CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 20

Question 1. Solve by the elimination method: \( x + y = 5 \), \( 2 x - 3 y = 4 \).
Answer:
\( x + y = 5 \) ...(i)    \( 2 x - 3 y = 4 \) ...(ii)
Multiplying (i) by 2: \( 2 x + 2 y = 10 \) ...(iii)
Subtracting (ii) from (iii), we get \( 5 y = 6 \Rightarrow y = \frac{6}{5} \).
Putting \( y = \frac{6}{5} \) in (i): \( \left(\frac{6}{5}\right) + x = 5 \Rightarrow x = \frac{19}{5} \).
Hence, \( x = \frac{19}{5},\ y = \frac{6}{5} \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Multiply so that one variable has equal coefficients, then add or subtract to remove it.
2. Substitute the value you found back into an original equation and verify in the other one.

Question 2. Solve by the elimination method: \( 2 x + 3 y = 8 \), \( 4 x + 6 y = 7 \).
Answer:
\( 2 x + 3 y - 8 = 0 \) and \( 4 x + 6 y - 7 = 0 \), so \( \frac{a_1}{a_2} = \frac{1}{2},\ \frac{b_1}{b_2} = \frac{1}{2},\ \frac{c_1}{c_2} = \frac{8}{7} \).
Here \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Hence the lines are parallel.
Since the lines are parallel, the pair has no solution. (Trying elimination: multiplying the first equation by 2 gives a contradiction.)
In simple words: Parallel lines never meet, so no pair of values can satisfy both equations.

Teacher's Note:
1. Check the ratios before eliminating; equal x and y ratios with a different constant ratio signal no solution.
2. Do not try to force an answer when the system is inconsistent.

Question 3. Solve by the elimination method: \( 11 x + 15 y + 23 = 0 \), \( 7 x - 2 y - 20 = 0 \).
Answer:
Simplifying the given equations, we get
\( 11 x + 15 y = -23 \) ...(i)    \( 7 x - 2 y = 20 \) ...(ii)
Multiplying (i) by 2: \( 22 x + 30 y = -46 \) ...(iii)
Multiplying (ii) by 15: \( 105 x - 30 y = 300 \) ...(iv)
Adding (iii) and (iv), we get \( 127 x = 254 \Rightarrow x = 2 \).
Putting \( x = 2 \) in (i): \( 11 \left(2\right) + 15 y = -23 \Rightarrow y = -3 \).
Hence, \( x = 2,\ y = -3 \).
In simple words: The answer is the pair of values that fits both equations, so always put it back to check.

Teacher's Note:
1. Multiply so that one variable has equal coefficients, then add or subtract to remove it.
2. Substitute the value you found back into an original equation and verify in the other one.

Question 4. Solve by the elimination method: \( 78 x + 91 y = 39 \), \( 65 x + 117 y = 42 \).
Answer:
Simplifying the given equations, we get
\( 6 x + 7 y = 3 \) ...(i)    \( 65 x + 117 y = 42 \) ...(ii)
Multiplying (i) by 65: \( 390 x + 455 y = 195 \) ...(iii)
Multiplying (ii) by 6: \( 390 x + 702 y = 252 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 247 y = -57 \Rightarrow y = \frac{3}{13} \).
Putting \( y = \frac{3}{13} \) in (i): \( 7 \left(\frac{3}{13}\right) + 6 x = 3 \Rightarrow x = \frac{3}{13} \).
Hence, \( x = \frac{3}{13},\ y = \frac{3}{13} \).
In simple words: We reduce two equations to one equation in a single variable, solve it, and then find the other variable.

Teacher's Note:
1. Multiply so that one variable has equal coefficients, then add or subtract to remove it.
2. Substitute the value you found back into an original equation and verify in the other one.

Question 5. Solve by the elimination method: \( \frac{x}{2} + \frac{2 y}{3} = -1 \), \( x - \frac{y}{3} = 3 \).
Answer:
Simplifying the given equations, we get
\( 3 x + 4 y = -6 \) ...(i)    \( 3 x - y = 9 \) ...(ii)
Subtracting (ii) from (i), we get \( 5 y = -15 \Rightarrow y = -3 \).
Putting \( y = -3 \) in (i): \( 4 \left(-3\right) + 3 x = -6 \Rightarrow x = 2 \).
Hence, \( x = 2,\ y = -3 \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Multiply so that one variable has equal coefficients, then add or subtract to remove it.
2. Substitute the value you found back into an original equation and verify in the other one.

