CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables VBQs Set 02

Multiple Choice Questions

Question. The pair of linear equations \( \frac{3x}{2} + \frac{5y}{3} = 7 \) and \( 9x + 10y = 14 \) is
(a) consistent
(b) inconsistent
(c) consistent with one solution
(d) consistent with many solutions
Answer: (b)
Explanation:
For the given pair of equations, we have:
\( \frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6} \)
\( \frac{b_1}{b_2} = \frac{5/3}{10} = \frac{1}{6} \)
\( \frac{c_1}{c_2} = \frac{7}{14} = \frac{1}{2} \)
\( \therefore \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Hence, the pair of equations is inconsistent.

Question. Graphically, the pair of equations \( 6x - 3y + 10 = 0 \) and \( 2x - y + 9 = 0 \) represents two lines which are:
(a) intersecting at exactly one point.
(b) intersecting at exactly two points.
(c) coincident
(d) parallel
Answer: (d)
Explanation: The given equations are:
\( 6x - 3y + 10 = 0 \) ...(i)
Also, \( 2x - y + 9 = 0 \) ...(ii)
Table for \( 6x - 3y + 10 = 0 \),
\( \begin{array}{|c|c|c|} \hline x & 0 & -5/3 \\ \hline y & 10/3 & 0 \\ \hline \end{array} \)
Table for \( 2x - y + 9 = 0 \),
\( \begin{array}{|c|c|c|} \hline x & 0 & -9/2 \\ \hline y & 9 & 0 \\ \hline \end{array} \)
Hence, the pair of equations represents two parallel lines.

Question. If a pair of linear equations is consistent, then the lines will be:
(a) parallel
(b) always coincident
(c) intersecting or coincident
(d) always intersecting
Answer: (c)
Explanation: The conditions for a pair of linear equations to be consistent are:

• Intersecting lines having unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
• OR Coincident or dependent lines having infinitely many solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)

Question. The value of \( k \) for which the system of linear equation \( x + 2y = 3, 5x + ky + 7 = 0 \) is inconsistent is
(a) \( -\frac{14}{3} \)
(b) \( \frac{2}{5} \)
(c) 5
(d) 10
Answer: (d)
Explanation: The system of equations will be inconsistent if \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Here, \( a_1 = 1, b_1 = 2, c_1 = 3, a_2 = 5, b_2 = k, c_2 = -7 \).
\( \frac{1}{5} = \frac{2}{k} \neq \frac{3}{-7} \)
i.e., when \( k = 10 \)

Question. The value of \( k \) for which the system of equations \( x + y - 4 = 0 \) and \( 2x + ky = 3 \), has no solution, is
(a) -2
(b) \( \neq 2 \)
(c) 3
(d) 2
Answer: (d)
Explanation: \( x + y - 4 = 0 \) and \( 2x + ky = 3 \) has no solution, when:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \therefore \frac{1}{2} = \frac{1}{k} \neq \frac{4}{3} \Rightarrow k = 2 \)

Question. The pair of equations \( y = 0 \) and \( y = -7 \) has:
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution
Answer: (d)
Explanation: We know that equation of the form \( y = 'a' \) is a line parallel to the x-axis at a distance 'a' from it. The given pair of equations are \( y = 0 \) and \( y = -7 \). \( y = 0 \) is the equation of the x-axis and \( y = -7 \) is the equation of the line parallel to the x-axis. So, these two equations represent two parallel lines. We know that parallel lines never intersect. So, there is no solution for these lines.

Question. The pair of equations \( x = a \) and \( y = b \) graphically represents lines which are:
(a) parallel
(b) intersecting at (b, a)
(c) coincident
(d) intersecting at (a, b)
Answer: (d)
Explanation: We know that \( x = a \) is the equation of a straight line parallel to the y-axis at a distance of 'a' from it. Again, \( y = b \) is the equation of a straight line parallel to the x-axis at a distance of 'b' from it. So, the pair of equations \( x = a \) and \( y = b \) graphically represents lines which are intersecting at (a, b). Hence, the two lines are intersecting at (a, b).

