CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set 07

The system of equations is given by
\( a_1x + b_1y + c_1 = 0 \) ....(1)
\( a_2x + b_2y + c_2 = 0 \) ....(2)
(a) It is consistent with unique solution, if \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). It shows that lines represented by equations (1) and (2) are not parallel.
(b) It is consistent with infinitely many solutions, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \). It shows that lines represented by equation (1) and (2) are coincident.
(c) It is inconsistent, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). It shows that lines represented by equation (1) and (2) are parallel and non–coincident.

EXAMPLES

Question. Show that the following system of equations has unique solution \( 2x – 3y = 6; x + y = 1 \).
Answer: The given system of equation can be written as
\( 2x – 3y – 6 = 0 \)
\( x + y – 1 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 2, b_1 = -3, c_1 = -6 \)
and \( a_2 = 1, b_2 = 1, c_2 = -1 \)
\( \frac{a_1}{a_2} = \frac{2}{1} = 2, \frac{b_1}{b_2} = \frac{-3}{1} = -3 \)
\( \frac{c_1}{c_2} = \frac{-6}{-1} = 6 \)
Clearly, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
So, the given system of equations has a unique solution. i.e., it is consistent.

Question. Show that the following system of equations has unique solution : \( x – 2y = 2 ; 4x – 2y = 5 \)
Answer: The given system of equations can be written as
\( x – 2y – 2 = 0 \)
\( 4x – 2y – 5 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 1, b_1 = -2, c_1 = -2 \)
and \( a_2 = 4, b_2 = -2, c_2 = -5 \)
\( \frac{a_1}{a_2} = \frac{1}{4}, \frac{b_1}{b_2} = \frac{-2}{-2} = 1, \frac{c_1}{c_2} = \frac{-2}{-5} = \frac{2}{5} \)
Clearly, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
So, the given system of equations has a unique solution i.e. It is consistent.

Question. For what value of k the following system of equations has a unique solution : \( x – ky = 2 ; 3x + 2y = -5 \)
Answer: The given system of equation can be written as
\( x – ky – 2 = 0 \)
\( 3x + 2y + 5 = 0 \)
The given system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
where, \( a_1 = 1, b_1 = -k, c_1 = -2 \)
and \( a_2 = 3, b_2 = 2, c_2 = 5 \)
Clearly, for unique solution \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \Rightarrow \frac{1}{3} \neq \frac{-k}{2} \Rightarrow k \neq \frac{-2}{3} \)

Question. Show that the following system has infinitely many solutions. \( x = 3y + 3 ; 9y = 3x – 9 \)
Answer: The given system of equations can be written as
\( x – 3y – 3 = 0 \)
\( 3x – 9y – 9 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 1, b_1 = -3, c_1 = -3 \)
and \( a_2 = 3, b_2 = -9, c_2 = -9 \)
\( \frac{a_1}{a_2} = \frac{1}{3}, \frac{b_1}{b_2} = \frac{-3}{-9} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{-3}{-9} = \frac{1}{3} \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) so the given system of equations has infinitely many solutions.

Question. Show that the following system has infinitely many solutions : \( 2y = 4x – 6 ; 2x = y + 3 \)
Answer: The given system of equations can be written as
\( 4x – 2y – 6 = 0 \)
\( 2x – y – 3 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 4, b_1 = -2, c_1 = -6 \)
and \( a_2 = 2, b_2 = -1, c_2 = -3 \)
\( \frac{a_1}{a_2} = \frac{4}{2} = 2, \frac{b_1}{b_2} = \frac{-2}{-1} = 2, \frac{c_1}{c_2} = \frac{-6}{-3} = 2 \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), so the given system of equations has infinitely many solutions.

