CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 21

Question 1. Solve by the cross-multiplication method: \( x - 2 y = 10 \), \( 4 x + y = 13 \).
Answer:
\( x - 2 y = 10 \) ...(i)    \( 4 x + y = 13 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( x - 2 y - 10 = 0 \) and \( 4 x + y - 13 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(-2)(-13) - (1)(-10)} = \frac{y}{(-10)(4) - (-13)(1)} = \frac{1}{(1)(1) - (4)(-2)} \]
\[ \frac{x}{36} = \frac{y}{-27} = \frac{1}{9} \]
\( x = \frac{36}{9} = 4 \) and \( y = \frac{-27}{9} = -3 \).
Hence, \( x = 4,\ y = -3 \).
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 2. Solve by the cross-multiplication method: \( 5 x + 3 y = 35 \), \( 2 x + 4 y = 28 \).
Answer:
Simplifying the given equations, we get
\( 5 x + 3 y = 35 \) ...(i)    \( x + 2 y = 14 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( 5 x + 3 y - 35 = 0 \) and \( x + 2 y - 14 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(3)(-14) - (2)(-35)} = \frac{y}{(-35)(1) - (-14)(5)} = \frac{1}{(5)(2) - (1)(3)} \]
\[ \frac{x}{28} = \frac{y}{35} = \frac{1}{7} \]
\( x = \frac{28}{7} = 4 \) and \( y = \frac{35}{7} = 5 \).
Hence, \( x = 4,\ y = 5 \).
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. 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 by the cross-multiplication method: \( 4 x - 3 y + 1 = 0 \), \( 2 x - 5 y + 11 = 0 \).
Answer:
Simplifying the given equations, we get
\( 4 x - 3 y = -1 \) ...(i)    \( 2 x - 5 y = -11 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( 4 x - 3 y + 1 = 0 \) and \( 2 x - 5 y + 11 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(-3)(11) - (-5)(1)} = \frac{y}{(1)(2) - (11)(4)} = \frac{1}{(4)(-5) - (2)(-3)} \]
\[ \frac{x}{-28} = \frac{y}{-42} = \frac{1}{-14} \]
\( x = \frac{-28}{-14} = 2 \) and \( y = \frac{-42}{-14} = 3 \).
Hence, \( x = 2,\ y = 3 \).
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. 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 4. Solve by the cross-multiplication method: \( 3 x + 4 y = 27 \), \( 5 x - 3 y = 16 \).
Answer:
\( 3 x + 4 y = 27 \) ...(i)    \( 5 x - 3 y = 16 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( 3 x + 4 y - 27 = 0 \) and \( 5 x - 3 y - 16 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(4)(-16) - (-3)(-27)} = \frac{y}{(-27)(5) - (-16)(3)} = \frac{1}{(3)(-3) - (5)(4)} \]
\[ \frac{x}{-145} = \frac{y}{-87} = \frac{1}{-29} \]
\( x = \frac{-145}{-29} = 5 \) and \( y = \frac{-87}{-29} = 3 \).
Hence, \( x = 5,\ y = 3 \).
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 5. Solve by the cross-multiplication method: \( 2 x + 3 y = 46 \), \( 3 x + 5 y = 74 \).
Answer:
\( 2 x + 3 y = 46 \) ...(i)    \( 3 x + 5 y = 74 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( 2 x + 3 y - 46 = 0 \) and \( 3 x + 5 y - 74 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(3)(-74) - (5)(-46)} = \frac{y}{(-46)(3) - (-74)(2)} = \frac{1}{(2)(5) - (3)(3)} \]
\[ \frac{x}{8} = \frac{y}{10} = \frac{1}{1} \]
\( x = \frac{8}{1} = 8 \) and \( y = \frac{10}{1} = 10 \).
Hence, \( x = 8,\ y = 10 \).
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. 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 6. Solve by the cross-multiplication method: \( 2 x - y + 4 = 0 \), \( x + y - 1 = 0 \).
Answer:
Simplifying the given equations, we get
\( 2 x - y = -4 \) ...(i)    \( x + y = 1 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( 2 x - y + 4 = 0 \) and \( x + y - 1 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(-1)(-1) - (1)(4)} = \frac{y}{(4)(1) - (-1)(2)} = \frac{1}{(2)(1) - (1)(-1)} \]
\[ \frac{x}{-3} = \frac{y}{6} = \frac{1}{3} \]
\( x = \frac{-3}{3} = -1 \) and \( y = \frac{6}{3} = 2 \).