Question. For which value(s) of \( p \), will the lines represented by the following pair of linear equations be parallel:
\( 3x - y - 5 = 0 \)
\( 6x - 2y - p = 0 \)
(a) all real values except 10
(b) 10
(c) \( \frac{5}{2} \)
(d) \( \frac{1}{2} \)
Answer: (a)

Question. If the lines given by \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \) are parallel, then the value of \( k \) is:
(a) \( -\frac{4}{5} \)
(b) \( \frac{5}{2} \)
(c) \( \frac{15}{4} \)
(d) \( \frac{3}{2} \)
Answer: (c)
Explanation: The given equation of lines are \( 3x + 2ky = 2 \) and \( 2x + 5y + 1 = 0 \). Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have \( a_1 = 3; b_1 = 2k; c_1 = -2; a_2 = 2; b_2 = 5; c_2 = 1 \). We know that the condition for parallel lines is \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
\( \Rightarrow \frac{3}{2} = \frac{2k}{5} \Rightarrow 15 = 4k \Rightarrow k = \frac{15}{4} \). Thus, \( k = \frac{15}{4} \).

Question. The pair of equations, \( x = 0 \) and \( x = -4 \) has:
(a) a unique solution
(b) no solution
(c) infinitely many solution
(d) only solution (0, 0)
Answer: (b)
Explanation: Since the lines represented by the given equations are parallel to each other, the pair of equations has no solution.

Question. One equation of a pair of dependent linear equations is \( -5x + 7y = 2 \). The second equation can be:
(a) \( 10x + 14y + 4 = 0 \)
(b) \( -10x - 14y + 4 = 0 \)
(c) \( -10x + 14y + 4 = 0 \)
(d) \( 10x - 14y = -4 \)
Answer: (d)
Explanation: In a pair of dependent linear equation, one equation is just a multiple of another equation. Thus, the second equation is \( k(-5x + 7y - 2) = 0 \). Putting \( k = -2 \), we get \( 10x - 14y + 4 = 0 \). On moving it to the other side, we get \( 10x - 14y = -4 \). \( \therefore \) (D) option is correct.

Question. A pair of linear equations which has a unique solution \( x = 2, y = -3 \) is:
(a) \( x + y = -1 \) and \( 2x - 3y = -5 \)
(b) \( 2x + 5y = -11 \) and \( 4x + 10y = -22 \)
(c) \( 2x - y = 1 \) and \( 3x + 2y = 0 \)
(d) \( x - 4y - 14 = 0 \) and \( 5x - y - 13 = 0 \)
Answer: (d)
Explanation: If \( x = 2 \) and \( y = -3 \) is a unique solution of any pair of equation, then these values must satisfy that pair of equations. Putting the values in the equations for every option and checking it - For case (D): The given equations are \( x - 4y - 14 = 0 \) and \( 5x - y - 13 = 0 \). Putting \( x = 2, y = -3 \) in the LHS of the equation \( x - 4y - 14 = 0 \), we get \( 2 - 4 \times (-3) - 14 = 2 + 12 - 14 = 0 = RHS \). Putting \( x = 2, y = -3 \) in the LHS of the equation \( 5x - y - 13 = 0 \), we get \( 5 \times 2 - (-3) - 13 = 10 + 3 - 13 = 0 = RHS \). Since \( x = 2, y = -3 \) is satisfying both the equations, option (D) is true. Hence, \( x = 2, y = -3 \) is the unique solution for these equations.

Question. If \( x = a, y = b \) is the solution of the equations \( x - y = 2 \) and \( x + y = 4 \), then the values of \( a \) and \( b \) respectively are:
(a) 3 and 5
(b) 5 and 3
(c) 3 and 1
(d) -1 and -3
Answer: (c)
Explanation: Since \( x = a, y = b \) is the solution of the equations \( x - y = 2 \) and \( x + y = 4 \), these values must satisfy the given pair of equations. Putting the values in the equations, we have \( a - b = 2 \) ...(i) and \( a + b = 4 \) ...(ii). Adding equations (i) and (ii), we get \( 2a = 6 \) or \( a = 3 \). Putting the value of \( a \) in equation (ii), we get \( 3 + b = 4 \) or \( b = 1 \). Hence, option (C) is correct.