Question. Find the value of k for which the following system of equations has infinitely many solutions. \( (k – 1) x + 3y = 7; (k + 1) x + 6y = (5k – 1) \)
Answer: The given system of equations can be written as
\( (k – 1)x + 3y – 7 = 0 \)
\( (k + 1) x + 6y – (5k – 1) = 0 \)
Here \( a_1 = (k – 1), b_1 = 3, c_1 = -7 \)
and \( a_2 = (k + 1), b_2 = 6, c_2 = - (5k – 1) \)
For the system of equations to have infinite number of solutions.
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{k - 1}{k + 1} = \frac{3}{6} = \frac{-7}{-(5k - 1)} \)
\( \Rightarrow \frac{k - 1}{k + 1} = \frac{1}{2} = \frac{7}{5k - 1} \)
Taking I and II
\( \frac{k - 1}{k + 1} = \frac{1}{2} \Rightarrow 2k – 2 = k + 1 \Rightarrow k = 3 \)
Taking II and III
\( \frac{1}{2} = \frac{7}{5k - 1} \Rightarrow 5k – 1 = 14 \Rightarrow 5k = 15 \Rightarrow k = 3 \)
Hence, k = 3.

Question. For what values of a and b, the following system of equations have an infinite number of solutions: \( 2x + 3y = 7; (a – b) x + (a + b) y = 3a + b – 2 \)
Answer: The given system of linear equations can be written as
\( 2x + 3y – 7 = 0 \)
\( (a – b) x + (a + b) y – (3a + b – 2) = 0 \)
The above system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \),
where \( a_1 = 2, b_1 = 3, c_1 = -7 \)
\( a_2 = (a – b), b_2 = (a + b), c_2 = - (3a + b – 2) \)
For the given system of equations to have an infinite number of solutions
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Here, \( \frac{a_1}{a_2} = \frac{2}{a - b}, \frac{b_1}{b_2} = \frac{3}{a + b} \) and \( \frac{c_1}{c_2} = \frac{-7}{-(3a + b - 2)} = \frac{7}{3a + b - 2} \)
\( \Rightarrow \frac{2}{a - b} = \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
\( \Rightarrow \frac{2}{a - b} = \frac{3}{a + b} \) and \( \frac{3}{a + b} = \frac{7}{3a + b - 2} \)
\( \Rightarrow 2a + 2b = 3a – 3b \) and \( 9a +3b – 6 = 7a + 7b \)
\( \Rightarrow 2a – 3a = -3b – 2b \) and \( 9a – 7a = 7b – 3b + 6 \)
\( \Rightarrow -a = -5b \) and \( 2a = 4b + 6 \)
\( \Rightarrow a = 5b \) .... (3) and \( a = 2b + 3 \) .... (4)
Solving (3) and (4) we get
\( 5b = 2b + 3 \Rightarrow b = 1 \)
Substituting b = 1 in (3), we get \( a = 5 \times 1 = 5 \)
Thus, a = 5 and b = 1
Hence, the given system of equations has infinite number of solutions when \( a = 5, b = 1 \)

Question. Show that the following system of equations is inconsistent. \( 2x + 7y = 11; 5x + \frac{35}{2} y = 25 \)
Answer: The given system of equations can be written as
\( 2x + 7y – 11 = 0 \)
\( 5x + \frac{35}{2} y – 25 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where, \( a_1 = 2, b_1 = 7, c_1 = -11 \)
and \( a_2 = 5, b_2 = \frac{35}{2}, c_2 = - 25 \)
\( \frac{a_1}{a_2} = \frac{2}{5}, \frac{b_1}{b_2} = \frac{7}{\frac{35}{2}} = \frac{2}{5}, \frac{c_1}{c_2} = \frac{-11}{-25} = \frac{11}{25} \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
So, the given system of equations has no solution, i.e. it is inconsistent. Proved.

Question. Show that the following system of equations has no solution : \( 2x + 4y = 10 ; 3x + 6y = 12 \)
Answer: The given system of equations can be written as
\( 2x + 4y – 10 = 0 \)
\( 3x + 6y – 12 = 0 \)
The given equations are of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
Where \( a_1 = 2, b_1 = 4, c_1 = - 10 \)
and \( a_2 = 3, b_2 = 6, c_2 = - 12 \)
\( \frac{a_1}{a_2} = \frac{2}{3}, \frac{b_1}{b_2} = \frac{4}{6} = \frac{2}{3}, \frac{c_1}{c_2} = \frac{-10}{-12} = \frac{5}{6} \)
Clearly, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
So, the given system of equations has no solution, i.e., it is inconsistent. Proved.

