + (-x^2 + 2x + 5) = 2x^2 - 3x + 5 \)

Now, subtract the second sum from the first:
\( (2x^2 - x + 9) - (2x^2 - 3x + 5) = 2x^2 - x + 9 - 2x^2 + 3x - 5 = 2x + 4 \)
In simple words: Find two separate sums, then take one sum away from the other by flipping the signs of the second sum.

Exam Tip: Show your working in stages - first add each pair of expressions separately, then subtract the two results. This makes it easier to check your work.

Question 9. What should be added to x² - y² + 2xy to obtain x² + y² + 5xy?
Answer: Let the expression to be added be denoted as A. Then:
\( (x^2 - y^2 + 2xy) + A = x^2 + y^2 + 5xy \)

To find A, subtract the first expression from the second:
\( A = (x^2 + y^2 + 5xy) - (x^2 - y^2 + 2xy) \)
\( = x^2 + y^2 + 5xy - x^2 + y^2 - 2xy \)
\( = (1 - 1)x^2 + (1 + 1)y^2 + (5 - 2)xy \)
\( = 2y^2 + 3xy \)
In simple words: To find what to add, subtract what you have from what you want to get.

Exam Tip: If you have an equation like "A + B = C", then "what to add" means A = C - B. Always rearrange the equation to isolate the unknown term.

Question 9. Find the expression that must be subtracted from \( x^2 + y^2 + 5xy \) to get \( x^2 - y^2 + 2xy \).
Answer: To find what needs to be subtracted, we work backwards. The expression to subtract equals the first expression minus the second expression. So we compute: \( (x^2 + y^2 + 5xy) - (x^2 - y^2 + 2xy) = x^2 + y^2 + 5xy - x^2 + y^2 - 2xy = (1 - 1)x^2 + (1 + 1)y^2 + (5 - 2)xy = 2y^2 + 3xy \).
In simple words: When you take away the second expression from the first, what's left is the answer. That answer is \( 2y^2 + 3xy \).

Exam Tip: Always arrange like terms together before combining them - this makes it harder to make arithmetic errors with the coefficients.

Question 10. What should be subtracted from \( -7mn + 2m^2 + 3n^2 \) to get \( m^2 + 2mn + n^2 \)?
Answer: The expression to subtract is found by calculating: \( (-7mn + 2m^2 + 3n^2) - (m^2 + 2mn + n^2) = -7mn + 2m^2 + 3n^2 - m^2 - 2mn - n^2 = (2 - 1)m^2 + (-7 - 2)mn + (3 - 1)n^2 = m^2 - 9mn + 2n^2 \).
In simple words: Line up all the similar terms and subtract their numbers. You'll get \( m^2 - 9mn + 2n^2 \) as your final answer.

Exam Tip: Watch out for negative signs when combining coefficients of like terms - this is where most calculation mistakes happen.

Question 11. How much is \( y^4 - 12y^2 + y + 14 \) greater than \( 17y^3 + 34y^2 - 51y + 68 \)?
Answer: To find how much greater, we subtract the second expression from the first: \( (y^4 - 12y^2 + y + 14) - (17y^3 + 34y^2 - 51y + 68) = y^4 - 12y^2 + y + 14 - 17y^3 - 34y^2 + 51y - 68 = y^4 - 17y^3 + (-12 - 34)y^2 + (1 + 51)y + (14 - 68) = y^4 - 17y^3 - 46y^2 + 52y - 54 \).
In simple words: Take the first expression and remove the second expression from it. Group matching terms together, then add up their numbers. The result is \( y^4 - 17y^3 - 46y^2 + 52y - 54 \).

Exam Tip: When subtracting one polynomial from another, remember to flip the sign of every term in the second polynomial before combining.

Question 12. How much does \( 93p^2 - 55p + 4 \) exceed \( 13p^3 - 5p^2 + 17p - 90 \)?
Answer: To find the excess amount, we calculate the difference: \( (93p^2 - 55p + 4) - (13p^3 - 5p^2 + 17p - 90) = 93p^2 - 55p + 4 - 13p^3 + 5p^2 - 17p + 90 = -13p^3 + (93 + 5)p^2 + (-55 - 17)p + (4 + 90) = -13p^3 + 98p^2 - 72p + 94 \).
In simple words: Subtract the second expression from the first. Collect all the same types of terms, then combine their coefficients. You get \( -13p^3 + 98p^2 - 72p + 94 \).

Exam Tip: Pay close attention to the highest degree term - in this case, the second expression has a \( p^3 \) term that the first doesn't, so it appears with a negative sign in the answer.

Question 13. What should be taken away from \( 3x^2 - 4y^2 + 5xy + 20 \) to obtain \( -x^2 - y^2 + 6xy + 20 \)?
Answer: The expression to remove is: \( (3x^2 - 4y^2 + 5xy + 20) - (-x^2 - y^2 + 6xy + 20) = 3x^2 - 4y^2 + 5xy + 20 + x^2 + y^2 - 6xy - 20 = (3 + 1)x^2 + (-4 + 1)y^2 + (5 - 6)xy + (20 - 20) = 4x^2 - 3y^2 - xy \).
In simple words: Subtract the final expression from the starting expression. Group the matching terms and add their coefficients. The answer is \( 4x^2 - 3y^2 - xy \).

Exam Tip: Notice that the constant terms (20 in each) cancel out - always check if terms vanish completely after subtraction.

Question 14. From the sum of \( 2y^2 + 3yz \), \( -y^2 - yz - z^2 \) and \( yz + 2z^2 \), subtract the sum of \( 3y^2 - z^2 \) and \( -y^2 + yz + z^2 \).
Answer: First, we find the sum of the first group: \( (2y^2 + 3yz) + (-y^2 - yz - z^2) + (yz + 2z^2) = (2 - 1)y^2 + (3 - 1 + 1)yz + (-1 + 2)z^2 = y^2 + 3yz + z^2 \). Next, we find the sum of the second group: \( (3y^2 - z^2) + (-y^2 + yz + z^2) = (3 - 1)y^2 + yz + (-1 + 1)z^2 = 2y^2 + yz \). Now we subtract the second sum from the first: \( (y^2 + 3yz + z^2) - (2y^2 + yz) = y^2 + 3yz + z^2 - 2y^2 - yz = (1 - 2)y^2 + (3 - 1)yz + z^2 = -y^2 + 2yz + z^2 \).
In simple words: First, add the first three expressions together. Then add the last two expressions together. Finally, take away the second result from the first result.

