CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 22

Question 1. Solve for x and y: \( 6 \left(a x + b y\right) = 3 a + 2 b \), \( 6 \left(- a y + b x\right) = - 2 a + 3 b \).
Answer:
\( 6 a x + 6 b y = 3 a + 2 b \) ...(i)    \( - 6 a y + 6 b x = - 2 a + 3 b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = 6 a,\ b_1 = 6 b,\ c_1 = - 3 a - 2 b \) and \( a_2 = 6 b,\ b_2 = - 6 a,\ c_2 = 2 a - 3 b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 18 \left(a^{2} + b^{2}\right)} = \frac{y}{- 12 \left(a^{2} + b^{2}\right)} = \frac{1}{- 36 \left(a^{2} + b^{2}\right)} \]
Therefore \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \).
Hence, \( x = \frac{1}{2},\ y = \frac{1}{3} \).
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 2. Solve for x and y: \( 3 \left(a y + b x\right) = a - 6 b \), \( 3 \left(a x - b y\right) = - 6 a - b \).
Answer:
\( 3 a y + 3 b x = a - 6 b \) ...(i)    \( 3 a x - 3 b y = - 6 a - b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = 3 b,\ b_1 = 3 a,\ c_1 = - a + 6 b \) and \( a_2 = 3 a,\ b_2 = - 3 b,\ c_2 = 6 a + b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{18 \left(a^{2} + b^{2}\right)} = \frac{y}{- 3 \left(a^{2} + b^{2}\right)} = \frac{1}{- 9 \left(a^{2} + b^{2}\right)} \]
Therefore \( x = -2 \) and \( y = \frac{1}{3} \).
Hence, \( x = -2,\ y = \frac{1}{3} \).
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 3. Solve for x and y: \( 3 y + \frac{4}{x} = 14 \), \( - 4 y + \frac{3}{x} = 23 \).
Answer:
Let \( \frac{1}{x} = u \) and \( y = v \). Then the given equations become linear in u and v.
\( 4 u + 3 v = 14 \) ...(i)    \( 3 u - 4 v = 23 \) ...(ii)
Multiplying (i) by 3: \( 12 u + 9 v = 42 \) ...(iii)
Multiplying (ii) by 4: \( 12 u - 16 v = 92 \) ...(iv)
Subtracting (iv) from (iii), we get \( 25 v = -50 \Rightarrow v = -2 \).
Putting \( v = -2 \) in (i): \( 3 \left(-2\right) + 4 u = 14 \Rightarrow u = 5 \).
Hence \( u = 5 \) and \( v = -2 \).
Therefore \( x = \frac{1}{5} \) and \( y = -2 \).
Hence, \( x = \frac{1}{5},\ y = -2 \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 4. Solve for x and y: \( \frac{a y}{b} + \frac{b x}{a} = a^{2} + b^{2} \), \( x + y = 2 a b \).
Answer:
\( a^{2} y + b^{2} x = - a b \left(- a^{2} - b^{2}\right) \) ...(i)    \( x + y = 2 a b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b^{2},\ b_1 = a^{2},\ c_1 = a b \left(- a^{2} - b^{2}\right) \) and \( a_2 = 1,\ b_2 = 1,\ c_2 = - 2 a b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a b \left(a - b\right) \left(a + b\right)} = \frac{y}{- a b \left(a - b\right) \left(a + b\right)} = \frac{1}{- \left(a - b\right) \left(a + b\right)} \]
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 5. Solve for x and y: \( \frac{3}{y} + \frac{2}{x} = 13 \), \( - \frac{4}{y} + \frac{5}{x} = -2 \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 2 u + 3 v = 13 \) ...(i)    \( 5 u - 4 v = -2 \) ...(ii)
Multiplying (i) by 5: \( 10 u + 15 v = 65 \) ...(iii)
Multiplying (ii) by 2: \( 10 u - 8 v = -4 \) ...(iv)
Subtracting (iv) from (iii), we get \( 23 v = 69 \Rightarrow v = 3 \).
Putting \( v = 3 \) in (i): \( 3 \left(3\right) + 2 u = 13 \Rightarrow u = 2 \).
Hence \( u = 2 \) and \( v = 3 \).