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. 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 by the cross-multiplication method: \( \frac{x}{2} - \frac{y}{3} + 4 = 0 \), \( \frac{x}{2} - \frac{5 y}{3} + 12 = 0 \).
Answer:
Simplifying the given equations, we get
\( 3 x - 2 y = -24 \) ...(i)    \( 3 x - 10 y = -72 \) ...(ii)
Writing in the form \( ax + by + c = 0 \): \( 3 x - 2 y + 24 = 0 \) and \( 3 x - 10 y + 72 = 0 \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{(-2)(72) - (-10)(24)} = \frac{y}{(24)(3) - (72)(3)} = \frac{1}{(3)(-10) - (3)(-2)} \]
\[ \frac{x}{96} = \frac{y}{-144} = \frac{1}{-24} \]
\( x = \frac{96}{-24} = -4 \) and \( y = \frac{-144}{-24} = 6 \).
Hence, \( x = -4,\ y = 6 \).
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 by the cross-multiplication method: \( a x + a + b y = 0 \), \( a y + b x + b = 0 \), where \( a \neq b \).
Answer:
\( a x + b y = - a \) ...(i)    \( a y + b x = - b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = a,\ b_1 = b,\ c_1 = a \) and \( a_2 = b,\ b_2 = a,\ c_2 = b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- \left(a - b\right) \left(a + b\right)} = \frac{y}{0} = \frac{1}{\left(a - b\right) \left(a + b\right)} \]
Therefore \( x = -1 \) and \( y = 0 \).
Hence, \( x = -1,\ y = 0 \).
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 9. Solve by the cross-multiplication method: \( \frac{y}{b} + \frac{x}{a} = 0 \), \( - \frac{y}{b} + \frac{x}{a} = 4 \), where \( a \neq b \).
Answer:
\( a y + b x = 0 \) ...(i)    \( - a y + b x = 4 a b \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = b,\ b_1 = a,\ c_1 = 0 \) and \( a_2 = b,\ b_2 = - a,\ c_2 = - 4 a b \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- 4 a^{2} b} = \frac{y}{4 a b^{2}} = \frac{1}{- 2 a b} \]
Therefore \( x = 2 a \) and \( y = - 2 b \).
Hence, \( x = 2 a,\ y = - 2 b \).
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 10. Solve by the cross-multiplication method: \( x + y = a + b \), \( a x - b y = a^{2} - b^{2} \), where \( a \neq b \).
Answer:
\( x + y = a + b \) ...(i)    \( a x - b y = a^{2} - b^{2} \) ...(ii)
In the form \( a_1x + b_1y + c_1 = 0 \): \( a_1 = 1,\ b_1 = 1,\ c_1 = - a - b \) and \( a_2 = a,\ b_2 = - b,\ c_2 = - a^{2} + b^{2} \).
\[ \frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1} \]
\[ \frac{x}{- a \left(a + b\right)} = \frac{y}{- b \left(a + b\right)} = \frac{1}{- a - b} \]
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 11. Solve for x and y: \( \frac{3}{y} + \frac{2}{x} = 2 \), \( - \frac{1}{2 y} + 1 \frac{1}{x} = 1 \cdot \frac{1}{3} \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 2 u + 3 v = 2 \) ...(i)    \( 6 u - 3 v = 2 \) ...(ii)
Adding (i) and (ii), we get \( 8 u = 4 \Rightarrow u = \frac{1}{2} \).
Putting \( u = \frac{1}{2} \) in (i): \( 2 \left(\frac{1}{2}\right) + 3 v = 2 \Rightarrow v = \frac{1}{3} \).
Hence \( u = \frac{1}{2} \) and \( v = \frac{1}{3} \).
Therefore \( x = 2 \) and \( y = 3 \).
Hence, \( x = 2,\ y = 3 \).