Exam Tip: For multi-step problems, always clearly show the intermediate sums - this helps you catch errors and makes your work easier to follow.

Exercise 8.3

Question 1. If \( m = 2 \), find the value of:
(i) \( 3m - 5 \)
(ii) \( 9 - 5m \)
(iii) \( 3m^2 - 2m - 7 \)
(iv) \( \frac{5}{2}m - 4 \)
Answer:
Given \( m = 2 \).
(i) Plugging in \( m = 2 \) into \( 3m - 5 \): \( 3m - 5 = 3(2) - 5 = 6 - 5 = 1 \).
(ii) Plugging in \( m = 2 \) into \( 9 - 5m \): \( 9 - 5m = 9 - 5(2) = 9 - 10 = -1 \).
(iii) Plugging in \( m = 2 \) into \( 3m^2 - 2m - 7 \): \( 3m^2 - 2m - 7 = 3(2)^2 - 2(2) - 7 = 3(4) - 4 - 7 = 12 - 4 - 7 = 1 \).
(iv) Plugging in \( m = 2 \) into \( \frac{5}{2}m - 4 \): \( \frac{5}{2}m - 4 = \frac{5}{2}(2) - 4 = 5 - 4 = 1 \).
In simple words: Replace the letter \( m \) with the number 2 in each expression, then do the arithmetic to find the answer.

Exam Tip: Always substitute the variable value carefully and evaluate in the correct order - powers first, then multiplication and division, then addition and subtraction.

Question 2. If \( p = -2 \), find the value of:
(i) \( 4p + 7 \)
(ii) \( -3p^2 + 4p + 7 \)
(iii) \( -2p^3 - 3p^2 + 4p + 7 \)
Answer:
Given \( p = -2 \).
(i) Plugging in \( p = -2 \) into \( 4p + 7 \): \( 4p + 7 = 4(-2) + 7 = -8 + 7 = -1 \).
(ii) Plugging in \( p = -2 \) into \( -3p^2 + 4p + 7 \): \( -3p^2 + 4p + 7 = -3(-2)^2 + 4(-2) + 7 = -3(4) - 8 + 7 = -12 - 8 + 7 = -13 \).
(iii) Plugging in \( p = -2 \) into \( -2p^3 - 3p^2 + 4p + 7 \): \( -2p^3 - 3p^2 + 4p + 7 = -2(-2)^3 - 3(-2)^2 + 4(-2) + 7 = -2(-8) - 3(4) - 8 + 7 = 16 - 12 - 8 + 7 = 3 \).
In simple words: Replace \( p \) with -2 everywhere you see it, being careful with the signs when dealing with negative numbers and their powers.

Exam Tip: When substituting negative numbers, always use parentheses around them - this prevents sign errors, especially when raising negative numbers to powers.

Question 3. If \( a = 2 \) and \( b = -2 \), find the value of:
(i) \( a^2 + b^2 \)
(ii) \( a^2 + ab + b^2 \)
(iii) \( a^2 - b^2 \)
Answer:
Given \( a = 2 \) and \( b = -2 \).
(i) Plugging in \( a = 2 \) and \( b = -2 \) into \( a^2 + b^2 \): \( a^2 + b^2 = (2)^2 + (-2)^2 = 4 + 4 = 8 \).
(ii) Plugging in \( a = 2 \) and \( b = -2 \) into \( a^2 + ab + b^2 \): \( a^2 + ab + b^2 = (2)^2 + (2)(-2) + (-2)^2 = 4 - 4 + 4 = 4 \).
(iii) Plugging in \( a = 2 \) and \( b = -2 \) into \( a^2 - b^2 \): \( a^2 - b^2 = (2)^2 - (-2)^2 = 4 - 4 = 0 \).
In simple words: Substitute both values into each expression and calculate step by step, keeping track of negative signs carefully.

Exam Tip: For expressions with squares of negative numbers, remember that \( (-2)^2 = 4 \) (positive), not -4. This is a common mistake.

Question 4. When \( a = 0 \) and \( b = -1 \), find the value of the given expressions:
(i) \( 2a^2 + b^2 + 1 \)
(ii) \( a^2 + ab + 2 \)
(iii) \( 2a^2b + 2ab^2 + ab \)
Answer:
Given \( a = 0 \) and \( b = -1 \).
(i) Plugging in \( a = 0 \) and \( b = -1 \) into \( 2a^2 + b^2 + 1 \): \( 2a^2 + b^2 + 1 = 2(0)^2 + (-1)^2 + 1 = 0 + 1 + 1 = 2 \).
(ii) Plugging in \( a = 0 \) and \( b = -1 \) into \( a^2 + ab + 2 \): \( a^2 + ab + 2 = (0)^2 + (0)(-1) + 2 = 0 + 0 + 2 = 2 \).
(iii) Plugging in \( a = 0 \) and \( b = -1 \) into \( 2a^2b + 2ab^2 + ab \): \( 2a^2b + 2ab^2 + ab = 2(0)^2(-1) + 2(0)(-1)^2 + (0)(-1) = 0 + 0 + 0 = 0 \).
In simple words: When \( a = 0 \), any term with \( a \) becomes zero. Substitute and simplify carefully.

Exam Tip: Expressions with a zero variable often simplify quickly - use this to your advantage to avoid unnecessary calculations.

Question 5. If \( p = -10 \), find the value of \( p^2 - 2p - 100 \).
Answer: Given \( p = -10 \). Plugging in \( p = -10 \) into \( p^2 - 2p - 100 \): \( p^2 - 2p - 100 = (-10)^2 - 2(-10) - 100 = 100 + 20 - 100 = 20 \).
In simple words: Replace \( p \) with -10, calculate the square (which gives 100), then handle the multiplication and subtraction carefully.

Exam Tip: Notice how \( -2(-10) \) becomes \( +20 \) because of the double negative - this is crucial to get right.

Question 6. If \( z = 10 \), find the value of \( z^3 - 3(z - 10) \).
Answer: Given \( z = 10 \). Plugging in \( z = 10 \) into \( z^3 - 3(z - 10) \): \( z^3 - 3(z - 10) = (10)^3 - 3(10 - 10) = 1000 - 3(0) = 1000 - 0 = 1000 \).
In simple words: Replace \( z \) with 10 everywhere. Notice that inside the parentheses you get \( 10 - 10 = 0 \), so that part disappears.