Therefore \( x = \frac{1}{2} \) and \( y = \frac{1}{3} \).
Hence, \( x = \frac{1}{2},\ y = \frac{1}{3} \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 6. Solve for x and y: \( a x - a + b y + b = 0 \), \( - a y - a + b x - b = 0 \).
Answer:
\( a x + b y = a - b \) ...(i)    \( - a y + b x = a + b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a,\ b_1 = b,\ c_1 = - a + b \) and \( a_2 = b,\ b_2 = - a,\ c_2 = - a - b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a^{2} - b^{2}} = \frac{y}{a^{2} + b^{2}} = \frac{1}{- a^{2} - b^{2}} \]
Therefore \( x = 1 \) and \( y = -1 \).
Hence, \( x = 1,\ y = -1 \).
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. 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 7. Solve for x and y: \( a x + b y = a - b \), \( - a y + b x = a + b \).
Answer:
\( a x + b y = a - b \) ...(i)    \( - a y + b x = a + b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a,\ b_1 = b,\ c_1 = - a + b \) and \( a_2 = b,\ b_2 = - a,\ c_2 = - a - b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a^{2} - b^{2}} = \frac{y}{a^{2} + b^{2}} = \frac{1}{- a^{2} - b^{2}} \]
Therefore \( x = 1 \) and \( y = -1 \).
Hence, \( x = 1,\ y = -1 \).
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. 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: \( \frac{y}{b} + \frac{x}{a} = 2 \), \( a x - b y = a^{2} - b^{2} \).
Answer:
\( a y + b x = 2 a b \) ...(i)    \( a x - b y = a^{2} - b^{2} \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b,\ b_1 = a,\ c_1 = - 2 a b \) and \( a_2 = a,\ b_2 = - b,\ c_2 = - a^{2} + b^{2} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a \left(a^{2} + b^{2}\right)} = \frac{y}{- b \left(a^{2} + b^{2}\right)} = \frac{1}{- a^{2} - b^{2}} \]
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 9. Solve for x and y: \( - \frac{y}{b} + \frac{x}{a} = a - b \), \( a x + b y = a^{3} + b^{3} \).
Answer:
\( - a y + b x = - a b \left(- a + b\right) \) ...(i)    \( a x + b y = a^{3} + b^{3} \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b,\ b_1 = - a,\ c_1 = a b \left(- a + b\right) \) and \( a_2 = a,\ b_2 = b,\ c_2 = - a^{3} - b^{3} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{a^{2} \left(a^{2} + b^{2}\right)} = \frac{y}{b^{2} \left(a^{2} + b^{2}\right)} = \frac{1}{a^{2} + b^{2}} \]
Therefore \( x = a^{2} \) and \( y = b^{2} \).
Hence, \( x = a^{2},\ y = b^{2} \).
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 10. Solve for x and y: \( \frac{2 a}{x} + \frac{3 b}{y} + 1 = 0 \), \( \frac{3 a}{x} - \frac{b}{y} - 4 = 0 \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 2 a u + 3 b v = -1 \) ...(i)    \( 3 a u - b v = 4 \) ...(ii)
In the form \( a_1u + b_1v + c_1 = 0 \): \( a_1 = 2 a,\ b_1 = 3 b,\ c_1 = 1 \) and \( a_2 = 3 a,\ b_2 = - b,\ c_2 = -4 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 11 b} = \frac{y}{11 a} = \frac{1}{- 11 a b} \]
Therefore \( u = \frac{1}{a} \) and \( v = - \frac{1}{b} \).
Hence \( u = \frac{1}{a} \) and \( v = - \frac{1}{b} \).
Therefore \( x = a \) and \( y = - b \).
Hence, \( x = a,\ y = - b \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 11. Solve for x and y: \( \frac{a x}{b} - \frac{b y}{a} = a + b \), \( a x - b y = 2 a b \).
Answer:
\( a^{2} x - b^{2} y = - a b \left(- a - b\right) \) ...(i)    \( a x - b y = 2 a b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a^{2},\ b_1 = - b^{2},\ c_1 = a b \left(- a - b\right) \) and \( a_2 = a,\ b_2 = - b,\ c_2 = - 2 a b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a b^{2} \left(a - b\right)} = \frac{y}{a^{2} b \left(a - b\right)} = \frac{1}{- a b \left(a - b\right)} \]
Therefore \( x = b \) and \( y = - a \).