Question. Aruna has only \( Rs 1 \) and \( Rs 2 \) coins with her. If the total number of coins that she has is 50 and the amount of money with her is \( Rs 75 \), then the number of \( Rs 1 \) and \( Rs 2 \) coins respectively are:
(a) 35 and 15
(b) 35 and 20
(c) 15 and 35
(d) 25 and 25
Answer: (d)
Explanation: Let number of \( Rs 1 \) coins \( = x \) and number of \( Rs 2 \) coins \( = y \). It is given that, total number of coins \( = x + y = 50 \) ...(i). Also, amount of money with her \( = (\text{Number of } Rs 1 \text{ coins} \times 1) + (\text{Number of } Rs 2 \text{ coins} \times 2) \). Now, by the given condition: \( x(1) + y(2) = 75 \Rightarrow x + 2y = 75 \) ...(ii). On subtracting eq. (i) from eq. (ii), we get \( (x + 2y) - (x + y) = (75 - 50) \Rightarrow y = 25 \). Putting \( y = 25 \) in eq. (i), we get \( x + 25 = 50 \Rightarrow x = 25 \). Hence, Aruna has 25 \( Rs 1 \) coins and 25 \( Rs 2 \) coins.

Fill in the Blanks/True False

Question. The value of \( k \) for which pair of linear equations \( 3x + 2y = 5 \) and \( x - ky = 2 \) has a unique solution is .................... .
Answer: \( k \neq -2/3 \)
Explanation: For unique solution \( \frac{3}{1} \neq \frac{2}{-k} \Rightarrow k \neq -2/3 \)

Question. The value of \( a \) so that the point (3, \( a \)), lies on the line represented by \( 2x - 3y = 5 \) is .................... .
Answer: \( \frac{1}{3} \)
Explanation: Given (3, \( a \)) lies on \( 2x - 3y = 5 \). \( \Rightarrow 2 \times 3 - 3a = 5 \Rightarrow 3a = 6 - 5 = 1 \Rightarrow a = \frac{1}{3} \)

Question. The co-ordinate where the line \( x - y = 8 \) will intersect \( y \)-axis is .................... .
Answer: (0, -8)
Explanation: At \( y \)-axis, \( x = 0 \). \( \therefore 0 - y = 8 \Rightarrow y = -8 \). Point \( = (0, -8) \)

Question. The value of \( k \) for which the pair of linear equations \( kx + 3y = k - 2 \) and \( 12x + ky = k \) has no solution is .................... .
Answer: \( k = \pm 6 \)
Explanation: Since, pair of linear equations has no solution. Then \( \frac{k}{12} = \frac{3}{k} \neq \frac{k-2}{k} \) i.e., \( k^2 = 36 \Rightarrow k = \pm 6 \)

Question. The graphical representation of the pair of equations \( x + 2y - 4 = 0 \) and \( 2x + 4y - 12 = 0 \) represents .................... .
Answer: Parallel lines
Explanation: \( x + 2y - 4 = 0 \) and \( 2x + 4y - 12 = 0 \). Here \( \frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{2}{4} = \frac{1}{2} \)

Question. If \(x - y = 2\) and \( \frac{1}{x+y} = \frac{2}{5} \), then \(x =\) ................... .
Answer: \(\frac{9}{4}\)
Explanation: \(x - y = 2\) and \( \frac{1}{x+y} = \frac{2}{5} \)
\(x - y = 2\) ...(i)
\(x + y = \frac{5}{2}\) ...(ii)
Adding equations (i) and (ii), we get
\((x - y) + (x + y) = 2 + \frac{5}{2}\)
\(2x = \frac{9}{2} \Rightarrow x = \frac{9}{4}\)

