CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 19

Question 1. Solve by the substitution method: \( 3 x + 11 y = 13 \), \( 8 x + 13 y = 2 \).
Answer:
\( 3 x + 11 y = 13 \) ...(i)    \( 8 x + 13 y = 2 \) ...(ii)
From equation (i), \( y = \frac{13}{11} - \frac{3 x}{11} \) ...(iii)
Substituting this in equation (ii): \( 13 \left(\frac{13}{11} - \frac{3 x}{11}\right) + 8 x = 2 \Rightarrow x = -3 \).
Putting \( x = -3 \) in (iii): \( y = 2 \).
Hence, \( x = -3,\ y = 2 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 2. Solve by the substitution method: \( x + 2 y = 1.6 \), \( 2 x + y = 1.4 \).
Answer:
Simplifying the given equations, we get
\( 5 x + 10 y = 8 \) ...(i)    \( 10 x + 5 y = 7 \) ...(ii)
From equation (i), \( y = \frac{4}{5} - \frac{x}{2} \) ...(iii)
Substituting this in equation (ii): \( 5 \left(\frac{4}{5} - \frac{x}{2}\right) + 10 x = 7 \Rightarrow x = \frac{2}{5} \).
Putting \( x = \frac{2}{5} \) in (iii): \( y = \frac{3}{5} \).
Hence, \( x = \frac{2}{5},\ y = \frac{3}{5} \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 3. Solve by the substitution method: \( 11 x - 8 y = 27 \), \( 3 x + 5 y = -7 \).
Answer:
\( 11 x - 8 y = 27 \) ...(i)    \( 3 x + 5 y = -7 \) ...(ii)
From equation (i), \( y = \frac{11 x}{8} - \frac{27}{8} \) ...(iii)
Substituting this in equation (ii): \( 5 \left(\frac{11 x}{8} - \frac{27}{8}\right) + 3 x = -7 \Rightarrow x = 1 \).
Putting \( x = 1 \) in (iii): \( y = -2 \).
Hence, \( x = 1,\ y = -2 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 4. Solve by the substitution method: \( 0.04 x + 0.02 y = 5 \), \( 0.5 x - 0.4 y = 30 \).
Answer:
Simplifying the given equations, we get
\( 2 x + y = 250 \) ...(i)    \( 5 x - 4 y = 300 \) ...(ii)
From equation (i), \( y = 250 - 2 x \) ...(iii)
Substituting this in equation (ii): \( - 4 \left(250 - 2 x\right) + 5 x = 300 \Rightarrow x = 100 \).
Putting \( x = 100 \) in (iii): \( y = 50 \).
Hence, \( x = 100,\ y = 50 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 5. Solve by the substitution method: \( 5 x + 8 y = -1 \), \( - x + 6 y = 4 y - 7 \).
Answer:
Simplifying the given equations, we get
\( 5 x + 8 y = -1 \) ...(i)    \( - x + 2 y = -7 \) ...(ii)
From equation (ii), \( x = 2 y + 7 \) ...(iii)
Substituting this in equation (i): \( 5 \left(2 y + 7\right) + 8 y = -1 \Rightarrow y = -2 \).
Putting \( y = -2 \) in (iii): \( x = 3 \).
Hence, \( x = 3,\ y = -2 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 6. Solve by the substitution method: \( 12 x - 16 y = 20 \), \( 8 x + 6 y = 30 \).
Answer:
Simplifying the given equations, we get
\( 3 x - 4 y = 5 \) ...(i)    \( 4 x + 3 y = 15 \) ...(ii)
From equation (i), \( y = \frac{3 x}{4} - \frac{5}{4} \) ...(iii)
Substituting this in equation (ii): \( 3 \left(\frac{3 x}{4} - \frac{5}{4}\right) + 4 x = 15 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (iii): \( y = 1 \).
Hence, \( x = 3,\ y = 1 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 7. Solve by the substitution method: \( 8 x - 5 y + 40 = 0 \), \( 7 x - 2 y = 0 \).
Answer:
Simplifying the given equations, we get
\( 8 x - 5 y = -40 \) ...(i)    \( 7 x - 2 y = 0 \) ...(ii)
From equation (i), \( y = \frac{8 x}{5} + 8 \) ...(iii)
Substituting this in equation (ii): \( - 2 \left(\frac{8 x}{5} + 8\right) + 7 x = 0 \Rightarrow x = \frac{80}{19} \).
Putting \( x = \frac{80}{19} \) in (iii): \( y = \frac{280}{19} \).
Hence, \( x = \frac{80}{19},\ y = \frac{280}{19} \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 8. Solve by the substitution method: \( \frac{9 x + 10 y}{2} = 23 \), \( \frac{5 x}{4} - 2 y = 3 \).
Answer:
Simplifying the given equations, we get
\( 9 x + 10 y = 46 \) ...(i)    \( 5 x - 8 y = 12 \) ...(ii)
From equation (i), \( y = \frac{23}{5} - \frac{9 x}{10} \) ...(iii)