Exam Tip: When evaluating expressions, watch for terms that become zero after substitution - they can make the problem much simpler.

Question 7. Simplify the following expressions and find their values when \( x = 2 \):
(i) \( x + 7 + 4(x - 5) \)
(ii) \( 3(x + 2) + 5x - 7 \)
(iii) \( 6x + 5(x - 2) \)
(iv) \( 4(2x - 1) + 3x + 11 \)
Answer:
(i) Simplifying \( x + 7 + 4(x - 5) \): Expand the brackets: \( x + 7 + 4x - 20 = 5x - 13 \). When \( x = 2 \): \( 5x - 13 = 5(2) - 13 = 10 - 13 = -3 \).
(ii) Simplifying \( 3(x + 2) + 5x - 7 \): Expand the brackets: \( 3x + 6 + 5x - 7 = 8x - 1 \). When \( x = 2 \): \( 8x - 1 = 8(2) - 1 = 16 - 1 = 15 \).
(iii) Simplifying \( 6x + 5(x - 2) \): Expand the brackets: \( 6x + 5x - 10 = 11x - 10 \). When \( x = 2 \): \( 11x - 10 = 11(2) - 10 = 22 - 10 = 12 \).
(iv) Simplifying \( 4(2x - 1) + 3x + 11 \): Expand the brackets: \( 8x - 4 + 3x + 11 = 11x + 7 \). When \( x = 2 \): \( 11x + 7 = 11(2) + 7 = 22 + 7 = 29 \).
In simple words: First, expand the brackets by multiplying. Then gather all \( x \) terms and all regular numbers together. Finally, put in \( x = 2 \) and compute the answer.

Exam Tip: Always simplify first before substituting the value - this reduces arithmetic errors and makes the final calculation faster.

Question 8. Simplify the following expressions and find their values when \( a = -1 \) and \( b = -2 \):
(i) \( 2a - 2b - 4 - 5 + a \)
(ii) \( 2(a^2 + ab) + 3 - ab \)
Answer:
(i) Simplifying \( 2a - 2b - 4 - 5 + a \): Gather like terms: \( (2a + a) - 2b + (-4 - 5) = 3a - 2b - 9 \). When \( a = -1 \) and \( b = -2 \): \( 3a - 2b - 9 = 3(-1) - 2(-2) - 9 = -3 + 4 - 9 = -8 \).
(ii) Simplifying \( 2(a^2 + ab) + 3 - ab \): Expand the brackets: \( 2a^2 + 2ab + 3 - ab = 2a^2 + ab + 3 \). When \( a = -1 \) and \( b = -2 \): \( 2a^2 + ab + 3 = 2(-1)^2 + (-1)(-2) + 3 = 2(1) + 2 + 3 = 2 + 2 + 3 = 7 \).
In simple words: Collect matching terms and simplify first. Then substitute the values and calculate step by step, watching the negative signs.

Exam Tip: When combining like terms, write them in parentheses to keep track - this helps prevent sign errors.

Objective Type Questions - Mental Maths

Question 1. Fill in the blanks:
(i) The terms with different algebraic factors are called ......
(ii) The number of terms in a monomial is ......
(iii) An algebraic expression having two unlike terms is called a ......
(iv) \( 3a^2b \) and \( -7ba^2 \) are ...... terms.
(v) \( -6a^2b \) and \( -6ab^2 \) are ...... terms.
(vi) The number of unlike terms in the algebraic expression \( 3x^2 - 2xy + 5x^2 \) is ......
(vii) The factors of the term \( -3p^2q^2 \) are ......
(viii) The perimeter of a triangle whose sides measure \( 2a \), \( b \) and \( a + b \) is ......
(ix) The value of the expression \( 2x^3 - 7x^2 + 5x - 3 \) when \( x = 1 \) is ......
(x) In the term \( -7a^2bc \), the coefficient of \( a \) is ......
(xi) The degree of the polynomial \( 3 - 5x^2 + 7x^3 - x^4 \) is ......
(xii) The degree of the polynomial \( 3x^2 - 2xy^2 + 5 \) is ......
Answer:
(i) Unlike terms.
(ii) One.
(iii) Binomial.
(iv) Since \( 3a^2b \) and \( -7ba^2 \) have the same variable parts \( a^2b \), they are like terms.
(v) Since \( -6a^2b \) and \( -6ab^2 \) have different variable parts (one has \( a^2b \) and the other has \( ab^2 \)), they are unlike terms.
(vi) Combining like terms: \( 3x^2 - 2xy + 5x^2 = 8x^2 - 2xy \), which gives 2 unlike terms.
(vii) The factors of \( -3p^2q^2 \) are -3, p, p, q, q.
(viii) Adding the sides: \( 2a + b + (a + b) = 3a + 2b \).
(ix) Substituting \( x = 1 \): \( 2(1)^3 - 7(1)^2 + 5(1) - 3 = 2 - 7 + 5 - 3 = -3 \).
(x) Rewriting: \( -7a^2bc = (-7abc) \cdot a \), so the coefficient of \( a \) is \( -7abc \).
(xi) The highest power of \( x \) in \( 3 - 5x^2 + 7x^3 - x^4 \) is 4, so the degree is 4.
(xii) The degree of each term is: 2 (for \( 3x^2 \)), 3 (for \( -2xy^2 \)), and 0 (for 5), so the overall degree is 3.
In simple words: Unlike terms have different variable parts. Like terms have matching variable parts. The degree is the highest power of any variable.

Exam Tip: For degree of a term with multiple variables, add all the exponents - for \( xy^2 \), the degree is 1+2=3, not just 2.

Question 2. State whether the following statements are true (T) or false (F):
(i) The expression \( 5x + 7 - 2x \) is a trinomial.
(ii) \( (7x - 10) - (3x - 5) = 4x - 15 \)
(iii) The coefficient of \( 3x \) in \( -3x^3y \) is \( -xy \)
Answer:
(i) False. The expression \( 5x + 7 - 2x \) simplifies to \( 3x + 7 \), which is a binomial (two terms), not a trinomial.
(ii) False. Calculating \( (7x - 10) - (3x - 5) = 7x - 10 - 3x + 5 = 4x - 5 \), not \( 4x - 15 \).
(iii) False. In the term \( -3x^3y \), if we extract \( 3x \) as a factor, we get \( -3x^3y = (3x) \cdot (-x^2y) \), so the coefficient of \( 3x \) is \( -x^2y \), not \( -xy \).
In simple words: For (i), always simplify before counting terms. For (ii), remember to distribute the negative sign. For (iii), understand what coefficient means - it's what you multiply by the factor to get the original term.