Hence, \( x = b,\ y = - a \).
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: \( - \frac{a^{2} y}{b} + \frac{b^{2} x}{a} = a b \left(a + b\right) \), \( - a^{2} y + b^{2} x = 2 a^{2} b^{2} \).
Answer:
\( - a^{3} y + b^{3} x = - a^{2} b^{2} \left(- a - b\right) \) ...(i)    \( - a^{2} y + b^{2} x = 2 a^{2} b^{2} \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b^{3},\ b_1 = - a^{3},\ c_1 = a^{2} b^{2} \left(- a - b\right) \) and \( a_2 = b^{2},\ b_2 = - a^{2},\ c_2 = - 2 a^{2} b^{2} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{a^{4} b^{2} \left(a - b\right)} = \frac{y}{- a^{2} b^{4} \left(a - b\right)} = \frac{1}{a^{2} b^{2} \left(a - b\right)} \]
Therefore \( x = a^{2} \) and \( y = - b^{2} \).
Hence, \( x = a^{2},\ y = - b^{2} \).
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 13. Solve for x and y: \( m x - n y = m^{2} + n^{2} \), \( x + y = 2 m \).
Answer:
\( m x - n y = m^{2} + n^{2} \) ...(i)    \( x + y = 2 m \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = m,\ b_1 = - n,\ c_1 = - m^{2} - n^{2} \) and \( a_2 = 1,\ b_2 = 1,\ c_2 = - 2 m \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{\left(m + n\right)^{2}} = \frac{y}{\left(m - n\right) \left(m + n\right)} = \frac{1}{m + n} \]
Therefore \( x = m + n \) and \( y = m - n \).
Hence, \( x = m + n,\ y = m - n \).
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: \( \frac{a^{2}}{x} - \frac{b^{2}}{y} = 0 \), \( \frac{a^{2} b}{x} + \frac{a b^{2}}{y} = a + b \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( a^{2} u - b^{2} v = 0 \) ...(i)    \( a^{2} b u + a b^{2} v = a + b \) ...(ii)
In the form \( a_1u + b_1v + c_1 = 0 \): \( a_1 = a^{2},\ b_1 = - b^{2},\ c_1 = 0 \) and \( a_2 = a^{2} b,\ b_2 = a b^{2},\ c_2 = - a - b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{b^{2} \left(a + b\right)} = \frac{y}{a^{2} \left(a + b\right)} = \frac{1}{a^{2} b^{2} \left(a + b\right)} \]
Therefore \( u = \frac{1}{a^{2}} \) and \( v = \frac{1}{b^{2}} \).
Hence \( u = \frac{1}{a^{2}} \) and \( v = \frac{1}{b^{2}} \).
Therefore \( x = a^{2} \) and \( y = b^{2} \).
Hence, \( x = a^{2},\ y = b^{2} \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 15. Solve for x and y: \( x + \frac{6}{y} = 6 \), \( 3 x - \frac{8}{y} = 5 \).
Answer:
Let \( \frac{1}{y} = u \) and \( x = v \). Then the given equations become linear in u and v.
\( 6 u + v = 6 \) ...(i)    \( - 8 u + 3 v = 5 \) ...(ii)
Multiplying (i) by 3: \( 18 u + 3 v = 18 \) ...(iii)
Subtracting (ii) from (iii), we get \( 26 u = 13 \Rightarrow u = \frac{1}{2} \).
Putting \( u = \frac{1}{2} \) in (i): \( 6 \left(\frac{1}{2}\right) + v = 6 \Rightarrow v = 3 \).
Hence \( u = \frac{1}{2} \) and \( v = 3 \).
Therefore \( y = 2 \) and \( x = 3 \).
Solving these two equations for x and y, we get \( x = 3,\ y = 2 \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 16. Solve for x and y: \( \frac{x + 1}{2} + \frac{y - 1}{3} = 8 \), \( \frac{x - 1}{3} + \frac{y + 1}{2} = 9 \).