Question. For what values of k will the following system of liner equations has no solution. \( 3x + y = 1; (2k – 1) x + (k – 1) y = 2k + 1 \)
Answer: The given system of equations may be written as
\( 3x + y – 1 = 0 \)
\( (2k – 1) x + (k – 1) y – (2k + 1) = 0 \)
The above system of equations is of the form
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
where \( a_1 = 3, b_1 = 1, c_1 = -1 \)
and \( a_2 = (2k – 1), b_2 = (k – 1), c_2 = -(2k+1) \)
\( \frac{a_1}{a_2} = \frac{3}{2k - 1}, \frac{b_1}{b_2} = \frac{1}{k - 1}, \frac{c_1}{c_2} = \frac{-1}{-(2k + 1)} = \frac{1}{2k+1} \)
Clearly, for no solution \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \Rightarrow \frac{3}{2k - 1} = \frac{1}{k - 1} \)
\( \Rightarrow 3k – 3 = 2k – 1 \Rightarrow k = 2 \)
and \( \frac{1}{k - 1} \neq \frac{1}{2k + 1} \Rightarrow 2k + 1 \neq k – 1 \Rightarrow k \neq -2 \)
and \( \frac{3}{2k - 1} \neq \frac{1}{2k + 1} \Rightarrow 6k + 3 \neq 2k – 1 \Rightarrow 4k \neq - 4 \Rightarrow k \neq - 1 \)
Hence the given system of linear equations has no solution, when k = 2 and \( k \neq -2 \) and \( k \neq - 1 \).

Question. Determine the value of k for each of the following given system of equations having unique/consistent solution. \( 2x + 3y – 5 = 0; kx – 6y = 8 \)
Answer: (i) The given system of equations may be written as
\( 2x + 3y – 5 = 0 \)
\( kx – 6y – 8 = 0 \)
Here, \( a_1 = 2, b_1 = 3, c_1 = -5 \),
\( a_2 = k, b_2 = -6, c_2 = -8 \)
As the given equations have unique solution, we get,
\( \frac{a_1}{a_2} = \frac{2}{k} \) and \( \frac{b_1}{b_2} = \frac{3}{-6} = \frac{-1}{2} \)
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \Rightarrow \frac{2}{k} \neq \frac{-1}{2} \Rightarrow k \neq - 4 \)
Thus the given system of equations have a unique solution for all real values of k except –4.

Question. Determine the value of k for each of the following given system of equations having unique/consistent solution. \( 2x + ky = 1; 5x – 7y – 5 = 0 \)
Answer: (ii) The given system of equations may be written as
\( 2x + ky – 1 = 0 \)
\( 5x – 7y – 5 = 0 \)
Here, \( a_1 = 2, b_1 = k, c_1 = -1 \),
\( a_2 = 5, b_2 = -7, c_2 = -5 \)
We have \( \frac{a_1}{a_2} = \frac{2}{5} \) and \( \frac{b_1}{b_2} = \frac{k}{-7} = \frac{-k}{7} \)
Here \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{2}{5} \neq \frac{-k}{7} \)
It satisfies the condition that the system of given solutions has a unique solution.
So, \( \frac{2}{5} \neq \frac{-k}{7} \Rightarrow k \neq \frac{-14}{5} \)
Thus, the given system of equations has a unique solution for all real values of k except \( \frac{-14}{5} \).

Question. Determine the value of k for each of the following given system of equations having unique/consistent solution. \( x – ky – 2 = 0; 3x + 2y + 5 = 0 \)
Answer: (i) We have,
\( x – ky – 2 = 0 \)
\( 3x + 2y + 5 = 0 \)
Here, \( a_1 = 1, b_1 = -k, c_1 = -2 \),
\( a_2 = 3, b_2 = 2, c_2 = 5 \)
Since, the system of equations has a unique solution, we have
\( \Rightarrow \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) or \( \frac{1}{3} \neq \frac{-k}{2} \Rightarrow k \neq \frac{-2}{3} \)
Thus, the given system of equations has a solution for all values of k except \( \frac{-2}{3} \).

