Table of Contents
Exercise 4.3 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on solving pairs of equations by reducing them to linear equations and solving word problems. Mathematical expressions are rendered using MathJax.
1. Solve each of the following pairs of equations by reducing them to a pair of linear equations.
(i) \( \frac{5}{x-1} + \frac{1}{y-2} = 2 \), \( \frac{6}{x-1} – \frac{3}{y-2} = 1 \)
Substitute \( u = \frac{1}{x-1} \), \( v = \frac{1}{y-2} \).
Rewrite equations: \( 5u + v = 2 \), \( 6u – 3v = 1 \).
Multiply the first by 3: \( 15u + 3v = 6 \).
Add to the second: \( (15u + 3v) + (6u – 3v) = 6 + 1 \implies 21u = 7 \implies u = \frac{1}{3} \).
Substitute \( u = \frac{1}{3} \) into \( 5u + v = 2 \): \( 5 \left(\frac{1}{3}\right) + v = 2 \implies \frac{5}{3} + v = 2 \implies v = 2 – \frac{5}{3} = \frac{1}{3} \).
Solve for \( x \), \( y \): \( u = \frac{1}{x-1} = \frac{1}{3} \implies x – 1 = 3 \implies x = 4 \).
\( v = \frac{1}{y-2} = \frac{1}{3} \implies y – 2 = 3 \implies y = 5 \).
Check: \( \frac{5}{4-1} + \frac{1}{5-2} = \frac{5}{3} + \frac{1}{3} = 2 \), \( \frac{6}{4-1} – \frac{3}{5-2} = \frac{6}{3} – \frac{3}{3} = 1 \), both true.
Solution: \( (x, y) = (4, 5) \)
(ii) \( \frac{x+y}{xy} = 2 \), \( \frac{x-y}{xy} = 6 \)
Simplify: First equation: \( \frac{x+y}{xy} = \frac{1}{y} + \frac{1}{x} = 2 \).
Second equation: \( \frac{x-y}{xy} = \frac{1}{y} – \frac{1}{x} = 6 \).
Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \). Then: \( u + v = 2 \), \( -u + v = 6 \).
Add the equations: \( (u + v) + (-u + v) = 2 + 6 \implies 2v = 8 \implies v = 4 \).
Substitute \( v = 4 \) into \( u + v = 2 \): \( u + 4 = 2 \implies u = -2 \).
Solve: \( u = \frac{1}{x} = -2 \implies x = -\frac{1}{2} \), \( v = \frac{1}{y} = 4 \implies y = \frac{1}{4} \).
Check: \( \frac{-\frac{1}{2} + \frac{1}{4}}{-\frac{1}{2} \cdot \frac{1}{4}} = \frac{-\frac{1}{4}}{-\frac{1}{8}} = 2 \), \( \frac{-\frac{1}{2} – \frac{1}{4}}{-\frac{1}{8}} = \frac{-\frac{3}{4}}{-\frac{1}{8}} = 6 \), both true.
Solution: \( (x, y) = \left(-\frac{1}{2}, \frac{1}{4}\right) \)
(iii) \( \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \), \( \frac{4}{\sqrt{x}} – \frac{9}{\sqrt{y}} = -1 \)
Substitute \( u = \frac{1}{\sqrt{x}} \), \( v = \frac{1}{\sqrt{y}} \). Then \( u^2 = \frac{1}{x} \), \( v^2 = \frac{1}{y} \).
Rewrite: \( 2u + 3v = 2 \), \( 4u – 9v = -1 \).
Multiply the first by 3: \( 6u + 9v = 6 \).
Add to the second: \( (6u + 9v) + (4u – 9v) = 6 – 1 \implies 10u = 5 \implies u = \frac{1}{2} \).
Substitute \( u = \frac{1}{2} \) into \( 2u + 3v = 2 \): \( 2 \left(\frac{1}{2}\right) + 3v = 2 \implies 1 + 3v = 2 \implies 3v = 1 \implies v = \frac{1}{3} \).
Solve: \( u = \frac{1}{\sqrt{x}} = \frac{1}{2} \implies \sqrt{x} = 2 \implies x = 4 \).
\( v = \frac{1}{\sqrt{y}} = \frac{1}{3} \implies \sqrt{y} = 3 \implies y = 9 \).
Check: \( \frac{2}{\sqrt{4}} + \frac{3}{\sqrt{9}} = 1 + 1 = 2 \), \( \frac{4}{\sqrt{4}} – \frac{9}{\sqrt{9}} = 2 – 3 = -1 \), both true.
Solution: \( (x, y) = (4, 9) \)
(iv) \( 6x + 3y = 6xy \), \( 2x + 4y = 5xy \)
Divide the first by \( xy \): \( \frac{6x}{xy} + \frac{3y}{xy} = \frac{6xy}{xy} \implies \frac{6}{y} + \frac{3}{x} = 6 \).
Divide the second by \( xy \): \( \frac{2}{y} + \frac{4}{x} = 5 \).
Let \( u = \frac{1}{x} \), \( v = \frac{1}{y} \). Then: \( 3u + 6v = 6 \implies u + 2v = 2 \), \( 4u + 2v = 5 \).
Subtract: \( (4u + 2v) – (u + 2v) = 5 – 2 \implies 3u = 3 \implies u = 1 \).
Substitute \( u = 1 \) into \( u + 2v = 2 \): \( 1 + 2v = 2 \implies 2v = 1 \implies v = \frac{1}{2} \).
Solve: \( u = \frac{1}{x} = 1 \implies x = 1 \), \( v = \frac{1}{y} = \frac{1}{2} \implies y = 2 \).