Exam Tip: When deciding True or False, work out the correct answer yourself before checking against the given statement - don't be misled by similar-looking but incorrect options.

Question. (iv) The constant term in the expression \( 2x^2 - 3xy - 7 \) is 7
Answer: This statement is false. The constant term in the expression \( 2x^2 - 3xy - 7 \) is actually -7, not 7. A constant term is the number that stands alone without any variables, and in this case that number is -7.
In simple words: The constant term is -7 because it's the number by itself with no x or y attached to it.

Exam Tip: Always pay attention to the sign in front of the constant term - the negative sign is part of the value, so -7 and 7 are different.

Question. (v) If \( x = 3 \) and \( y = \frac{1}{3} \) then the value of \( xy(x^2 + y^2) \) is \( 9\frac{1}{9} \)
Answer: This statement is true. When \( x = 3 \) and \( y = \frac{1}{3} \), we substitute these values into the expression. First, \( xy = 3 \times \frac{1}{3} = 1 \). Next, we find \( x^2 + y^2 = 9 + \frac{1}{9} = \frac{82}{9} \). Therefore, \( xy(x^2 + y^2) = 1 \times \frac{82}{9} = \frac{82}{9} = 9\frac{1}{9} \).
In simple words: When you put the numbers in and work through the steps, you get \( 9\frac{1}{9} \), so the statement is correct.

Exam Tip: Substitute variables step by step and simplify each part before combining - this reduces mistakes in calculations.

Question. (vi) \( (3x - y + 5) - (x + y) \) is a binomial.
Answer: This statement is false. When you expand \( (3x - y + 5) - (x + y) = 3x - y + 5 - x - y = 2x - 2y + 5 \), the result has 3 terms, making it a trinomial, not a binomial.
In simple words: After simplifying, the answer has three different terms (2x, -2y, and 5), so it's a trinomial.

Exam Tip: Distribute the negative sign carefully across all terms in the second bracket - a common mistake is forgetting to change signs, which leads to the wrong number of terms.

Question. (vii) Sum of 2 and p is 2p
Answer: This statement is false. The sum of 2 and p means you add them together, which gives \( 2 + p \), not \( 2p \). The expression \( 2p \) represents 2 multiplied by p, not 2 plus p.
In simple words: "Sum" means adding, so 2 plus p is written as \( 2 + p \), not \( 2p \).

Exam Tip: Remember that "sum" always means addition, while a coefficient written next to a variable (like 2p) means multiplication.

Question. (viii) Sum of \( x^2 + x \) and \( y^2 + y \) is \( 2x^2 + 2y^2 \)
Answer: This statement is false. When you add \( x^2 + x \) and \( y^2 + y \), you combine all the terms together: \( (x^2 + x) + (y^2 + y) = x^2 + x + y^2 + y \). The correct sum contains all four terms, not just \( 2x^2 + 2y^2 \).
In simple words: You can't combine different types of terms. x² and y² don't add together, and x and y don't add together.

Exam Tip: Only like terms (same variable and same power) can be combined - unlike terms must be kept separate in the final answer.

Question. (ix) In like terms, variables and their powers are same.
Answer: This statement is true. Like terms are terms that have the same variables raised to the same powers. For example, \( 3xy \) and \( -5xy \) are like terms because both contain the variables x and y with the same exponents, even though their coefficients are different.
In simple words: Like terms have exactly the same letters with the same powers - only the numbers in front can be different.

Exam Tip: When combining like terms, add or subtract only the numerical coefficients while keeping the variable part unchanged.

Question. (x) Every polynomial is a monomial.
Answer: This statement is false. A monomial is a specific type of polynomial that contains exactly one term. However, a polynomial can have one, two, three, or many terms. For instance, \( 3x + 5 \) is a binomial with two terms, and \( x^2 + 2x + 1 \) is a trinomial with three terms. Not all polynomials fit the definition of a monomial.
In simple words: A monomial has only one term, but a polynomial can have one or more terms, so not every polynomial is a monomial.

Exam Tip: Remember the hierarchy: monomials, binomials, and trinomials are all types of polynomials, but polynomials are broader and include expressions with more than three terms.

Question. (xi) If we add a monomial and a binomial, then answer can never be a monomial.
Answer: This statement is false. If you add a monomial and a binomial, the result can sometimes be a monomial if the monomial and one term of the binomial are like terms and cancel out. For example, if you add \( -x \) (a monomial) and \( (x + 5) \) (a binomial), you get \( -x + x + 5 = 5 \), which is a monomial.
In simple words: If the monomial cancels out one of the terms in the binomial, you might end up with just one term left, which is a monomial.

Exam Tip: Always check if terms can be combined or cancelled before deciding how many terms the final answer has.

Question. (xii) If we subtract a monomial from a binomial, then the answer is atleast a binomial.
Answer: This statement is false. When you subtract a monomial from a binomial, it is possible to get just one term if the monomial is like one of the terms in the binomial. For example, if you subtract \( x \) (a monomial) from \( (x + 5) \) (a binomial), you get \( (x + 5) - x = 5 \), which is a monomial, not a binomial.
In simple words: If the monomial cancels with one of the binomial's terms, you end up with just one term, which is a monomial, not at least a binomial.

Exam Tip: Before concluding that subtraction always produces a binomial or trinomial, look for opportunities where like terms might cancel completely.

Question. (xiii) If we add a monomial and a trinomial, then the answer can be a monomial.
Answer: This statement is false. A trinomial has three unlike terms. When you add a monomial to it, that monomial can combine with at most one of the three terms if they are like terms. Even in the best case, this leaves you with at least two remaining terms, so the result must be at least a binomial, never a monomial.
In simple words: A monomial can only cancel one term from the trinomial, leaving two or more terms behind, so the answer is always a binomial or trinomial.

Exam Tip: Count carefully - when adding or subtracting expressions, the maximum number of terms that can cancel is limited to one at a time.

Question. (xiv) If we add a monomial and a binomial, then the answer can be a trinomial.
Answer: This statement is true. When you add a monomial and a binomial, if none of them are like terms, they remain separate. For example, \( z + (x + y) = x + y + z \), which is a trinomial with three terms.
In simple words: If the monomial doesn't match any term in the binomial, all three terms stay separate and form a trinomial.