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 12. Solve for x and y: \( \frac{7}{y} + \frac{4}{x} = 29 \), \( \frac{3}{y} + 1 \frac{1}{x} = 11 \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 4 u + 7 v = 29 \) ...(i)    \( u + 3 v = 11 \) ...(ii)
Multiplying (ii) by 4: \( 4 u + 12 v = 44 \) ...(iv)
Subtracting (iv) from (i), we get \( - 5 v = -15 \Rightarrow v = 3 \).
Putting \( v = 3 \) in (i): \( 7 \left(3\right) + 4 u = 29 \Rightarrow u = 2 \).
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 13. Solve for x and y: \( - \frac{1}{7 y} + 1 \frac{1}{3 x} = \frac{2}{3} \), \( - \frac{1}{3 y} + 1 \frac{1}{2 x} = 1 \cdot \frac{1}{6} \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 7 u - 3 v = 14 \) ...(i)    \( 3 u - 2 v = 1 \) ...(ii)
Multiplying (i) by 2: \( 14 u - 6 v = 28 \) ...(iii)
Multiplying (ii) by 3: \( 9 u - 6 v = 3 \) ...(iv)
Subtracting (iv) from (iii), we get \( 5 u = 25 \Rightarrow u = 5 \).
Putting \( u = 5 \) in (i): \( 7 \left(5\right) - 3 v = 14 \Rightarrow v = 7 \).
Hence \( u = 5 \) and \( v = 7 \).
Therefore \( x = \frac{1}{5} \) and \( y = \frac{1}{7} \).
Hence, \( x = \frac{1}{5},\ y = \frac{1}{7} \).
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 14. Solve for x and y: \( \frac{9}{y} + 1 \frac{1}{5 x} = 4 \), \( \frac{27}{y} + \frac{3}{x} = 24 \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( u + 45 v = 20 \) ...(i)    \( u + 9 v = 8 \) ...(ii)
Subtracting (ii) from (i), we get \( 36 v = 12 \Rightarrow v = \frac{1}{3} \).
Putting \( v = \frac{1}{3} \) in (i): \( 45 \left(\frac{1}{3}\right) + u = 20 \Rightarrow u = 5 \).
Hence \( u = 5 \) and \( v = \frac{1}{3} \).
Therefore \( x = \frac{1}{5} \) and \( y = 3 \).
Hence, \( x = \frac{1}{5},\ y = 3 \).
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 15. Solve for x and y: \( - \frac{9}{2 y} + \frac{11}{2 x} = - \frac{23}{2} \), \( \frac{7}{15 y} + \frac{3}{4 x} = \frac{23}{6} \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 11 u - 9 v = -23 \) ...(i)    \( 45 u + 28 v = 230 \) ...(ii)
Multiplying (i) by 28: \( 308 u - 252 v = -644 \) ...(iii)
Multiplying (ii) by 9: \( 405 u + 252 v = 2070 \) ...(iv)
Adding (iii) and (iv), we get \( 713 u = 1426 \Rightarrow u = 2 \).
Putting \( u = 2 \) in (i): \( 11 \left(2\right) - 9 v = -23 \Rightarrow v = 5 \).
Hence \( u = 2 \) and \( v = 5 \).
Therefore \( x = \frac{1}{2} \) and \( y = \frac{1}{5} \).
Hence, \( x = \frac{1}{2},\ y = \frac{1}{5} \).
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 16. Solve for x and y: \( x + y = 2 x y \), \( x - y = x y \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( u + v = 2 \) ...(i)    \( - u + v = 1 \) ...(ii)
Adding (i) and (ii), we get \( 2 v = 3 \Rightarrow v = \frac{3}{2} \).
Putting \( v = \frac{3}{2} \) in (i): \( \left(\frac{3}{2}\right) + u = 2 \Rightarrow u = \frac{1}{2} \).
Hence \( u = \frac{1}{2} \) and \( v = \frac{3}{2} \).
Therefore \( x = 2 \) and \( y = \frac{2}{3} \).
Hence, \( x = 2,\ y = \frac{2}{3} \).
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 17. Solve for x and y: \( 4 x + 3 y = 8 x y \), \( 6 x + 5 y = 13 x y \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 3 u + 4 v = 8 \) ...(i)    \( 5 u + 6 v = 13 \) ...(ii)
Multiplying (i) by 3: \( 9 u + 12 v = 24 \) ...(iii)
Multiplying (ii) by 2: \( 10 u + 12 v = 26 \) ...(iv)
Subtracting (iv) from (iii), we get \( - u = -2 \Rightarrow u = 2 \).