Question. The value of \(p\) for the following pair of linear equations \((p - 3)x + 3y = p; px + py = 12\) have infinitely many solutions is .................... .
Answer: \(p = 6\)
Explanation: Given equations are
\((p - 3)x + 3y = p\) and \(px + py = 12\)
For infinitely many solution
\(\frac{p - 3}{p} = \frac{3}{p} = \frac{p}{12}\)
\(\Rightarrow \frac{p - 3}{p} = \frac{3}{p}\) and \(\frac{3}{p} = \frac{p}{12}\)
\(\Rightarrow p - 3 = 3\) and \(p^2 = 36\)
\(\Rightarrow p = 6\) and \(p = \pm 6\)
Common value is
So, \(p = 6\)

Question. If \(x = a, y = b\) is the solution of the pair of equation \(x - y = 2\) and \(x + y = 4\) then the value of \(3a + 4b\) is .................... .
Answer: 13
Explanation: \(x - y = 2\) ...(i)
\(x + y = 4\) ...(ii)
On adding (i) and (ii), we get
\(2x = 6 \Rightarrow x = 3\)
and putting \(x = 3\) in (i),
\(3 - y = 2 \Rightarrow y = 1\)
\(3a + 4b = 3x + 4y = 3 \times 3 + 4 \times 1 = 13\)

Question. For the pair of equations \( \lambda x + 3y = -7 \) and \( 2x + 6y = 14 \) to have infinitely many solutions, the value of \( \lambda \) should be 1. Is this statement true? Give reasons.
Answer: No, for no value of \( \lambda \) will the given pair of linear equations has infinitely many solutions. The given pair of linear equations is \( \lambda x + 3y + 7 = 0 \) and \( 2x + 6y - 14 = 0 \). Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have \( a_1 = \lambda, b_1 = 3, c_1 = 7 \); \( a_2 = 2, b_2 = 6, c_2 = -14 \);
\( \frac{a_1}{a_2} = \frac{\lambda}{2}; \frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}; \frac{c_1}{c_2} = \frac{7}{-14} = -\frac{1}{2} \)
For infinitely many solutions
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{\lambda}{2} = \frac{1}{2} \) and \( \frac{\lambda}{2} = -\frac{1}{2} \)
\( \Rightarrow \lambda = 1 \) and \( \lambda = -1 \)
Since, \( \lambda \) does not have a unique value so for no value of \( \lambda \) will the given pair of linear equations have infinitely many solutions.

Question. For all real values of \(c\), the pair of equations \(x - 2y = 8\) and \(5x - 10y = c\) have a unique solution. Justify whether it is true or false. 
Answer: False
The given pair of equations will not have a unique solution for any value of \(c\).
The given pair of linear equations is \(x - 2y - 8 = 0\) and \(5x - 10y - c = 0\). Comparing with \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\), we have \(a_1 = 1, b_1 = -2, c_1 = -8\); \(a_2 = 5, b_2 = -10, c_2 = -c\);
\( \frac{a_1}{a_2} = \frac{1}{5}; \frac{b_1}{b_2} = \frac{-2}{-10} = \frac{1}{5}; \frac{c_1}{c_2} = \frac{-8}{-c} = \frac{8}{c} \)
But for \(c = 40\) (any real value), the ratio will be \( \frac{c_1}{c_2} = \frac{8}{40} = \frac{1}{5} \). When \(c = 40\), \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{1}{5} \). Thus, the given pair of linear equations will have infinitely many solutions for \(c = 40\). Also, when \(c \neq 40\), \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). Thus, the given pair of linear equations will have no solution for \(c \neq 40\). Hence, for any value of \(c\), the system of linear equations does not have a unique solution.

Very short Questions

Question. Write the relationship between the coefficients, if the following pair of equations are inconsistent. \(ax + by + c = 0; a'x + b'y + c' = 0\).
Answer: The required relationship is: \( \frac{a}{a'} = \frac{b}{b'} \neq \frac{c}{c'} \)

Question. When will the system \(kx - y = 2\) and \(6x - 2y = 3\) has a unique solution only? 
Answer: A pair of linear pair has unique solution only when, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \). Then, \( \frac{k}{6} \neq \frac{-1}{-2} \). So, \( k \neq 3 \).