Question 6. Solve by the elimination method: \( \frac{x}{3} + \frac{y}{4} = 11 \), \( \frac{5 x}{6} - \frac{y}{3} + 7 = 0 \).
Answer:
Simplifying the given equations, we get
\( 4 x + 3 y = 132 \) ...(i)    \( 5 x - 2 y = -42 \) ...(ii)
Multiplying (i) by 2: \( 8 x + 6 y = 264 \) ...(iii)
Multiplying (ii) by 3: \( 15 x - 6 y = -126 \) ...(iv)
Adding (iii) and (iv), we get \( 23 x = 138 \Rightarrow x = 6 \).
Putting \( x = 6 \) in (i): \( 4 \left(6\right) + 3 y = 132 \Rightarrow y = 36 \).
Hence, \( x = 6,\ y = 36 \).
In simple words: The answer is the pair of values that fits both equations, so always put it back to check.

Teacher's Note:
1. Multiply so that one variable has equal coefficients, then add or subtract to remove it.
2. Substitute the value you found back into an original equation and verify in the other one.

Question 7. Solve for x and y: \( a x - b y = a^{2} + b^{2} \), \( x + y = 2 a \), where \( a + b \neq 0 \).
Answer:
\( a x - b y = a^{2} + b^{2} \) ...(i)    \( x + y = 2 a \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a,\ b_1 = - b,\ c_1 = - a^{2} - b^{2} \) and \( a_2 = 1,\ b_2 = 1,\ c_2 = - 2 a \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{\left(a + b\right)^{2}} = \frac{y}{\left(a - b\right) \left(a + b\right)} = \frac{1}{a + b} \]
Therefore \( x = a + b \) and \( y = a - b \).
Hence, \( x = a + b,\ y = a - b \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 8. Solve for x and y: \( a - \frac{a y}{b} + b + \frac{b x}{a} = 0 \), \( 2 a b - a y + b x = 0 \).
Answer:
\( - a^{2} y + b^{2} x = - a b \left(a + b\right) \) ...(i)    \( - a y + b x = - 2 a b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b^{2},\ b_1 = - a^{2},\ c_1 = a b \left(a + b\right) \) and \( a_2 = b,\ b_2 = - a,\ c_2 = 2 a b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a^{2} b \left(a - b\right)} = \frac{y}{a b^{2} \left(a - b\right)} = \frac{1}{a b \left(a - b\right)} \]
Therefore \( x = - a \) and \( y = b \).
Hence, \( x = - a,\ y = b \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 9. Solve for x and y: \( a x - 2 a + b y + 3 b = 0 \), \( - a y - 3 a + b x - 2 b = 0 \).
Answer:
\( a x + b y = 2 a - 3 b \) ...(i)    \( - a y + b x = 3 a + 2 b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a,\ b_1 = b,\ c_1 = - 2 a + 3 b \) and \( a_2 = b,\ b_2 = - a,\ c_2 = - 3 a - 2 b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 2 \left(a^{2} + b^{2}\right)} = \frac{y}{3 \left(a^{2} + b^{2}\right)} = \frac{1}{- a^{2} - b^{2}} \]
Therefore \( x = 2 \) and \( y = -3 \).
Hence, \( x = 2,\ y = -3 \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 10. Solve for x and y: \( a + 4 b + 2 \left(a x - b y\right) = 0 \), \( - 4 a + b + 2 \left(a y + b x\right) = 0 \).
Answer:
\( 2 a x - 2 b y = - a - 4 b \) ...(i)    \( 2 a y + 2 b x = 4 a - b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = 2 a,\ b_1 = - 2 b,\ c_1 = a + 4 b \) and \( a_2 = 2 b,\ b_2 = 2 a,\ c_2 = - 4 a + b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 2 \left(a^{2} + b^{2}\right)} = \frac{y}{8 \left(a^{2} + b^{2}\right)} = \frac{1}{4 \left(a^{2} + b^{2}\right)} \]
Therefore \( x = - \frac{1}{2} \) and \( y = 2 \).
Hence, \( x = - \frac{1}{2},\ y = 2 \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 11. Solve for x and y: \( a y + b x = 0 \), \( x \left(a + b\right) + y \left(a - b\right) = a^{2} + b^{2} \).
Answer:
\( a y + b x = 0 \) ...(i)    \( x \left(a + b\right) + y \left(a - b\right) = a^{2} + b^{2} \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b,\ b_1 = a,\ c_1 = 0 \) and \( a_2 = a + b,\ b_2 = a - b,\ c_2 = - a^{2} - b^{2} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a \left(a^{2} + b^{2}\right)} = \frac{y}{b \left(a^{2} + b^{2}\right)} = \frac{1}{- a^{2} - b^{2}} \]