Putting \( u = 2 \) in (i): \( 3 \left(2\right) + 4 v = 8 \Rightarrow v = \frac{1}{2} \).
Hence \( u = 2 \) and \( v = \frac{1}{2} \).
Therefore \( x = \frac{1}{2} \) and \( y = 2 \).
Hence, \( x = \frac{1}{2},\ y = 2 \).
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 18. Solve for x and y: \( 6 x + 5 y = 8 x y \), \( 8 x + 3 y = 7 x y \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 5 u + 6 v = 8 \) ...(i)    \( 3 u + 8 v = 7 \) ...(ii)
Multiplying (i) by 3: \( 15 u + 18 v = 24 \) ...(iii)
Multiplying (ii) by 5: \( 15 u + 40 v = 35 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 22 v = -11 \Rightarrow v = \frac{1}{2} \).
Putting \( v = \frac{1}{2} \) in (i): \( 6 \left(\frac{1}{2}\right) + 5 u = 8 \Rightarrow u = 1 \).
Hence \( u = 1 \) and \( v = \frac{1}{2} \).
Therefore \( x = 1 \) and \( y = 2 \).
Hence, \( x = 1,\ y = 2 \).
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 19. Solve for x and y: \( \frac{x - y}{x y} = 9 \), \( \frac{x + y}{x y} = 5 \).
Answer:
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( - u + v = 9 \) ...(i)    \( u + v = 5 \) ...(ii)
Adding (i) and (ii), we get \( 2 v = 14 \Rightarrow v = 7 \).
Putting \( v = 7 \) in (i): \( \left(7\right) - u = 9 \Rightarrow u = -2 \).
Hence \( u = -2 \) and \( v = 7 \).
Therefore \( x = - \frac{1}{2} \) and \( y = \frac{1}{7} \).
Hence, \( x = - \frac{1}{2},\ y = \frac{1}{7} \).
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: \( 25 x y + 9 = 53 x \), \( 27 - 4 x y = x \).
Answer:
Let \( \frac{1}{x} = u \) and \( y = v \). Then the given equations become linear in u and v.
\( 9 u + 25 v = 53 \) ...(i)    \( 27 u - 4 v = 1 \) ...(ii)
Multiplying (i) by 3: \( 27 u + 75 v = 159 \) ...(iii)
Subtracting (ii) from (iii), we get \( 79 v = 158 \Rightarrow v = 2 \).
Putting \( v = 2 \) in (i): \( 25 \left(2\right) + 9 u = 53 \Rightarrow u = \frac{1}{3} \).
Hence \( u = \frac{1}{3} \) and \( v = 2 \).
Therefore \( x = 3 \) and \( y = 2 \).
Hence, \( x = 3,\ y = 2 \).
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 21. Solve for x and y: \( \frac{3}{y - 2} + \frac{16}{x + 3} = 5 \), \( - \frac{1}{y - 2} + \frac{8}{x + 3} = 0 \).
Answer:
Let \( \frac{1}{x + 3} = u \) and \( \frac{1}{y - 2} = v \). Then the given equations become linear in u and v.
\( 16 u + 3 v = 5 \) ...(i)    \( 8 u - v = 0 \) ...(ii)
Multiplying (ii) by 3: \( 24 u - 3 v = 0 \) ...(iv)
Adding (i) and (iv), we get \( 40 u = 5 \Rightarrow u = \frac{1}{8} \).
Putting \( u = \frac{1}{8} \) in (i): \( 16 \left(\frac{1}{8}\right) + 3 v = 5 \Rightarrow v = 1 \).
Hence \( u = \frac{1}{8} \) and \( v = 1 \).
Therefore \( x + 3 = 8 \) and \( y - 2 = 1 \).
Solving these two equations for x and y, we get \( x = 5,\ y = 3 \).
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 22. Solve for x and y: \( \frac{13}{x + y} - \frac{56}{x - y} = 21 \), \( \frac{11}{x + y} - \frac{23}{x - y} = \frac{100}{7} \).