Question. Find the solution of \(x + y = 3\) and \(7x + 6y = 2\).
Answer: \(x + y = 3\) gives, \(y = 3 - x\) ...(i)
So, \(7x + 6y = 2\) gives \(7x + 6(3 - x) = 2 \Rightarrow 7x + 18 - 6x = 2 \Rightarrow x = -16\)
From (i), \(y = 3 + 16 = 19\)
Thus, \(x = -16\) and \(y = 19\) is the required solution.

SHORT ANSWER Type Questions 

Question. Find the value(s) of \(k\) for which the pair of equations \( \begin{cases} kx + 2y = 3 \\ 3x + 6y = 10 \end{cases} \) has a unique solution. 
Answer: Given: pair of equation is \(kx + 2y = 3\) and \(3x + 6y = 10\). For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
Here, \(a_1 = k, b_1 = 2; a_2 = 3, b_2 = 6\)
\( \therefore \frac{k}{3} \neq \frac{2}{6} \Rightarrow k \neq 1 \)
Hence, the pair of equation has a unique solution for all real values of \(k\) except 1.

Question. The larger of two supplementary angles exceeds the smaller by 18°. Find the angles. 
Answer: Let, the smaller angle be \(x\) and the larger angle be \(y\). According to the given conditions:
\(y = x + 18° \Rightarrow -x + y = 18°\) ...(i)
and \(x + y = 180°\) ...(ii) (Sum of supplementary angles is 180°)
Now, on adding equation (i) and (ii), we get:
\(2y = 198° \Rightarrow y = 99°\)
Put the value of \(y\) in equation (i), we get \( -x + 99° = 18° \Rightarrow x = 99° - 18° = 81° \)
Hence, the two supplementary angles are 81° and 99°.

Question. In a \(\triangle ABC, \angle A = x°, \angle B = 3x°\) and \( \angle C = y°\). If \(3y° - 5x° = 30°\) prove that the triangle is right angled. 
Answer: We know that, \( \angle A + \angle B + \angle C = 180° \) [Sum of interior angles of triangle ABC is 180°]
\( \Rightarrow x + 3x + y = 180° \Rightarrow 4x + y = 180° \) ...(i)
and \(3y - 5x = 30°\) [Given] ...(ii)
Multiply equation (i) by 3: \(12x + 3y = 540°\) ...(iii)
Subtracting (ii) from (iii), we get \(17x = 510 \Rightarrow x = 30\).
Putting value of \(x\) in equation (i), we get \( 4 \times 30° + y = 180° \Rightarrow y = 60° \).
\( \therefore \angle A = 30°, \angle B = 3 \times 30° = 90° \) and \( \angle C = 60° \).
Hence, \(\triangle ABC\) is right angled triangle at B.

Question. Find \(c\) if the system of equations \(cx + 3y + (3 - c) = 0; 12x + cy - c = 0\) has infinitely many solutions? 
Answer: Given equation is: \(cx + 3y + (3 - c) = 0\) and \(12x + cy - c = 0\). Condition for equations to have infinitely many solutions is: \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
Here, \(a_1 = c, b_1 = 3, c_1 = 3 - c; a_2 = 12, b_2 = c, c_2 = -c \)
\( \therefore \frac{c}{12} = \frac{3}{c} = \frac{3 - c}{-c} \)
\( \Rightarrow c^2 = 36 \Rightarrow c = 6 \) or \( c = -6 \) ...(i)
Also, \( -3c = 3c - c^2 \Rightarrow c^2 = 6c \Rightarrow c = 6 \) or \( c = 0 \) ...(ii)
From (i) and (ii), we get, \(c = 6\).
Hence, the value of \(c = 6\).