Question. Determine the value of k for each of the following given system of equations having unique/consistent solution. \( 2x – 3y – 1 = 0; kx + 5y – 7 = 0 \)
Answer: (ii) We have
\( 2x – 3y – 1 = 0 \)
\( kx + 5y – 7 = 0 \)
Here, \( a_1 = 2, b_1 = -3, c_1 = -1 \),
\( a_2 = k, b_2 = 5, c_2 = -7 \)
Since, the given system of equations has a unique solution, we get
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{2}{k} \neq \frac{-3}{5} \Rightarrow k \neq \frac{-10}{3} \)
Thus, the given system of equation has a unique solution for all value of k except \( \frac{-10}{3} \).

Question. Find the value of k for each of the following systems of equations having infinitely many solutions. \( 2x + 3y = k; (k – 1) x + (k + 2) y = 3k \)
Answer: (i) We have
\( 2x + 3y – k = 0 \)
\( (k – 1) x + (k + 2) y – 3k = 0 \)
Here \( a_1 = 2, b_1 = 3, c_1 = -k \),
\( a_2 = k – 1, b_2 = k + 2, c_2 = -3k \)
Since, the given system of equations has infinitely many solutions, we get
\( \frac{a_1}{a_2} = \frac{2}{k - 1}, \frac{b_1}{b_2} = \frac{3}{k + 2}, \frac{c_1}{c_2} = \frac{-k}{-3k} = \frac{1}{3} \)
and \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{1}{3} \)
\( \Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2} \) or \( \frac{3}{k + 2} = \frac{1}{3} \)
\( \Rightarrow 2k + 4 = 3k – 3 \) or \( k + 2 = 9 \)
\( \Rightarrow 3k – 2k = 4 + 3 \) or \( k = 7 \Rightarrow k = 7 \)
It shows that the given system of equations has infinitely many solutions at k = 7.

Question. Find the value of k for each of the following systems of equations having infinitely many solutions. \( 2x + 3y = 2; (k + 2) x + (2k + 1) y = 2 (k – 1) \)
Answer: (ii) We have
\( 2x + 3y – 2 = 0 \)
\( (k + 2) x + (2k + 1) y – 2 (k – 1) = 0 \)
Here, \( a_1 = 2, b_1 = 3, c_1 = -2 \),
\( a_2 = k + 2, b_2 = 2k + 1, c_2 = -2 (k – 1) \)
Since, the given system of equations has infinitely many solutions, we get
\( \frac{a_1}{a_2} = \frac{2}{k + 2}, \frac{b_1}{b_2} = \frac{3}{2k + 1}, \frac{c_1}{c_2} = \frac{-2}{-2(k - 1)} = \frac{1}{k-1} \)
and \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{2}{k + 2} = \frac{3}{2k + 1} = \frac{1}{k - 1} \)
\( \Rightarrow \frac{2}{k + 2} = \frac{3}{2k + 1} \) or \( \frac{3}{2k + 1} = \frac{1}{k - 1} \)
\( \Rightarrow 4k + 2 = 3k + 6 \) or \( 3k – 3 = 2k + 1 \)
\( \Rightarrow 4k – 3k = 6 – 2 \) or \( 3k – 2k = 1 + 3 \)
\( \Rightarrow k = 4 \) or \( k = 4 \Rightarrow k = 4 \)

Question. Determine the values of k for the following system of equations having no solution. \( x + 2y = 0; 2x + ky = 5 \)
Answer: The given system of equations may be written as
\( x + 2y = 0 \)
\( 2x + ky – 5 = 0 \)
Here, \( a_1 = 1, b_1 = 2, c_1 = 0 \),
\( a_2 = 2, b_2 = k, c_2 = -5 \)
As the given system of equations has no solution, we get
\( \frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{2}{k}, \frac{c_1}{c_2} = \frac{0}{-5} \)
We must write \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{1}{2} = \frac{2}{k} \Rightarrow k = 4 \)
Here, for this value of k, we get \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).

Question. Find the value of k of the following system of equations having infinitely many solutions. \( 2x – 3y = 7; (k + 2) x – (2k + 1) y = 3(2k – 1) \)
Answer: A given system of equations has infinitely many solutions, if
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
So, we get \( \Rightarrow \frac{2}{k + 2} = \frac{-3}{-(2k + 1)} = \frac{7}{3(2k - 1)} \)
\( \Rightarrow \frac{2}{k + 2} = \frac{3}{2k + 1} \) or \( \frac{3}{2k + 1} = \frac{7}{6k - 3} \)
\( \Rightarrow 4k + 2 = 3k + 6 \) or \( 18k – 9 = 14k + 7 \)
\( \Rightarrow k = 4 \) or \( k = 4 \Rightarrow k = 4 \)
Thus, the given system of equations has infinitely many solutions at k = 4.