Therefore \( x = a \) and \( y = - b \).
Hence, \( x = a,\ y = - b \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 12. Solve for x and y: \( x \left(a + c\right) - y \left(a - c\right) = 2 a b \), \( x \left(a + b\right) - y \left(a - b\right) = 2 a b \).
Answer:
\( x \left(a + c\right) + y \left(- a + c\right) = 2 a b \) ...(i)    \( x \left(a + b\right) + y \left(- a + b\right) = 2 a b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a + c,\ b_1 = - a + c,\ c_1 = - 2 a b \) and \( a_2 = a + b,\ b_2 = - a + b,\ c_2 = - 2 a b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{2 a b \left(b - c\right)} = \frac{y}{- 2 a b \left(b - c\right)} = \frac{1}{2 a \left(b - c\right)} \]
Therefore \( x = b \) and \( y = - b \).
Hence, \( x = b,\ y = - b \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 13. Solve for x and y: \( x \left(a + 2 b\right) + y \left(2 a - b\right) = 2 \), \( x \left(a - 2 b\right) + y \left(2 a + b\right) = 3 \).
Answer:
\( x \left(a + 2 b\right) + y \left(2 a - b\right) = 2 \) ...(i)    \( x \left(a - 2 b\right) + y \left(2 a + b\right) = 3 \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a + 2 b,\ b_1 = 2 a - b,\ c_1 = -2 \) and \( a_2 = a - 2 b,\ b_2 = 2 a + b,\ c_2 = -3 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 2 a + 5 b} = \frac{y}{a + 10 b} = \frac{1}{10 a b} \]
Therefore \( x = - \frac{2 a - 5 b}{10 a b} \) and \( y = \frac{a + 10 b}{10 a b} \).
Hence, \( x = - \frac{2 a - 5 b}{10 a b},\ y = \frac{a + 10 b}{10 a b} \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 14. Solve for x and y: \( \sqrt{2} x - \sqrt{3} y = 0 \), \( \sqrt{5} x + \sqrt{2} y = 0 \).
Answer:
\( \sqrt{2} x - \sqrt{3} y = 0 \) ...(i)    \( \sqrt{5} x + \sqrt{2} y = 0 \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = \sqrt{2},\ b_1 = - \sqrt{3},\ c_1 = 0 \) and \( a_2 = \sqrt{5},\ b_2 = \sqrt{2},\ c_2 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{0} = \frac{y}{0} = \frac{1}{2 + \sqrt{15}} \]
Therefore \( x = 0 \) and \( y = 0 \).
Hence, \( x = 0,\ y = 0 \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 15. Solve for x and y: \( \sqrt{7} x + \sqrt{11} y = 0 \), \( \sqrt{3} x - \sqrt{5} y = 0 \).
Answer:
\( \sqrt{7} x + \sqrt{11} y = 0 \) ...(i)    \( \sqrt{3} x - \sqrt{5} y = 0 \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = \sqrt{7},\ b_1 = \sqrt{11},\ c_1 = 0 \) and \( a_2 = \sqrt{3},\ b_2 = - \sqrt{5},\ c_2 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{0} = \frac{y}{0} = \frac{1}{- \sqrt{35} - \sqrt{33}} \]
Therefore \( x = 0 \) and \( y = 0 \).
Hence, \( x = 0,\ y = 0 \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Write both equations as ax + by + c = 0 first, because the formula needs all terms on the left.
2. The denominator a1b2 - a2b1 must not be zero for a unique solution.

Question 16. Solve by the elimination method: \( 0.5 x + 0.7 y = 0.74 \), \( 0.3 x + 0.5 y = 0.5 \).
Answer:
Simplifying the given equations, we get
\( 25 x + 35 y = 37 \) ...(i)    \( 3 x + 5 y = 5 \) ...(ii)
Multiplying (ii) by 7: \( 21 x + 35 y = 35 \) ...(iv)
Subtracting (iv) from (i), we get \( 4 x = 2 \Rightarrow x = \frac{1}{2} \).
Putting \( x = \frac{1}{2} \) in (i): \( 25 \left(\frac{1}{2}\right) + 35 y = 37 \Rightarrow y = \frac{7}{10} \).
Hence, \( x = \frac{1}{2},\ y = \frac{7}{10} \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. Multiply so that one variable has equal coefficients, then add or subtract to remove it.
2. Substitute the value you found back into an original equation and verify in the other one.