Question. For what value of \(k\), does the system of linear equations \(2x + 3y = 7\) and \((k - 1)x + (k + 2)y = 3k\) have an infinite number of solutions? 
Answer: The given system of linear equation is \(2x + 3y = 7\) and \((k - 1)x + (k + 2)y = 3k\). For infinitely many solutions, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
\(a_1 = 2, b_1 = 3, c_1 = 7\) and \(a_2 = (k - 1), b_2 = (k + 2), c_2 = 3k\).
\( \Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{7}{3k} \)
\( \Rightarrow 2(k + 2) = 3(k - 1) \Rightarrow 2k + 4 = 3k - 3 \Rightarrow k = 7 \)
and \( 3(3k) = 7(k + 2) \Rightarrow 9k = 7k + 14 \Rightarrow 2k = 14 \Rightarrow k = 7 \)
Hence, the value of \(k\) is 7.

Question. If \( 2x + y = 23 \) and \( 4x - y = 19 \), find the values of \( 5y - 2x \) and \( \frac{y}{x} - 2 \). 
Answer: The given equations are
\( 2x + y = 23 \dots (i) \)
\( 4x - y = 19 \dots (ii) \)
On adding both equations, we get
\( \Rightarrow 6x = 42 \)
\( \Rightarrow x = 7 \)
Putting the value of \( x \) in eq. (i), we get
\( \Rightarrow 2(7) + y = 23 \)
\( \Rightarrow y = 23 - 14 \)
\( \Rightarrow y = 9 \)
We have \( 5y - 2x = 5(9) - 2(7) \)
\( = 45 - 14 \)
\( = 31 \)
and \( \frac{y}{x} - 2 = \frac{9}{7} - 2 \)
\( = -\frac{5}{7} \)
Hence, the values of \( (5y - 2x) \) and \( \frac{y}{x} - 2 \) are \( 31 \) and \( -\frac{5}{7} \) respectively.

Question. Write an equation for a line passing through the point representing solution of the pair of linear equations \( x + y = 2 \) and \( 2x - y = 1 \). How many such lines can we find? 
Answer: The given equations are
\( x + y = 2 \dots (i) \)
\( 2x - y = 1 \dots (ii) \)
Adding eq. (i) and (ii), we have
\( 3x = 3 \Rightarrow x = 1 \)
Substituting \( x = 1 \) in eq. (i), we have
\( y = 1 \)
So, the solution is \( x = 1 \) and \( y = 1 \) and the point that represents the solution is \( (1, 1) \).
We also know that an infinite number of lines can pass through a given point, say \( (1, 1) \).
Hence, infinite lines can pass through the intersection point of the linear equations \( x + y = 2 \) and \( 2x - y = 1 \) i.e., \( P(1, 1) \).

Question. A fraction becomes \( \frac{1}{4} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction. 
Answer: Let the fraction be \( \frac{a}{b} \).
Then, according to the question,
\( \frac{a-1}{b} = \frac{1}{4} \) and \( \frac{a}{b+8} = \frac{1}{4} \)
\( \Rightarrow 3a - b = 3 \) and \( 4a - b = 8 \)
On solving these equations, we get:
\( a = 5, b = 12 \).
So, the fraction is \( \frac{5}{12} \).

Question. In the figure, ABCDE is a pentagon with \( BE \parallel CD \) and \( BC \parallel DE \). \( BC \) is perpendicular to \( CD \). \( AB = 5 \) cm, \( AE = 5 \) cm, \( BE = 7 \) cm, \( BC = x - y \) and \( CD = x + y \). If the perimeter of ABCDE is 27 cm. Find the value of \( x \) and \( y \), given \( x, y \neq 0 \). 
Answer: \( x + y = 7 \) and \( 2(x - y) + x + y + 5 + 5 = 27 \)
\( \therefore x + y = 7 \) and \( 3x - y = 17 \)
Solving, we get, \( x = 6 \) and \( y = 1 \)
[CBSE Marking Scheme 2019]