Exam Tip: The result is a trinomial only when the monomial is unlike both terms of the binomial - if they share a like term, fewer terms will remain.

Multiple Choice Questions

Question 3. The algebraic expression for the statement 'Thrice square of a number x subtracted from five times the sum of y and 2' is
(a) \( 5y + 2 - 3x^2 \)
(b) \( 3x^2 - (5y + 2) \)
(c) \( 5(y + 2) - 3x^2 \)
(d) \( 5(y + 2) - (3x)^2 \)
Answer: (c) \( 5(y + 2) - 3x^2 \)
In simple words: "Five times the sum of y and 2" means \( 5(y + 2) \). "Thrice square of x" means \( 3x^2 \). When you subtract one from the other, you get option (c).

Exam Tip: Translate phrase by phrase - "thrice" means 3 times, "square" means power 2, and "subtracted from" means the first part is what you subtract, not what you subtract from.

Question 4. The expression \( 7x - 5(x^2 + y^2) \) is a
(a) monomial
(b) binomial
(c) trinomial
(d) none of these
Answer: (c) trinomial
In simple words: When you expand \( 7x - 5(x^2 + y^2) \) into \( 7x - 5x^2 - 5y^2 \), you have three different terms, so it's a trinomial.

Exam Tip: Always expand and simplify an expression before counting terms - don't count terms in the factored form.

Question 5. The coefficient of \( 5a^2 \) in \( -5a^3bc \) is
(a) \( -bc \)
(b) \( a^2bc \)
(c) \( -a^2bc \)
(d) \( -abc \)
Answer: (d) \( -abc \)
In simple words: You can rewrite \( -5a^3bc \) as \( (5a^2) \times (-abc) \), so the coefficient of \( 5a^2 \) is \( -abc \).

Exam Tip: The coefficient of a part is the remaining factor left after removing that part from the whole term.

Question 6. Which of the following is a pair of like terms?
(a) \( -5xy, 5x \)
(b) \( -5xy, 3yz \)
(c) \( -5xy, -5y \)
(d) \( -5xy, 7yx \)
Answer: (d) \( -5xy, 7yx \)
In simple words: Like terms must have the same variables in the same powers. Since \( xy = yx \), the terms \( -5xy \) and \( 7yx \) are like terms.

Exam Tip: The order of variables doesn't matter - \( xy \) and \( yx \) represent the same thing and can be combined.

Question 7. The like terms in the expressions \( 3x(3 - 2y) \) and \( 2(xy + x^2) \) are
(a) \( 9x \) and \( 2x^2 \)
(b) \( -6xy \) and \( 2xy \)
(c) \( 9x \) and \( 2xy \)
(d) \( -6xy \) and \( 2x^2 \)
Answer: (b) \( -6xy \) and \( 2xy \)
In simple words: Expand the first expression to get \( 9x - 6xy \) and the second to get \( 2xy + 2x^2 \). The terms \( -6xy \) and \( 2xy \) both contain xy, so they are like terms.

Exam Tip: Always expand expressions fully before identifying like terms - they may not be obvious in factored form.

Question 8. Identify the binomial out of the following:
(a) \( 3xy^2 + 5y - x^2y \)
(b) \( 2x^2y - 5y - 2x^2y \)
(c) \( 3xy^2 + 5y - xy^2 \)
(d) \( xy + yz + zx \)
Answer: (c) \( 3xy^2 + 5y - xy^2 \)
In simple words: When you simplify \( 3xy^2 + 5y - xy^2 = 2xy^2 + 5y \), you get exactly two terms, which makes it a binomial.

Exam Tip: Combine like terms before determining the type of expression - the others become a monomial, trinomial, or trinomial after simplification.

Question 9. The number of (unlike) terms in the expression \( 3xy^2 + 2y^2z - y^2x + y(xz + yz) - 5 \) is
(a) 3
(b) 4
(c) 5
(d) 6
Answer: (b) 4
In simple words: Expand and combine: \( 3xy^2 + 2y^2z - xy^2 + xyz + y^2z - 5 = 2xy^2 + 3y^2z + xyz - 5 \). This gives 4 distinct terms.

Exam Tip: Group and combine like terms systematically to avoid missing or double-counting any unlike terms.

Question 10. The value of the expression \( x^3 + y^3 \) when \( x = 2 \) and \( y = -2 \) is
(a) 0
(b) 8
(c) 16
(d) -16
Answer: (a) 0
In simple words: Substitute: \( (2)^3 + (-2)^3 = 8 + (-8) = 0 \).

Exam Tip: Pay close attention to negative numbers - \( (-2)^3 = -8 \), not 8, because odd powers preserve the sign.

Question 11. \( -xy - (-5xy) \) is equal to
(a) \( -6xy \)
(b) \( 6xy \)
(c) \( -4xy \)
(d) \( 4xy \)
Answer: (d) \( 4xy \)
In simple words: Simplify: \( -xy - (-5xy) = -xy + 5xy = 4xy \). Subtracting a negative becomes adding a positive.

Exam Tip: Double-negative signs become positive - always distribute the negative carefully across all terms in brackets.

Question 12. On subtracting \( 7x + 5y - 3 \) from \( 5y - 3x - 9 \), we get
(a) \( 10x + 6 \)
(b) \( -10x - 6 \)
(c) \( 10x + 10y - 12 \)
(d) \( -10x - 12 \)
Answer: (b) \( -10x - 6 \)
In simple words: Set up \( (5y - 3x - 9) - (7x + 5y - 3) \). Distribute the negative: \( 5y - 3x - 9 - 7x - 5y + 3 = -10x - 6 \).

Exam Tip: Write out the subtraction carefully with brackets, then distribute the negative sign to each term before combining like terms.

Question 13. The value of the expression \( \frac{5}{3}x^2 + 1 \) when \( x = -2 \) is
(a) \( -\frac{17}{3} \)
(b) \( -\frac{7}{3} \)
(c) \( \frac{21}{3} \)
(d) \( \frac{23}{3} \)
Answer: (d) \( \frac{23}{3} \)
In simple words: Substitute \( x = -2 \): \( \frac{5}{3}(-2)^2 + 1 = \frac{5}{3}(4) + 1 = \frac{20}{3} + \frac{3}{3} = \frac{23}{3} \).

Exam Tip: When substituting negative values, remember that \( (-2)^2 = 4 \) (positive), and convert whole numbers to fractions with a common denominator before adding.