Question 17. Solve by the elimination method: \( 23 x - 29 y = 98 \), \( 29 x - 23 y = 110 \).
Answer:
\( 23 x - 29 y = 98 \) ...(i)    \( 29 x - 23 y = 110 \) ...(ii)
Adding (i) and (ii), we get \( 52 x - 52 y = 208 \), i.e., \( x - y = 4 \) ...(iii)
Subtracting (ii) from (i), we get \( - 6 x - 6 y = -12 \), i.e., \( x + y = 2 \) ...(iv)
Solving (iii) and (iv): \( x = 3 \) and \( y = -1 \).
Hence, \( x = 3,\ y = -1 \).
In simple words: The answer is the pair of values that fits both equations, so always put it back to check.

Teacher's Note:
1. When the coefficients of x and y are interchanged between the two equations, adding and subtracting gives easy equations.
2. Solve x + y and x - y first and then find x and y.

Question 18. Solve by the elimination method: \( 217 x + 131 y = 913 \), \( 131 x + 217 y = 827 \).
Answer:
\( 217 x + 131 y = 913 \) ...(i)    \( 131 x + 217 y = 827 \) ...(ii)
Adding (i) and (ii), we get \( 348 x + 348 y = 1740 \), i.e., \( x + y = 5 \) ...(iii)
Subtracting (ii) from (i), we get \( 86 x - 86 y = 86 \), i.e., \( x - y = 1 \) ...(iv)
Solving (iii) and (iv): \( x = 3 \) and \( y = 2 \).
Hence, \( x = 3,\ y = 2 \).
In simple words: We reduce two equations to one equation in a single variable, solve it, and then find the other variable.

Teacher's Note:
1. When the coefficients of x and y are interchanged between the two equations, adding and subtracting gives easy equations.
2. Solve x + y and x - y first and then find x and y.

Question 19. Solve by the elimination method: \( 23 x + 37 y = 32 \), \( 37 x + 23 y = 88 \).
Answer:
\( 23 x + 37 y = 32 \) ...(i)    \( 37 x + 23 y = 88 \) ...(ii)
Adding (i) and (ii), we get \( 60 x + 60 y = 120 \), i.e., \( x + y = 2 \) ...(iii)
Subtracting (ii) from (i), we get \( - 14 x + 14 y = -56 \), i.e., \( x - y = 4 \) ...(iv)
Solving (iii) and (iv): \( x = 3 \) and \( y = -1 \).
Hence, \( x = 3,\ y = -1 \).
In simple words: Find the one pair of numbers x and y that makes both equations true at the same time.

Teacher's Note:
1. When the coefficients of x and y are interchanged between the two equations, adding and subtracting gives easy equations.
2. Solve x + y and x - y first and then find x and y.

Question 20. Solve by the elimination method: \( 65 x - 33 y = 97 \), \( 33 x - 65 y = 1 \).
Answer:
\( 65 x - 33 y = 97 \) ...(i)    \( 33 x - 65 y = 1 \) ...(ii)
Adding (i) and (ii), we get \( 98 x - 98 y = 98 \), i.e., \( x - y = 1 \) ...(iii)
Subtracting (ii) from (i), we get \( 32 x + 32 y = 96 \), i.e., \( x + y = 3 \) ...(iv)
Solving (iii) and (iv): \( x = 2 \) and \( y = 1 \).
Hence, \( x = 2,\ y = 1 \).
In simple words: The answer is the pair of values that fits both equations, so always put it back to check.

Teacher's Note:
1. When the coefficients of x and y are interchanged between the two equations, adding and subtracting gives easy equations.
2. Solve x + y and x - y first and then find x and y.

Question 21. Solve by the elimination method: \( 99 x + 101 y = 499 \), \( 101 x + 99 y = 501 \).
Answer:
\( 99 x + 101 y = 499 \) ...(i)    \( 101 x + 99 y = 501 \) ...(ii)
Adding (i) and (ii), we get \( 200 x + 200 y = 1000 \), i.e., \( x + y = 5 \) ...(iii)
Subtracting (ii) from (i), we get \( - 2 x + 2 y = -2 \), i.e., \( x - y = 1 \) ...(iv)
Solving (iii) and (iv): \( x = 3 \) and \( y = 2 \).
Hence, \( x = 3,\ y = 2 \).
In simple words: We reduce two equations to one equation in a single variable, solve it, and then find the other variable.

Teacher's Note:
1. When the coefficients of x and y are interchanged between the two equations, adding and subtracting gives easy equations.
2. Solve x + y and x - y first and then find x and y.