SHORT ANSWER (SA-II) Type Questions 

Question. For which value(s) of \( \lambda \) do the pair of linear equations \( \lambda x + y = \lambda^2 \) and \( x + \lambda y = 1 \) have
(a) no solution?
(b) infinitely many solutions?
(c) a unique solution? 
Answer: The given pair of linear equations is
\( \lambda x + y - \lambda^2 = 0 \)
and \( x + \lambda y - 1 = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = \lambda, b_1 = 1, c_1 = -\lambda^2 \);
\( a_2 = 1, b_2 = \lambda, c_2 = -1 \);
\( \frac{a_1}{a_2} = \frac{\lambda}{1} \); \( \frac{b_1}{b_2} = \frac{1}{\lambda} \); \( \frac{c_1}{c_2} = \frac{-\lambda^2}{-1} = \frac{\lambda^2}{1} \)
(A) For no solution,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} \neq \frac{\lambda^2}{1} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} \) and \( \frac{1}{\lambda} \neq \frac{\lambda^2}{1} \)
\( \lambda^2 - 1 = 0 \) and \( \lambda^3 \neq 1 \)
\( (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda \neq 1 \)
\( \lambda = 1, -1 \) and \( \lambda \neq 1 \)
Here, we take only \( \lambda = -1 \).
Hence for \( \lambda = -1 \), the pair of linear equations has no solution.
(B) For infinitely many solutions,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} = \frac{\lambda^2}{1} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} \) and \( \frac{1}{\lambda} = \frac{\lambda^2}{1} \)
\( \lambda^2 - 1 = 0 \) and \( \lambda^3 = 1 \)
\( (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda = 1 \)
\( \lambda = 1, -1 \) and \( \lambda = 1 \)
\( \lambda = 1 \) satisfies both the equations.
Hence, for \( \lambda = 1 \), the pair of linear equations has infinitely many solutions.
(C) For a unique solution,
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \frac{\lambda}{1} \neq \frac{1}{\lambda} \)
\( \lambda^2 - 1 \neq 0 \)
\( (\lambda - 1)(\lambda + 1) \neq 0 \)
\( \lambda \neq 1, -1 \)
Hence, for all real values of \( \lambda \) except \( \pm 1 \), the given pair of equations has a unique solution.

Question. For which values of \( a \) and \( b \) will the following pair of linear equations have infinitely many solutions?
\( x + 2y = 1 \)
\( (a - b)x + (a + b)y = a + b - 2 \) 
Answer: The given pair of linear equations is
\( x + 2y = 1 \)
and \( (a - b)x + (a + b)y = a + b - 2 \)
\( x + 2y - 1 = 0 \)
and \( (a - b)x + (a + b)y - (a + b - 2) = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = 1, b_1 = 2, c_1 = -1 \)
\( a_2 = (a - b), b_2 = (a + b), c_2 = -(a + b - 2) \)
\( \frac{a_1}{a_2} = \frac{1}{a - b} \); \( \frac{b_1}{b_2} = \frac{2}{a + b} \); \( \frac{c_1}{c_2} = \frac{-1}{-(a + b - 2)} = \frac{1}{a + b - 2} \)
For infinitely many solutions,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{1}{a - b} = \frac{2}{a + b} = \frac{1}{a + b - 2} \)
Taking the first two parts
\( \frac{1}{a - b} = \frac{2}{a + b} \)
\( \Rightarrow a + b = 2(a - b) \)
\( \Rightarrow 2a - a = 2b + b \)
\( \Rightarrow a = 3b \dots (i) \)
Taking the last two parts,
\( \frac{2}{a + b} = \frac{1}{a + b - 2} \)
\( \Rightarrow 2(a + b - 2) = (a + b) \)
\( \Rightarrow 2a + 2b - 4 = a + b \)
\( \Rightarrow a + b = 4 \dots (ii) \)
Putting the value of \( a \) from eq. (i) in eq. (ii), we get
\( \Rightarrow 3b + b = 4 \)
\( \Rightarrow 4b = 4 \)
\( \Rightarrow b = 1 \)
Putting the value of \( b \) in eq. (i), we get
\( a = 3(1) = 3 \)
The values \( (a, b) = (3, 1) \) satisfies all the parts.
Hence, the required values of \( a \) and \( b \) are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