Question 14. The degree of the polynomial \( 3x^3y - 5xy^4 - 2x + 1 \) is
(a) 5
(b) 4
(c) 3
(d) 2
Answer: (a) 5
In simple words: Find the highest total power of variables in any single term. The term \( -5xy^4 \) has degree \( 1 + 4 = 5 \), which is the highest.

Exam Tip: In multivariable terms, add all the exponents together to find that term's degree - the overall degree is the maximum degree among all terms.

Statement I-II Type Questions

Question 15. Statement I: The algebraic expression \( 5x^2 \times 3y^2 + 6z^2 \) is a trinomial. Statement II: A trinomial is an algebraic expression having three unlike terms.
(a) Statement I is true but statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (b) Statement I is false but Statement II is true
In simple words: When you multiply \( 5x^2 \times 3y^2 = 15x^2y^2 \), so the full expression becomes \( 15x^2y^2 + 6z^2 \), which has only 2 terms (a binomial). But Statement II correctly defines a trinomial.

Exam Tip: Always simplify multiplication first before counting the final number of terms - this is where most students make errors on this type of question.

Question 16. Statement I: On subtracting \( -2x^2 + 5y^3 \) from \( 4x^2 - 3y^3 + 6z \), we get \( 6x^2 - 8y^3 + 6z \) Statement II: On subtracting a monomial from a trinomial, it is possible to get a binomial.
(a) Statement I is true but statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (c) Both Statement I and Statement II are true
In simple words: For Statement I, expand: \( (4x^2 - 3y^3 + 6z) - (-2x^2 + 5y^3) = 4x^2 - 3y^3 + 6z + 2x^2 - 5y^3 = 6x^2 - 8y^3 + 6z \) ✓. For Statement II, if the monomial matches one term of the trinomial, you can cancel it to get a binomial, like \( (x^2 + x + 5) - x = x^2 + 5 \).

Exam Tip: Distribute the negative sign carefully across the entire expression being subtracted - missing any sign change will lead to an incorrect answer.

Question 17. Statement I: If \( x = 2 \) and \( y = -2 \), we can say that \( -\frac{1}{2}x + 5 > \frac{2}{3}y^2 \) Statement II: The value of an algebraic expression depends on the value of the variable(s) involved.
(a) Statement I is true but statement II is false.
(b) Statement I is false but Statement II is true.
(c) Both Statement I and Statement II are true.
(d) Both Statement I and Statement II are false.
Answer: (c) Both Statement I and Statement II are true
In simple words: For Statement I, calculate: \( -\frac{1}{2}(2) + 5 = -1 + 5 = 4 \) and \( \frac{2}{3}(-2)^2 = \frac{2}{3}(4) = \frac{8}{3} \approx 2.67 \). Since \( 4 > 2.67 \), the inequality is true ✓. Statement II is simply a fact - different variable values produce different results.

Exam Tip: For inequality problems, calculate each side separately and carefully, paying attention to negative numbers and fractions.

Check Your Progress

Question 1. Consider the expression \( \frac{3}{2}x^2y - \frac{1}{2}xy^2 + 6x^2y^2 \)
(i) How many terms are there? What do you call such an expression?
(ii) List out the terms.
(iii) In the term \( -\frac{1}{2}xy^2 \), write the numerical coefficient and the literal coefficient.
(iv) In the term \( -\frac{1}{2}xy^2 \), what is the coefficient of x?
Answer:
(i) There are 3 terms in the given expression. Because it has three terms, we call it a trinomial.

(ii) The three terms in this expression are: \( \frac{3}{2}x^2y \), \( -\frac{1}{2}xy^2 \), and \( 6x^2y^2 \)

(iii) In the term \( -\frac{1}{2}xy^2 \):
Numerical coefficient = \( -\frac{1}{2} \)
Literal coefficient = \( xy^2 \)

(iv) In the term \( -\frac{1}{2}xy^2 \), find the coefficient of x by identifying what remains after removing x. Since \( -\frac{1}{2}xy^2 = -\frac{1}{2} \times x \times y^2 \), the coefficient of x is \( -\frac{1}{2}y^2 \)
In simple words: The expression has 3 parts. For the part \( -\frac{1}{2}xy^2 \), the number is \( -\frac{1}{2} \), the letters are \( xy^2 \), and if you take out x, you're left with \( -\frac{1}{2}y^2 \).

Exam Tip: Numerical coefficient is just the number; literal coefficient includes all the variables; the coefficient of a specific variable is what's left when you remove that variable from the term.

Question 2. Write the degree of the following polynomials:
(i) \( \frac{2}{5}x^3 - 7x^2 - \frac{1}{2}x + 3 \)
(ii) \( \frac{2}{3}xy^2 - 5xy + \frac{3}{5}y^2x^2 + 2x \)
Answer:
(i) In the polynomial \( \frac{2}{5}x^3 - 7x^2 - \frac{1}{2}x + 3 \), examine the power of x in each term. The highest power of x is 3. Therefore, the degree of this polynomial is 3.

(ii) In the polynomial \( \frac{2}{3}xy^2 - 5xy + \frac{3}{5}y^2x^2 + 2x \), find the degree of each term by adding the exponents when there are multiple variables. The term \( \frac{3}{5}y^2x^2 \) has degree \( 2 + 2 = 4 \), which is the highest. Therefore, the degree of this polynomial is 4.
In simple words: Look at each term and find the highest power of the variables. For single-variable terms, take the power. For multi-variable terms, add all the powers together.

Exam Tip: When variables are multiplied, their exponents add together to give the term's degree - don't just look at one variable's exponent.

Question 3. Identify monomials, binomials and trinomials from the following algebraic expressions:
(i) 5x × y
(ii) 3 - 5x
(iii) \( \frac{1}{2}(7x - 3y + 5z) \)
(iv) 3x² - 1.2xy
(v) -3x³y⁴z⁵
(vi) 5x(2x - 3y) + 7x²
Answer:
(i) 5x × y = 5xy has one term, making it a monomial.
(ii) 3 − 5x contains two terms, making it a binomial.
(iii) \( \frac{1}{2}(7x - 3y + 5z) = \frac{7}{2}x - \frac{3}{2}y + \frac{5}{2}z \) has three terms, making it a trinomial.
(iv) 3x² − 1.2xy contains two terms, making it a binomial.
(v) −3x³y⁴z⁵ has one term, making it a monomial.
(vi) 5x(2x − 3y) + 7x² = 10x² − 15xy + 7x² = 17x² − 15xy has two terms, making it a binomial.
In simple words: Count how many separate parts are in each expression. One part is a monomial, two parts make a binomial, and three parts make a trinomial.