Answer:
Simplifying the given equations, we get
\( 3 x + 2 y = 47 \) ...(i)    \( 2 x + 3 y = 53 \) ...(ii)
Multiplying (i) by 2: \( 6 x + 4 y = 94 \) ...(iii)
Multiplying (ii) by 3: \( 6 x + 9 y = 159 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 5 y = -65 \Rightarrow y = 13 \).
Putting \( y = 13 \) in (i): \( 2 \left(13\right) + 3 x = 47 \Rightarrow x = 7 \).
Hence, \( x = 7,\ y = 13 \).
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 for x and y: \( 8 x - 9 y = 6 x y \), \( 10 x + 6 y = 19 x y \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( - 9 u + 8 v = 6 \) ...(i)    \( 6 u + 10 v = 19 \) ...(ii)
Multiplying (i) by 2: \( - 18 u + 16 v = 12 \) ...(iii)
Multiplying (ii) by 3: \( 18 u + 30 v = 57 \) ...(iv)
Adding (iii) and (iv), we get \( 46 v = 69 \Rightarrow v = \frac{3}{2} \).
Putting \( v = \frac{3}{2} \) in (i): \( 8 \left(\frac{3}{2}\right) - 9 u = 6 \Rightarrow u = \frac{2}{3} \).
Hence \( u = \frac{2}{3} \) and \( v = \frac{3}{2} \).
Therefore \( x = \frac{3}{2} \) and \( y = \frac{2}{3} \).
Hence, \( x = \frac{3}{2},\ y = \frac{2}{3} \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 18. Solve for x and y: \( 4 x + \frac{y}{3} = \frac{8}{3} \), \( \frac{x}{2} + \frac{3 y}{4} = - \frac{5}{2} \).
Answer:
Simplifying the given equations, we get
\( 12 x + y = 8 \) ...(i)    \( 2 x + 3 y = -10 \) ...(ii)
Multiplying (i) by 3: \( 36 x + 3 y = 24 \) ...(iii)
Subtracting (ii) from (iii), we get \( 34 x = 34 \Rightarrow x = 1 \).
Putting \( x = 1 \) in (i): \( 12 \left(1\right) + y = 8 \Rightarrow y = -4 \).
Hence, \( x = 1,\ y = -4 \).
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 19. Solve for x and y: \( 5 y + \frac{4}{x} = 7 \), \( 4 y + \frac{3}{x} = 5 \).
Answer:
Let \( \frac{1}{x} = u \) and \( y = v \). Then the given equations become linear in u and v.
\( 4 u + 5 v = 7 \) ...(i)    \( 3 u + 4 v = 5 \) ...(ii)
Multiplying (i) by 3: \( 12 u + 15 v = 21 \) ...(iii)
Multiplying (ii) by 4: \( 12 u + 16 v = 20 \) ...(iv)
Subtracting (iv) from (iii), we get \( - v = 1 \Rightarrow v = -1 \).
Putting \( v = -1 \) in (i): \( 5 \left(-1\right) + 4 u = 7 \Rightarrow u = 3 \).
Hence \( u = 3 \) and \( v = -1 \).
Therefore \( x = \frac{1}{3} \) and \( y = -1 \).
Hence, \( x = \frac{1}{3},\ y = -1 \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 20. Solve for x and y: \( \frac{a}{x} - \frac{b}{y} = 0 \), \( \frac{a^{2} b}{y} + \frac{a b^{2}}{x} = a^{2} + b^{2} \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( a u - b v = 0 \) ...(i)    \( a^{2} b v + a b^{2} u = a^{2} + b^{2} \) ...(ii)
In the form \( a_1u + b_1v + c_1 = 0 \): \( a_1 = a,\ b_1 = - b,\ c_1 = 0 \) and \( a_2 = a b^{2},\ b_2 = a^{2} b,\ c_2 = - a^{2} - b^{2} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{b \left(a^{2} + b^{2}\right)} = \frac{y}{a \left(a^{2} + b^{2}\right)} = \frac{1}{a b \left(a^{2} + b^{2}\right)} \]
Therefore \( u = \frac{1}{a} \) and \( v = \frac{1}{b} \).
Hence \( u = \frac{1}{a} \) and \( v = \frac{1}{b} \).