Question. Write a pair of linear equations which has the unique solution \( x = -1, y = 3 \). How many such pairs can you write? 
Answer: We know that the condition for the pair of system to have a unique solution is
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Let the equations be
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
It is given that \( x = -1 \) and \( y = 3 \) is the unique solution of these two equations, then it must satisfy the above equations.
\( \Rightarrow a_1(-1) + b_1(3) + c_1 = 0 \)
\( \Rightarrow - a_1 + 3b_1 + c_1 = 0 \dots (i) \)
and \( a_2(-1) + b_2(3) + c_2 = 0 \)
\( \Rightarrow - a_2 + 3b_2 + c_2 = 0 \dots (ii) \)
The restricted values of \( a_1, a_2 \) and \( b_1, b_2 \) are only
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \dots (iii) \)
So, all the real values of \( a_1, a_2, b_1, b_2 \) except those which satisfy eq. (iii) and satisfy eq. (i) and eq. (ii) will have the solution \( x = -1 \) and \( y = 3 \).
Hence, infinitely many pairs of linear equations are possible.

Question. Solve the pair of equations:
\( \frac{2}{x} + \frac{3}{y} = 11 \), \( \frac{5}{x} - \frac{4}{y} = -7 \)
Hence, find the value of \( 5x - 3y \). 
Answer: Given equations are
\( \frac{2}{x} + \frac{3}{y} = 11 \dots (i) \)
and \( \frac{5}{x} - \frac{4}{y} = -7 \dots (ii) \)
Eq (i) \( \times 5 \) and eq (ii) \( \times 2 \) give
\( \frac{10}{x} + \frac{15}{y} = 55 \dots (iii) \)
and \( \frac{10}{x} - \frac{8}{y} = -14 \dots (iv) \)
On subtracting eq (iv) from eq (iii), we have
\( \frac{23}{y} = 69 \)
\( \Rightarrow y = \frac{1}{3} \)
On substituting this value \( y = \frac{1}{3} \) in eq (i), we have:
\( \frac{2}{x} + 9 = 11 \)
i.e., \( \frac{2}{x} = 2 \)
or \( x = 1 \)
Thus, \( x = 1, y = \frac{1}{3} \) is the required solution.
Hence, \( 5x - 3y = 5(1) - 3(\frac{1}{3}) = 5 - 1 = 4 \)

Question. Find the solution of the pair of equations \( \frac{x}{10} + \frac{y}{5} - 1 = 0 \) and \( \frac{x}{8} + \frac{y}{6} = 15 \). Hence, find \( \lambda \), if \( y = \lambda x + 5 \). 
Answer: The given pair of equations is
\( \frac{x}{10} + \frac{y}{5} - 1 = 0 \)
\( \Rightarrow x + 2y - 10 = 0 \)
\( \Rightarrow x + 2y = 10 \dots (i) \)
and \( \frac{x}{8} + \frac{y}{6} = 15 \)
\( \Rightarrow 3x + 4y - 360 = 0 \)
\( \Rightarrow 3x + 4y = 360 \dots (ii) \)
Multiplying eq. (i) by 2 and subtracting it from eq. (ii), we get
\( (3x + 4y) - (2x + 4y) = 360 - 20 \)
\( \Rightarrow x = 340 \)
Putting the value of \( x \) in eq. (i), we get
\( 340 + 2y = 10 \)
\( \Rightarrow 2y = -330 \)
\( \Rightarrow y = -165 \)
It is given that \( y = \lambda x + 5 \)
Putting the values of \( x \) and \( y \) in the above equation, we get
\( y = \lambda x + 5 \)
\( \Rightarrow -165 = \lambda(340) + 5 \)
\( \Rightarrow -\lambda(340) = 5 + 165 \)
\( \Rightarrow -\lambda(340) = 170 \)
\( \Rightarrow \lambda = \frac{-170}{340} = \frac{-1}{2} \)
Hence, the solution of the pair of equations is \( x = 340, y = -165 \) and the required value of \( \lambda \) is \( \frac{-1}{2} \).