Question. Determine the values of a and b so that the following given system of linear equations has infinitely many solutions. \( 2x – (2a + 5) y = 5 ; (2b + 1) x – 9y = 15 \)
Answer: We have \( 2x – (2a + 5) y – 5 = 0 \) and \( (2b + 1) x – 9y – 15 = 0 \).
Hence, \( a_1 = 2, b_1 = -(2a + 5), c_1 = -5 \),
\( a_2 = 2b + 1, b_2 = -9, c_2 = - 15 \).
The given system of equations has infinitely many solutions, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) such that
\( \frac{2}{2b + 1} = \frac{-(2a + 5)}{-9} = \frac{-5}{-15} \Rightarrow \frac{2}{2b + 1} = \frac{2a + 5}{9} = \frac{1}{3} \)
\( \Rightarrow \frac{2}{2b + 1} = \frac{1}{3} \) and \( \frac{2a + 5}{9} = \frac{1}{3} \)
\( \Rightarrow 2b + 1 = 6 \) or \( 6a + 15 = 9 \)
\( \Rightarrow b = \frac{5}{2} \) and \( a = -1 \)
Thus, the given system of equations has infinitely many solutions at \( a = -1, b = \frac{5}{2} \).

Question. Find the value of c if the following system of equation has no solution. \( cx + 3y = 3; 12x + cy = 5 \)
Answer: We have \( cx + 3y - 3 = 0 \) and \( 12x + cy - 6 = 0 \).
The given system of equations has no solution, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) such that \( \frac{c}{12} = \frac{3}{c} \neq \frac{-3}{-6} \)
\( \Rightarrow \frac{c}{12} = \frac{3}{c} \) and \( \frac{3}{c} \neq \frac{1}{2} \)
\( \Rightarrow c^2 = 36 \Rightarrow c = \pm 6 \)
Thus, the given system of equation has no solution at \( c = \pm 6 \).

Question. For what value of p, the system of equations will have no solution ? \( px – (p – 3) = –3y; py = p – 12x \)
Answer: The given system of equations may be written as
\( px + 3y – (p – 3) = 0 \)
\( 12x + py – p = 0 \)
Here, \( a_1 = p, b_1 = 3, c_1 = -(p – 3), a_2 = 12, b_2 = p, c_2 = -p \)
The given system of equations will have no solution, if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
For it we get, \( \frac{p}{12} = \frac{3}{p} \) and \( \frac{3}{p} \neq \frac{-(p - 3)}{-p} \Rightarrow p^2 = 36 \Rightarrow p = \pm 6 \)
When \( p = 6 \), \( \frac{a_1}{a_2} = \frac{6}{12} = \frac{3}{6} = \frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{p - 3}{p} = \frac{6 - 3}{6} = \frac{1}{2} \)
So, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{2} \). Thus, p = 6 does not satisfy the equation \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
When \( p = -6 \), \( \frac{a_1}{a_2} = \frac{-6}{12} = \frac{3}{-6} = -\frac{1}{2} \) and \( \frac{c_1}{c_2} = \frac{p - 3}{p} = \frac{-6 - 3}{-6} = \frac{-9}{-6} = \frac{3}{2} \)
So, \( \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Thus, \( p = -6 \) satisfy the equation. Thus, the given system of equations will have no solution, if p = - 6.