Substituting this in equation (ii): \( - 8 \left(\frac{23}{5} - \frac{9 x}{10}\right) + 5 x = 12 \Rightarrow x = 4 \).
Putting \( x = 4 \) in (iii): \( y = 1 \).
Hence, \( x = 4,\ y = 1 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 9. Solve by the substitution method: \( \frac{3 x - y}{5} = 2 y - 1 \), \( \frac{3 x}{8} - \frac{y}{4} = 1 \cdot \frac{1}{2} \).
Answer:
Simplifying the given equations, we get
\( 3 x - 11 y = -5 \) ...(i)    \( 3 x - 2 y = 4 \) ...(ii)
From equation (i), \( y = \frac{3 x}{11} + \frac{5}{11} \) ...(iii)
Substituting this in equation (ii): \( - 2 \left(\frac{3 x}{11} + \frac{5}{11}\right) + 3 x = 4 \Rightarrow x = 2 \).
Putting \( x = 2 \) in (iii): \( y = 1 \).
Hence, \( x = 2,\ y = 1 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 10. Solve by the substitution method: \( 3 x + 15 = 4 y \), \( 3 y + 17 = 3 x + 2 \).
Answer:
Simplifying the given equations, we get
\( 3 x - 4 y = -15 \) ...(i)    \( - x + y = -5 \) ...(ii)
From equation (ii), \( y = x - 5 \) ...(iii)
Substituting this in equation (i): \( - 4 \left(x - 5\right) + 3 x = -15 \Rightarrow x = 35 \).
Putting \( x = 35 \) in (iii): \( y = 30 \).
Hence, \( x = 35,\ y = 30 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 11. Solve by the substitution method: \( x + 6 y = 2 x - 16 \), \( 3 x - 2 y = 24 \).
Answer:
Simplifying the given equations, we get
\( - x + 6 y = -16 \) ...(i)    \( 3 x - 2 y = 24 \) ...(ii)
From equation (i), \( x = 6 y + 16 \) ...(iii)
Substituting this in equation (ii): \( 3 \left(6 y + 16\right) - 2 y = 24 \Rightarrow y = - \frac{3}{2} \).
Putting \( y = - \frac{3}{2} \) in (iii): \( x = 7 \).
Hence, \( x = 7,\ y = - \frac{3}{2} \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 12. Solve by the substitution method: \( x = 3 y - 19 \), \( y = 3 x - 23 \).
Answer:
Simplifying the given equations, we get
\( x - 3 y = -19 \) ...(i)    \( - 3 x + y = -23 \) ...(ii)
From equation (i), \( x = 3 y - 19 \) ...(iii)
Substituting this in equation (ii): \( - 3 \left(3 y - 19\right) + y = -23 \Rightarrow y = 10 \).
Putting \( y = 10 \) in (iii): \( x = 11 \).
Hence, \( x = 11,\ y = 10 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 13. Solve by the substitution method: \( 5 x + 2 y = 14 \), \( x + 3 y = 8 \).
Answer:
\( 5 x + 2 y = 14 \) ...(i)    \( x + 3 y = 8 \) ...(ii)
From equation (ii), \( x = 8 - 3 y \) ...(iii)
Substituting this in equation (i): \( 5 \left(8 - 3 y\right) + 2 y = 14 \Rightarrow y = 2 \).
Putting \( y = 2 \) in (iii): \( x = 2 \).
Hence, \( x = 2,\ y = 2 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 14. Solve by the substitution method: \( x + y = 27 \), \( \frac{3 x}{4} + \frac{2 y}{3} = 19 \).
Answer:
Simplifying the given equations, we get
\( x + y = 27 \) ...(i)    \( 9 x + 8 y = 228 \) ...(ii)
From equation (i), \( y = 27 - x \) ...(iii)
Substituting this in equation (ii): \( 8 \left(27 - x\right) + 9 x = 228 \Rightarrow x = 12 \).
Putting \( x = 12 \) in (iii): \( y = 15 \).
Hence, \( x = 12,\ y = 15 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 15. Solve by the substitution method: \( 2 y + \frac{x + 11}{7} = 10 \), \( 3 x = \frac{y + 7}{11} + 8 \).
Answer:
Simplifying the given equations, we get
\( x + 14 y = 59 \) ...(i)    \( 33 x - y = 95 \) ...(ii)
From equation (i), \( x = 59 - 14 y \) ...(iii)
Substituting this in equation (ii): \( 33 \left(59 - 14 y\right) - y = 95 \Rightarrow y = 4 \).
Putting \( y = 4 \) in (iii): \( x = 3 \).
Hence, \( x = 3,\ y = 4 \).
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. Pick the equation in which a variable has coefficient 1 to keep the algebra simple.
2. After substituting, you get an equation in one variable only. Check the answer in both original equations.