Exam Tip: Always expand and simplify each expression before counting the terms - combine like terms to get the final count.

Question 4. Using horizontal method:
(i) Add x² + y² - 2xy, -2x² - y² - 2xy and 3x² + y² + xy
(ii) Subtract -x² + y² + 2xy from 2x² - 3y²
Answer:
(i) Group and combine the three expressions:
(x² + y² − 2xy) + (−2x² − y² − 2xy) + (3x² + y² + xy)
Collecting coefficients of each variable: (1 − 2 + 3)x² + (1 − 1 + 1)y² + (−2 − 2 + 1)xy
= 2x² + y² − 3xy

(ii) Subtract the second expression from the first:
(2x² − 3y²) − (−x² + y² + 2xy)
Remove parentheses and reverse signs: 2x² − 3y² + x² − y² − 2xy
Collect like terms: (2 + 1)x² + (−3 − 1)y² − 2xy
= 3x² − 4y² − 2xy
In simple words: Write all the expressions in a row, group the same terms together, add or subtract their numbers, and write your final answer.

Exam Tip: Keep track of signs carefully when subtracting - flip every sign in the expression being subtracted before combining.

Question 5. Using column method, add ab + 2bc - ca and 2ab - bc - ca and subtract 4ab + 5bc - 3ca.
Answer:
First, add ab + 2bc − ca and 2ab − bc − ca by the column method:

\[ \begin{array}{r} ab + 2bc - ca \\ + \quad 2ab - bc - ca \\ \hline 3ab + bc - 2ca \end{array} \]

Next, subtract 4ab + 5bc − 3ca from this sum using the column method. Reverse the signs of terms in the expression being subtracted, then add:

\[ \begin{array}{r} 3ab + bc - 2ca \\ - \quad (4ab + 5bc - 3ca) \\ \hline -ab - 4bc + ca \end{array} \]

The operation works as: (3ab + bc − 2ca) − (4ab + 5bc − 3ca) = 3ab + bc − 2ca − 4ab − 5bc + 3ca = −ab − 4bc + ca
In simple words: Stack the expressions vertically with the same terms in the same column, add or subtract each column separately, and write the result.

Exam Tip: Line up matching terms in their own columns - this prevents mixing terms and reduces sign errors.

Question 6. The sides of a triangle are 5a - 3b, 3a + 2b and 5b - 2a, find its perimeter.
Answer: The perimeter of a triangle is found by adding all three sides together.

Perimeter = (5a − 3b) + (3a + 2b) + (5b − 2a)

Collect all terms with the same variable: (5 + 3 − 2)a + (−3 + 2 + 5)b

= 6a + 4b
In simple words: Add up all three side lengths. Group the a terms together and the b terms together, then find your answer.

Exam Tip: Always identify and collect like terms (same variable and power) before combining - this gives you the correct perimeter.

Question 7. If two adjacent sides of a rectangle are 4x + 7y and 3y - x, find its perimeter.
Answer: The perimeter of a rectangle equals twice the sum of length and breadth.

Perimeter = 2[(4x + 7y) + (3y − x)]

Add the sides: (4 − 1)x + (7 + 3)y = 3x + 10y

Multiply by 2: 2(3x + 10y) = 6x + 20y
In simple words: Add the two given sides together, then multiply the result by 2 to get the full distance around the rectangle.

Exam Tip: Remember that a rectangle has two pairs of equal sides - that is why you use the formula 2(length + breadth).

Question 8. Subtract the sum of 3x² + 2xy - 2y² and 5y² - 7xy from 5x² + 2y² - 3xy
Answer: First, find the sum of the first two expressions:

(3x² + 2xy − 2y²) + (5y² − 7xy) = 3x² + (2 − 7)xy + (−2 + 5)y² = 3x² − 5xy + 3y²

Now subtract this sum from 5x² + 2y² − 3xy:

(5x² + 2y² − 3xy) − (3x² − 5xy + 3y²)

Remove parentheses: 5x² + 2y² − 3xy − 3x² + 5xy − 3y²

Collect like terms: (5 − 3)x² + (2 − 3)y² + (−3 + 5)xy = 2x² − y² + 2xy
In simple words: First add the first two expressions together. Then take that result away from the third expression.

Exam Tip: Perform the addition in the first step completely before moving to the subtraction - this avoids sign errors.

Question 9. What must be added to 5x³ - 2x² + 3x + 7 to get 7x³ + 7x - 5?
Answer: The expression that must be added is found by subtracting the given expression from the target expression.

Required expression = (7x³ + 7x − 5) − (5x³ − 2x² + 3x + 7)

= 7x³ + 7x − 5 − 5x³ + 2x² − 3x − 7

Collect like terms: (7 − 5)x³ + 2x² + (7 − 3)x + (−5 − 7)

= 2x³ + 2x² + 4x − 12
In simple words: To find what must be added, subtract the starting expression from the ending expression.

Exam Tip: Check your work by adding your answer to the original expression - you should get the target expression if correct.

Question 10. How much is 3p - 4q + r less than 4p + 3q - 5r?
Answer: "Less than" means we subtract the first expression from the second expression.

Required difference = (4p + 3q − 5r) − (3p − 4q + r)

Remove parentheses: 4p + 3q − 5r − 3p + 4q − r

Combine like terms: (4 − 3)p + (3 + 4)q + (−5 − 1)r = p + 7q − 6r
In simple words: To find how much less one expression is than another, subtract the smaller one from the larger one.

Exam Tip: Read "A is less than B" carefully - it means B - A, not A - B.

Question 11. How much is 3a² - 5ab + 7b² + 3 greater than 2a² + 2ab + 5?
Answer: "Greater than" means we subtract the second expression from the first expression.

Required difference = (3a² − 5ab + 7b² + 3) − (2a² + 2ab + 5)

Remove parentheses: 3a² − 5ab + 7b² + 3 − 2a² − 2ab − 5

Combine like terms: (3 − 2)a² + (−5 − 2)ab + 7b² + (3 − 5) = a² − 7ab + 7b² − 2
In simple words: To find how much more one expression is than another, subtract the smaller from the larger one.

Exam Tip: Read "A is greater than B" carefully - it means A - B, not B - A.