Question. Find the value of k for the following system of equations has no solution. \( (3k + 1) x + 3y = 2 ; (k^2 + 1) x – 5 = – (k – 2)y \)
Answer: The given system of equations may be written as
\( (3k + 1) x + 3y – 2 = 0 \)
\( (k^2 + 1) x + (k – 2) y – 5 = 0 \)
Here, \( a_1 = 3k + 1, b_1 = 3, c_1 = -2, a_2 = k^2 + 1, b_2 = k – 2, c_2 = - 5 \)
Since the given system of equations has no solution therefore, we can write;
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{3k + 1}{k^2 + 1} = \frac{3}{k - 2} \neq \frac{-2}{-5} \)
So, \( \frac{3k + 1}{k^2 + 1} = \frac{3}{k - 2} \Rightarrow 3k^2 - 6k + k - 2 = 3k^2 + 3 \Rightarrow - 5k = 5 \Rightarrow k = - 1 \)
Putting k = - 1 in the equation \( \frac{3}{k - 2} \neq \frac{2}{5} \), we get \( \frac{3}{-1 - 2} = -1 \neq \frac{2}{5} \). Thus, k = -1 satisfy the condition. Thus, the given system of equation has no solution at k = -1.

Question. Determine the values of a and b so that the following system of equations has infinite number of solutions. \( 3x + 4y – 12 = 0; 2(a – b) y – (5a – 1) = – (a + b) x \)
Answer: The given system of equations may be written as
\( 3x + 4y – 12 = 0 \)
\( (a + b) x + 2 (a – b) y – (5a – 1) = 0 \)
Here, \( a_1 = 3, b_1 = 4, c_1 = - 12, a_2 = a + b, b_2 = 2 (a – b), c_2 = - (5a – 1) \)
Since, the given system of equations has infinite number of solutions therefore, we get \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{3}{a + b} = \frac{4}{2(a - b)} = \frac{-12}{-(5a - 1)} \Rightarrow \frac{3}{a + b} = \frac{2}{a - b} \) and \( \frac{2}{a - b} = \frac{12}{5a - 1} \)
\( \Rightarrow 3a – 3b = 2a + 2b \) and \( 10a – 2 = 12a – 12b \)
\( \Rightarrow a – 5b = 0 \) and \( 2a – 12b = -2 \Rightarrow a – 6b = -1 \)
Adding/Subtracting we get, \( -5b + 6b = 1 \Rightarrow b = 1 \).
Putting b = 1 in \( a – 5b = 0 \), we get \( a = 5 \).
Thus, the given system of equations has infinitely many solutions at a = 5, b = 1.

HOMOGENEOUS EQUATIONS

The system of equations \( a_1x + b_1y = 0; a_2x + b_2y = 0 \) called homogeneous equations has only solution \( x = 0, y = 0 \), when \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
(i) when \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the system of equations has only one solution, and the system is consistent.
(ii) When \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), the system of equations has infinitely many solutions and the system is consistent.

Question. Find the value of k for which the system of equations \( 4x + 5y = 0; kx + 10y = 0 \) has infinitely many solutions.
Answer: The given system is of the form \( a_1x + b_1y = 0; a_2x + b_2y = 0 \). Here \( a_1 = 4, b_1 = 5 \) and \( a_2 = k, b_2 = 10 \). If \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \), the system has infinitely many solutions. \( \Rightarrow \frac{4}{k} = \frac{5}{10} \Rightarrow k = 8 \).

WORD PROBLEMS ON SIMULTANEOUS LINEAR EQUATION

Problems Based on Articles

Question. The coach of a cricket team buys 7 bats and 6 balls for ₹ 3800. Later, he buys 3 bats and 5 balls for ₹ 1750. Find the cost of each bat and each ball.
Answer: Let the cost of one bat be ₹ x and cost of one ball be ₹ y. Then
\( 7x + 6y = 3800 \) ....(1)
\( 3x + 5y = 1750 \) ....(2)
From (1) \( y = \frac{3800 - 7x}{6} \). Putting this in (2), we get \( 3x + 5 (\frac{3800 - 7x}{6}) = 1750 \).
Multiplying by 6, we get \( 18x + 5(3800 – 7x) = 10500 \Rightarrow 18x + 19000 – 35x = 10500 \)
\( \Rightarrow -17x = 10500 – 19000 \Rightarrow -17x = -8500 \Rightarrow x = 500 \).
Putting x = 500 in (1), we get \( 7(500) + 6y = 3800 \Rightarrow 3500 + 6y = 3800 \Rightarrow 6y = 300 \Rightarrow y = 50 \).
Hence, the cost of one bat = ₹ 500 and the cost of one ball = ₹ 50.