Question 16. Solve \( 2 x - y = 12 \), \( x + 3 y + 1 = 0 \) and hence find the value of m for which \( y = mx + 3 \).
Answer:
\( 2 x - y = 12 \) ...(i)    \( x + 3 y = -1 \) ...(ii)
From equation (i), \( y = 2 x - 12 \) ...(iii)
Substituting this in equation (ii): \( 3 \left(2 x - 12\right) + x = -1 \Rightarrow x = 5 \).
Putting \( x = 5 \) in (iii): \( y = -2 \).
Hence, \( x = 5,\ y = -2 \).
The line \( y = mx + 3 \) passes through the point (5, -2), so \( -2 = 5m + 3 \Rightarrow m = -1 \).
In simple words: Solve the pair first, then use the solution in the extra equation to find the missing constant.

Teacher's Note:
1. Substitute the solution into y = mx + 3.
2. Check that all three equations hold for the solution.

Question 17. Solve \( 4 x - 3 y + 17 = 0 \), \( 5 x + y + 7 = 0 \) and hence find the value of n for which \( y = nx - 1 \).
Answer:
Simplifying the given equations, we get
\( 4 x - 3 y = -17 \) ...(i)    \( 5 x + y = -7 \) ...(ii)
From equation (ii), \( y = - 5 x - 7 \) ...(iii)
Substituting this in equation (i): \( - 3 \left(- 5 x - 7\right) + 4 x = -17 \Rightarrow x = -2 \).
Putting \( x = -2 \) in (iii): \( y = 3 \).
Hence, \( x = -2,\ y = 3 \).
The line \( y = nx - 1 \) passes through the point (-2, 3), so \( 3 = -2n - 1 \Rightarrow n = -2 \).
In simple words: Solve the pair first, then use the solution in the extra equation to find the missing constant.

Teacher's Note:
1. Substitute the solution into y = nx - 1.
2. Check that all three equations hold for the solution.