Answer:
Let \( \frac{1}{x + y} = u \) and \( \frac{1}{x - y} = v \). Then the given equations become linear in u and v.
\( 13 u - 56 v = 21 \) ...(i)    \( 77 u - 161 v = 100 \) ...(ii)
Multiplying (i) by 77: \( 1001 u - 4312 v = 1617 \) ...(iii)
Multiplying (ii) by 13: \( 1001 u - 2093 v = 1300 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 2219 v = 317 \Rightarrow v = - \frac{1}{7} \).
Putting \( v = - \frac{1}{7} \) in (i): \( - 56 \left(- \frac{1}{7}\right) + 13 u = 21 \Rightarrow u = 1 \).
Hence \( u = 1 \) and \( v = - \frac{1}{7} \).
Therefore \( x + y = 1 \) and \( x - y = -7 \).
Solving these two equations for x and y, we get \( x = -3,\ y = 4 \).
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 23. Solve for x and y: \( - \frac{13}{3 x + 2 y} + \frac{24}{2 x + y} = 2 \), \( \frac{26}{3 x + 2 y} + \frac{8}{2 x + y} = 3 \).
Answer:
Let \( \frac{1}{2 x + y} = u \) and \( \frac{1}{3 x + 2 y} = v \). Then the given equations become linear in u and v.
\( 24 u - 13 v = 2 \) ...(i)    \( 8 u + 26 v = 3 \) ...(ii)
Multiplying (ii) by 3: \( 24 u + 78 v = 9 \) ...(iv)
Subtracting (iv) from (i), we get \( - 91 v = -7 \Rightarrow v = \frac{1}{13} \).
Putting \( v = \frac{1}{13} \) in (i): \( - 13 \left(\frac{1}{13}\right) + 24 u = 2 \Rightarrow u = \frac{1}{8} \).
Hence \( u = \frac{1}{8} \) and \( v = \frac{1}{13} \).
Therefore \( 2 x + y = 8 \) and \( 3 x + 2 y = 13 \).
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 24. Solve for x and y: \( - \frac{81}{y + 1} + \frac{29}{x - 1} = -26 \), \( \frac{19}{y + 1} - \frac{4}{x - 1} = \frac{15}{2} \).
Answer:
Let \( \frac{1}{x - 1} = u \) and \( \frac{1}{y + 1} = v \). Then the given equations become linear in u and v.
\( 29 u - 81 v = -26 \) ...(i)    \( - 8 u + 38 v = 15 \) ...(ii)
Multiplying (i) by 8: \( 232 u - 648 v = -208 \) ...(iii)
Multiplying (ii) by 29: \( - 232 u + 1102 v = 435 \) ...(iv)
Adding (iii) and (iv), we get \( 454 v = 227 \Rightarrow v = \frac{1}{2} \).
Putting \( v = \frac{1}{2} \) in (i): \( - 81 \left(\frac{1}{2}\right) + 29 u = -26 \Rightarrow u = \frac{1}{2} \).
Hence \( u = \frac{1}{2} \) and \( v = \frac{1}{2} \).
Therefore \( x - 1 = 2 \) and \( y + 1 = 2 \).
Solving these two equations for x and y, we get \( x = 3,\ y = 1 \).
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 25. Solve for x and y: \( \frac{y - 2}{4} + \frac{2}{x - 1} = 2 \), \( \frac{2 \left(y - 2\right)}{5} + \frac{3}{2 \left(x - 1\right)} = \frac{47}{20} \).
Answer:
Let \( \frac{1}{x - 1} = u \) and \( y - 2 = v \). Then the given equations become linear in u and v.
\( 8 u + v = 8 \) ...(i)    \( 30 u + 8 v = 47 \) ...(ii)
Multiplying (i) by 8: \( 64 u + 8 v = 64 \) ...(iii)
Subtracting (ii) from (iii), we get \( 34 u = 17 \Rightarrow u = \frac{1}{2} \).
Putting \( u = \frac{1}{2} \) in (i): \( 8 \left(\frac{1}{2}\right) + v = 8 \Rightarrow v = 4 \).
Hence \( u = \frac{1}{2} \) and \( v = 4 \).
Therefore \( x - 1 = 2 \) and \( y - 2 = 4 \).
Solving these two equations for x and y, we get \( x = 3,\ y = 6 \).