Question. Meena went to a bank to withdraw ₹ 2000. She asked the cashier to give ₹ 50 and ₹ 100 notes only. Meena got 25 notes in all. Find how many notes of ₹ 50 and ₹ 100 she received?
Answer: Let the number of notes of ₹ 50 be x, and the number of notes of ₹ 100 be y, then according to the question, \( x + y = 25 \) ....(1) and \( 50x + 100y = 2000 \) ....(2).
Multiplying (1) by 50, we get \( 50x + 50y = 1250 \) ....(3).
Subtracting (3) from (2), we have \( 50y = 750 \Rightarrow y = 15 \).
Putting y = 15 in (1), we get \( x + 15 = 25 \Rightarrow x = 10 \).
Hence, the number of notes of ₹ 50 was 10 and that of ₹ 100 was 15.

Question. Yash scored 40 marks in a test, receiving 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Answer: Let the number of correct answers of Yash be x and number of wrong answers be y. Then according to question :
Case I. \( 3x – y = 40 \) ....(1)
Case II. \( 4x – 2y = 50 \) ....(2)
Multiplying (1) by 2, we get \( 6x – 2y = 80 \) ....(3).
Subtracting (2) from (3), we get \( 2x = 30 \Rightarrow x = 15 \).
Putting x = 15 in (1); we get \( 3 \times 15 – y = 40 \Rightarrow 45 – y = 40 \Rightarrow y = 5 \).
Total number of questions = 15 + 5 = 20.

Problems Based on Numbers

Question. What number must be added to each of the numbers, 5, 9, 17, 27 to make the numbers in proportion?
Answer: Let x be added to each of the given numbers. Then, \( (5 + x) (27 + x) = (9 + x) (17 + x) \).
\( \Rightarrow 135 + 32x + x^2 = 153 + 26x + x^2 \Rightarrow 32x – 26x = 153 – 135 \Rightarrow 6x = 18 \Rightarrow x = 3 \).

Question. The average score of boys in an examination of a school is 71 and that of girls is 73. The average score of the school in the examination is 71.8. Find the ratio of the number of boys to the number of girls that appeared in the examination.
Answer: Let the number of boys = x and number of girls = y. Total score of boys = 71x, Total score of girls = 73y. According to the question, \( 71.8 = \frac{71x + 73y}{x + y} \Rightarrow 71.8x + 71.8y = 71x + 73y \Rightarrow 0.8x = 1.2y \Rightarrow \frac{x}{y} = \frac{1.2}{0.8} = \frac{3}{2} \).
Hence, the ratio is 3 : 2.

Question. The difference between two numbers is 26 and one number is three times the other. Find them.
Answer: Let the numbers be x and y. \( x – y = 26 \) ....(1) and \( x = 3y \) ....(2). Putting (2) in (1), \( 3y – y = 26 \Rightarrow 2y = 26 \Rightarrow y = 13 \). Putting y = 13 in (2), \( x = 3 \times 13 = 39 \). The numbers are 39 and 13.

Problems Based on Ages

Question. Father’s age is three times the sum of ages of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.
Answer: Let the age of father = x years and sum of ages of children = y years. \( x = 3y \) ....(1). After 5 years, Father = \( x + 5 \), children sum = \( y + 10 \). \( x + 5 = 2(y + 10) \Rightarrow x – 2y = 15 \) ....(2). Putting (1) in (2), \( 3y – 2y = 15 \Rightarrow y = 15 \). \( x = 3 \times 15 = 45 \). Father's age is 45 years.

Question. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages.
Answer: Let Jacob = x, son = y. Case I: \( x + 5 = 3(y + 5) \Rightarrow x – 3y = 10 \) ....(1). Case II: \( x – 5 = 7(y – 5) \Rightarrow x = 7y – 30 \) ....(2). Putting (2) in (1), \( 7y – 30 – 3y = 10 \Rightarrow 4y = 40 \Rightarrow y = 10 \). \( x = 7(10) - 30 = 40 \). Ages are 40 and 10 years.

Problems Based on two digit numbers

Question. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
Answer: Let units = x, tens = y. Original = \( 10y + x \), Reversed = \( 10x + y \). \( (10y + x) + (10x + y) = 99 \Rightarrow 11x + 11y = 99 \Rightarrow x + y = 9 \) ....(1). Digit difference \( x – y = 3 \) ....(2). Adding (1) and (2), \( 2x = 12 \Rightarrow x = 6 \). \( y = 3 \). Number is 36.