Therefore \( x = a \) and \( y = b \).
Hence, \( x = a,\ y = b \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 21. Solve for x and y: \( x \left(a - b\right) + y \left(a + b\right) = a^{2} - 2 a b - b^{2} \), \( \left(a + b\right) \left(x + y\right) = a^{2} + b^{2} \).
Answer:
\( x \left(a - b\right) + y \left(a + b\right) = a^{2} - 2 a b - b^{2} \) ...(i)    \( x \left(a + b\right) + y \left(a + b\right) = a^{2} + b^{2} \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a - b,\ b_1 = a + b,\ c_1 = - a^{2} + 2 a b + b^{2} \) and \( a_2 = a + b,\ b_2 = a + b,\ c_2 = - a^{2} - b^{2} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 2 b \left(a + b\right)^{2}} = \frac{y}{4 a b^{2}} = \frac{1}{- 2 b \left(a + b\right)} \]
Therefore \( x = a + b \) and \( y = - \frac{2 a b}{a + b} \).
Hence, \( x = a + b,\ y = - \frac{2 a b}{a + b} \).
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 22. Solve for x and y: \( 37 x + 43 y = 123 \), \( 43 x + 37 y = 117 \).
Answer:
\( 37 x + 43 y = 123 \) ...(i)    \( 43 x + 37 y = 117 \) ...(ii)
Adding (i) and (ii), we get \( 80 x + 80 y = 240 \), i.e., \( x + y = 3 \) ...(iii)
Subtracting (ii) from (i), we get \( - 6 x + 6 y = 6 \), i.e., \( x - y = -1 \) ...(iv)
Solving (iii) and (iv): \( x = 1 \) and \( y = 2 \).
Hence, \( x = 1,\ y = 2 \).
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. 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 23. Solve for x and y: \( 1 \frac{1}{y - 2} + \frac{5}{x - 1} = 2 \), \( - \frac{3}{y - 2} + \frac{6}{x - 1} = 1 \).
Answer:
Let \( \frac{1}{x - 1} = u \) and \( \frac{1}{y - 2} = v \). Then the given equations become linear in u and v.
\( 5 u + v = 2 \) ...(i)    \( 6 u - 3 v = 1 \) ...(ii)
Multiplying (i) by 3: \( 15 u + 3 v = 6 \) ...(iii)
Adding (iii) and (ii), we get \( 21 u = 7 \Rightarrow u = \frac{1}{3} \).
Putting \( u = \frac{1}{3} \) in (i): \( 5 \left(\frac{1}{3}\right) + v = 2 \Rightarrow v = \frac{1}{3} \).
Hence \( u = \frac{1}{3} \) and \( v = \frac{1}{3} \).
Therefore \( x - 1 = 3 \) and \( y - 2 = 3 \).
Solving these two equations for x and y, we get \( x = 4,\ y = 5 \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.

Question 24. Solve for x and y: \( \frac{10}{x + y} + \frac{2}{x - y} = 4 \), \( \frac{15}{x + y} - \frac{5}{x - y} = -2 \).
Answer:
Let \( \frac{1}{x + y} = u \) and \( \frac{1}{x - y} = v \). Then the given equations become linear in u and v.
\( 5 u + v = 2 \) ...(i)    \( 15 u - 5 v = -2 \) ...(ii)
Multiplying (i) by 5: \( 25 u + 5 v = 10 \) ...(iii)
Adding (iii) and (ii), we get \( 40 u = 8 \Rightarrow u = \frac{1}{5} \).
Putting \( u = \frac{1}{5} \) in (i): \( 5 \left(\frac{1}{5}\right) + v = 2 \Rightarrow v = 1 \).
Hence \( u = \frac{1}{5} \) and \( v = 1 \).
Therefore \( x + y = 5 \) and \( x - y = 1 \).
Solving these two equations for x and y, we get \( x = 3,\ y = 2 \).
In simple words: Replace each reciprocal by a new letter to get an ordinary pair of linear equations, solve it, and convert back.

Teacher's Note:
1. Substitute u and v for the reciprocals only when the equations contain x and y in the denominators.
2. Do not forget the last step of converting u and v back to x and y.