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 26. Solve for x and y: \( \frac{x - 4}{x - 3} = \frac{y + 4}{y + 7} \), \( \frac{x + 5}{x + 2} = \frac{y - 1}{y - 2} \).
Answer:
Cross-multiplying each equation and simplifying (the terms with xy cancel), we get
\( 3 x - y = 16 \) ...(i)    \( - x + 3 y = 8 \) ...(ii)
Multiplying (ii) by 3: \( - 3 x + 9 y = 24 \) ...(iv)
Adding (i) and (iv), we get \( 8 y = 40 \Rightarrow y = 5 \).
Putting \( y = 5 \) in (i): \( - \left(5\right) + 3 x = 16 \Rightarrow x = 7 \).
Hence, \( x = 7,\ y = 5 \).
In simple words: Remove the fractions by cross-multiplying, simplify to two linear equations, and solve.

Teacher's Note:
1. Cross-multiplication works because a/b = c/d means ad = bc.
2. Check that your answer does not make any denominator zero.

Question 27. Solve for x and y: \( \frac{3 x - 2}{3 y + 7} = \frac{5 x - 1}{5 y + 16} \), \( \frac{3 x - 15}{x - 9} = \frac{6 y - 5}{2 y + 3} \).
Answer:
Cross-multiplying each equation and simplifying (the terms with xy cancel), we get
\( 13 x - 7 y = 25 \) ...(i)    \( 7 x + 12 y = 45 \) ...(ii)
Multiplying (i) by 12: \( 156 x - 84 y = 300 \) ...(iii)
Multiplying (ii) by 7: \( 49 x + 84 y = 315 \) ...(iv)
Adding (iii) and (iv), we get \( 205 x = 615 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (i): \( 13 \left(3\right) - 7 y = 25 \Rightarrow y = 2 \).
Hence, \( x = 3,\ y = 2 \).
In simple words: Remove the fractions by cross-multiplying, simplify to two linear equations, and solve.

Teacher's Note:
1. Cross-multiplication works because a/b = c/d means ad = bc.
2. Check that your answer does not make any denominator zero.

Question 28. Solve for x and y: \( \frac{x + y - 1}{x - y + 1} = 7 \), \( \frac{- x + y + 1}{x - y + 1} = 35 \).
Answer:
Cross-multiplying each equation and simplifying (the terms with xy cancel), we get
\( - 3 x + 4 y = 4 \) ...(i)    \( - 18 x + 18 y = 17 \) ...(ii)
Multiplying (i) by 6: \( - 18 x + 24 y = 24 \) ...(iii)
Subtracting (ii) from (iii), we get \( 6 y = 7 \Rightarrow y = \frac{7}{6} \).
Putting \( y = \frac{7}{6} \) in (i): \( 4 \left(\frac{7}{6}\right) - 3 x = 4 \Rightarrow x = \frac{2}{9} \).
Hence, \( x = \frac{2}{9},\ y = \frac{7}{6} \).
In simple words: Remove the fractions by cross-multiplying, simplify to two linear equations, and solve.

Teacher's Note:
1. Cross-multiplication works because a/b = c/d means ad = bc.
2. Check that your answer does not make any denominator zero.

Question 29. Solve for x and y: \( \frac{x + 2 y + 1}{2 x - y + 1} = 2 \), \( \frac{3 x - y + 1}{x - y + 3} = 5 \).
Answer:
Cross-multiplying each equation and simplifying (the terms with xy cancel), we get
\( - 3 x + 4 y = 1 \) ...(i)    \( - x + 2 y = 7 \) ...(ii)
Multiplying (ii) by 3: \( - 3 x + 6 y = 21 \) ...(iv)
Subtracting (iv) from (i), we get \( - 2 y = -20 \Rightarrow y = 10 \).
Putting \( y = 10 \) in (i): \( 4 \left(10\right) - 3 x = 1 \Rightarrow x = 13 \).
Hence, \( x = 13,\ y = 10 \).
In simple words: Remove the fractions by cross-multiplying, simplify to two linear equations, and solve.

Teacher's Note:
1. Cross-multiplication works because a/b = c/d means ad = bc.
2. Check that your answer does not make any denominator zero.