Question. The sum of a two-digit number and the number obtained y reversing the order of its digits is 165. If the digits differ by 3, find the number.
Answer: \( 11x + 11y = 165 \Rightarrow x + y = 15 \) ....(1). \( x – y = 3 \) ....(2). Adding gives \( 2x = 18 \Rightarrow x = 9, y = 6 \). Number is 69.

Question. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number.
Answer: Let tens = x, units = y. \( x + y = 9 \) ....(1). \( 9(10x + y) = 2(10y + x) \Rightarrow 90x + 9y = 20y + 2x \Rightarrow 88x – 11y = 0 \Rightarrow 8x – y = 0 \) ....(2). Adding (1) and (2), \( 9x = 9 \Rightarrow x = 1 \). \( y = 8 \). Number is 18.

Problems Based on Fraction

Question. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.
Answer: Let fraction = \( x/y \). \( x + y = 2x + 4 \Rightarrow y = x + 4 \) ....(1). \( \frac{x + 3}{y + 3} = \frac{2}{3} \Rightarrow 3x + 9 = 2y + 6 \Rightarrow 3x – 2y + 3 = 0 \) ....(2). Substituting (1) in (2): \( 3x – 2(x + 4) + 3 = 0 \Rightarrow x = 5, y = 9 \). Fraction is 5/9.

Question. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Answer: \( x + y = 2y – 3 \Rightarrow y = 3 + x \) ....(1). \( x – 1 = \frac{1}{2}(y – 1) \Rightarrow 2x – y = 1 \) ....(2). Substituting (1) in (2): \( 2x – (3 + x) = 1 \Rightarrow x = 4, y = 7 \). Fraction is 4/7.

Question. A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.
Answer: Case 1: \( \frac{x+2}{y+2} = \frac{9}{11} \Rightarrow 11x – 9y = -4 \) ....(1). Case 2: \( \frac{x+3}{y+3} = \frac{5}{6} \Rightarrow 6x – 5y = -3 \) ....(2). Solving gives \( x = 7, y = 9 \). Fraction is 7/9.

Question. A fraction becomes 4/5 if 1 is added to each of the numerator and denominator. However, if we subtract 5 from each, the fraction becomes 1/2. Find the fraction.
Answer: \( \frac{x+1}{y+1} = \frac{4}{5} \Rightarrow 5x – 4y = -1 \) ....(1). \( \frac{x-5}{y-5} = \frac{1}{2} \Rightarrow 2x – y = 5 \) ....(2). Solving gives \( x = 7, y = 9 \). Fraction is 7/9.

Problem on Fixed Charges & Running Charges

Question. A Taxi charges consist of fixed charges and the remaining depending upon the distance travelled in kilometers. If a persons travels 10 km, he pays ₹ 68 and for travelling 15 km, he pays ₹ 98. Find the fixed charges and the rate per km.
Answer: Let fixed = x, rate = y. \( x + 10y = 68 \) ....(1); \( x + 15y = 98 \) ....(2). Subtracting gives \( 5y = 30 \Rightarrow y = 6 \). \( x = 8 \). Fixed charge = ₹ 8, Rate = ₹ 6/km.

Question. A lending library has a fixed charge for the first three days and an addition charge for each day thereafter. Sarika paid ₹ 27 for a book kept for seven days. While Susy paid ₹ 21 for the book the kept for five days. Find the fixed charge and the charge for each extra day.
Answer: Let fixed = x (for 3 days), extra = y. Sarika: \( x + 4y = 27 \); Susy: \( x + 2y = 21 \). Subtracting gives \( 2y = 6 \Rightarrow y = 3 \). \( x = 15 \). Fixed = ₹ 15, Extra = ₹ 3/day.

Question. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charges per kilometer? How much does a person have to pay for travelling a distance of 25 km?
Answer: \( x + 10y = 105 \); \( x + 15y = 155 \). Subtracting gives \( 5y = 50 \Rightarrow y = 10 \). \( x = 5 \). For 25 km: \( 5 + 25(10) = Rs 255 \).