Question 12. How much should 5x³ + 3x² - 2x + 1 be increased to get 6x² + 7?
Answer: To find what must be added (the increase), subtract the starting expression from the final expression.

Required expression = (6x² + 7) − (5x³ + 3x² − 2x + 1)

Remove parentheses: 6x² + 7 − 5x³ − 3x² + 2x − 1

Collect terms by degree: −5x³ + (6 − 3)x² + 2x + (7 − 1) = −5x³ + 3x² + 2x + 6
In simple words: Subtract the starting expression from what you want to end up with to find the increase needed.

Exam Tip: Arrange the final answer in standard form, with highest degree terms first.

Question 13. Subtract the sum of 12ab - 10b² - 18a² and 9ab + 12b² + 14a² from the sum of ab + 2b² and 3b² - a²
Answer: First, find the sum of 12ab − 10b² − 18a² and 9ab + 12b² + 14a²:

(12ab − 10b² − 18a²) + (9ab + 12b² + 14a²)
= (12 + 9)ab + (−10 + 12)b² + (−18 + 14)a²
= 21ab + 2b² − 4a²

Next, find the sum of ab + 2b² and 3b² − a²:

(ab + 2b²) + (3b² − a²) = ab + (2 + 3)b² − a² = ab + 5b² − a²

Now subtract the first sum from the second sum:

(ab + 5b² − a²) − (21ab + 2b² − 4a²)
= ab + 5b² − a² − 21ab − 2b² + 4a²
= (1 − 21)ab + (5 − 2)b² + (−1 + 4)a²
= −20ab + 3b² + 3a²
= 3a² − 20ab + 3b²
In simple words: Add the first pair of expressions, add the second pair, then subtract the first total from the second total.

Exam Tip: Work step-by-step and arrange your final answer in standard form with a-terms first, then ab-terms, then b-terms.

Question 14. When a = 3, b = 0, c = -2, find the values of:
(i) ab + 2bc + 3ca + 4abc
(ii) a³ + b³ + c³ - 3abc
Answer:
(i) Given: a = 3, b = 0, c = −2

Substitute into ab + 2bc + 3ca + 4abc:
= (3)(0) + 2(0)(−2) + 3(−2)(3) + 4(3)(0)(−2)
= 0 + 0 − 18 + 0
= −18

(ii) Substitute into a³ + b³ + c³ − 3abc:
= (3)³ + (0)³ + (−2)³ − 3(3)(0)(−2)
= 27 + 0 − 8 − 0
= 19
In simple words: Replace each letter with its value and work through the calculation step-by-step, being careful with negative numbers.

Exam Tip: Always substitute carefully and watch for terms that become zero - any term with b = 0 in it will equal zero.

Question 15. The length of a rectangle is 3x - 4y + 6z and the perimeter is 7x + 8y + 17z, find the breadth of the rectangle.
Answer: Using the rectangle perimeter formula:

Perimeter = 2(length + breadth)

So: length + breadth = \( \frac{\text{Perimeter}}{2} \)

Therefore: breadth = \( \frac{\text{Perimeter}}{2} \) − length

Substitute the given values:

breadth = \( \frac{7x + 8y + 17z}{2} \) − (3x − 4y + 6z)

= \( \frac{7x}{2} + 4y + \frac{17z}{2} \) − 3x + 4y − 6z

Group like terms: \( \left(\frac{7x}{2} - 3x\right) + (4y + 4y) + \left(\frac{17z}{2} - 6z\right) \)

= \( \frac{7x - 6x}{2} \) + 8y + \( \frac{17z - 12z}{2} \)

= \( \frac{x}{2} \) + 8y + \( \frac{5z}{2} \)

= \( \frac{1}{2}x + 8y + \frac{5}{2}z \)
In simple words: Divide the perimeter by 2, then subtract the length from that result to get the breadth.

Exam Tip: Always divide the full perimeter by 2 before subtracting - do not try to divide individual terms separately.

Question 16. Simplify: \( \frac{3x}{5} + \frac{2x}{3} - \left(\frac{x}{2} + \frac{2x}{5}\right) \)
Answer: First, expand the expression by removing parentheses:

\( \frac{3x}{5} + \frac{2x}{3} - \frac{x}{2} - \frac{2x}{5} \)

Group fractions with the same denominator:

\( \left(\frac{3x}{5} - \frac{2x}{5}\right) + \frac{2x}{3} - \frac{x}{2} \)

= \( \frac{x}{5} + \frac{2x}{3} - \frac{x}{2} \)

Find the least common denominator (LCD = 30):

= \( \frac{6x}{30} + \frac{20x}{30} - \frac{15x}{30} \)

= \( \frac{6x + 20x - 15x}{30} \)

= \( \frac{11x}{30} \)
In simple words: Combine terms with the same denominator first, then find a common denominator for all remaining fractions before adding and subtracting.

Exam Tip: Always find the LCD for all fractions before combining - this avoids arithmetic errors.

Question 17. If a = 3, b = -1, then find the value of each of the following:
(i) a^b
(ii) b^a
(iii) (ab)^b
(iv) (a + b)^b
(v) \( \left(\frac{b}{a}\right)^b \)
(vi) \( \left(\frac{a}{b} + \frac{b}{a}\right)^b \)
Answer:
Given: a = 3, b = −1

(i) a^b = 3^(-1) = \( \frac{1}{3} \)

(ii) b^a = (−1)³ = −1

(iii) (ab)^b = (3 × (−1))^(-1) = (−3)^(-1) = −\( \frac{1}{3} \)

(iv) (a + b)^b = (3 + (−1))^(-1) = (2)^(-1) = \( \frac{1}{2} \)

(v) \( \left(\frac{b}{a}\right)^b = \left(\frac{-1}{3}\right)^{-1} = \frac{3}{-1} = -3 \)

(vi) \( \left(\frac{a}{b} + \frac{b}{a}\right)^b = \left(\frac{3}{-1} + \frac{-1}{3}\right)^{-1} = \left(-3 - \frac{1}{3}\right)^{-1} = \left(\frac{-9 - 1}{3}\right)^{-1} = \left(\frac{-10}{3}\right)^{-1} = -\frac{3}{10} \)
In simple words: Replace a with 3 and b with −1 in each expression, then use the rules of exponents (a negative exponent means reciprocal) to simplify.

Exam Tip: Remember that a negative exponent means you take the reciprocal (flip) of the base - for example, 2^(-1) = 1/2.