Check: \( 6(1) + 3(2) = 6 \cdot 1 \cdot 2 \implies 12 = 12 \), \( 2(1) + 4(2) = 5 \cdot 1 \cdot 2 \implies 10 = 10 \), both true.
Solution: \( (x, y) = (1, 2) \)
(v) \( \frac{5}{x+y} – \frac{2}{x-y} = -1 \), \( \frac{15}{x+y} + \frac{7}{x-y} = 10 \)
Substitute \( u = \frac{1}{x+y} \), \( v = \frac{1}{x-y} \).
Rewrite: \( 5u – 2v = -1 \), \( 15u + 7v = 10 \).
Multiply the first by 7 and the second by 2: \( 35u – 14v = -7 \), \( 30u + 14v = 20 \).
Add: \( 65u = 13 \implies u = \frac{1}{5} \).
Substitute \( u = \frac{1}{5} \) into \( 5u – 2v = -1 \): \( 5 \left(\frac{1}{5}\right) – 2v = -1 \implies 1 – 2v = -1 \implies 2v = 2 \implies v = 1 \).
Solve: \( u = \frac{1}{x+y} = \frac{1}{5} \implies x + y = 5 \).
\( v = \frac{1}{x-y} = 1 \implies x – y = 1 \).
Add the resulting equations: \( 2x = 6 \implies x = 3 \), \( y = 5 – 3 = 2 \).
Check: \( \frac{5}{3+2} – \frac{2}{3-2} = 1 – 2 = -1 \), \( \frac{15}{5} + \frac{7}{1} = 3 + 7 = 10 \), both true.
Solution: \( (x, y) = (3, 2) \)
(vi) \( \frac{2}{x} + \frac{3}{y} = 13 \), \( \frac{5}{x} – \frac{4}{y} = -2 \)
Substitute \( u = \frac{1}{x} \), \( v = \frac{1}{y} \).
Rewrite: \( 2u + 3v = 13 \), \( 5u – 4v = -2 \).
Multiply the first by 4 and the second by 3: \( 8u + 12v = 52 \), \( 15u – 12v = -6 \).
Add: \( 23u = 46 \implies u = 2 \).
Substitute \( u = 2 \) into \( 2u + 3v = 13 \): \( 2(2) + 3v = 13 \implies 4 + 3v = 13 \implies 3v = 9 \implies v = 3 \).
Solve: \( u = \frac{1}{x} = 2 \implies x = \frac{1}{2} \), \( v = \frac{1}{y} = 3 \implies y = \frac{1}{3} \).
Check: \( \frac{2}{\frac{1}{2}} + \frac{3}{\frac{1}{3}} = 4 + 9 = 13 \), \( \frac{5}{\frac{1}{2}} – \frac{4}{\frac{1}{3}} = 10 – 12 = -2 \), both true.
Solution: \( (x, y) = \left(\frac{1}{2}, \frac{1}{3}\right) \)
(vii) \( \frac{10}{x+y} + \frac{2}{x-y} = 4 \), \( \frac{15}{x+y} – \frac{5}{x-y} = -2 \)
Substitute \( u = \frac{1}{x+y} \), \( v = \frac{1}{x-y} \).
Rewrite: \( 10u + 2v = 4 \implies 5u + v = 2 \), \( 15u – 5v = -2 \).
Multiply the first by 5: \( 25u + 5v = 10 \).
Add to the second: \( (25u + 5v) + (15u – 5v) = 10 – 2 \implies 40u = 8 \implies u = \frac{1}{5} \).
Substitute \( u = \frac{1}{5} \) into \( 5u + v = 2 \): \( 5 \left(\frac{1}{5}\right) + v = 2 \implies 1 + v = 2 \implies v = 1 \).
Solve: \( u = \frac{1}{x+y} = \frac{1}{5} \implies x + y = 5 \), \( v = \frac{1}{x-y} = 1 \implies x – y = 1 \).
Solve the linear system: \( x = 3 \), \( y = 2 \) (same as in (v), confirming consistency).
Solution: \( (x, y) = (3, 2) \)
(viii) \( \frac{1}{3x+y} + \frac{1}{3x-y} = \frac{3}{4} \), \( \frac{1}{2(3x+y)} – \frac{1}{2(3x-y)} = \frac{-1}{8} \)
Substitute \( u = 3x + y \), \( v = 3x – y \). Then \( u + v = 6x \), \( u – v = 2y \).
First equation: \( \frac{1}{u} + \frac{1}{v} = \frac{3}{4} \implies \frac{u+v}{uv} = \frac{3}{4} \implies 4(u + v) = 3uv \).
Second equation: \( \frac{1}{2u} – \frac{1}{2v} = \frac{-1}{8} \implies \frac{v – u}{2uv} = \frac{-1}{8} \implies v – u = -\frac{uv}{4} \).
Let \( p = \frac{1}{u} \), \( q = \frac{1}{v} \). Then: \( p + q = \frac{3}{4} \), \( \frac{q – p}{2} = \frac{-1}{8} \implies q – p = \frac{-1}{4} \).
Solve: Add the equations: \( 2q = \frac{3}{4} – \frac{1}{4} = \frac{1}{2} \implies q = \frac{1}{4} \). Then \( p = \frac{3}{4} – \frac{1}{4} = \frac{1}{2} \).
Solve: \( p = \frac{1}{u} = \frac{1}{2} \implies u = 2 \), \( q = \frac{1}{v} = \frac{1}{4} \implies v = 4 \).
Then: \( u + v = 6x \implies 6x = 2 + 4 = 6 \implies x = 1 \).
\( u – v = 2y \implies 2 – 4 = 2y \implies -2 = 2y \implies y = -1 \).
Check: \( 3x + y = 3(1) – 1 = 2 \), \( 3x – y = 3 + 1 = 4 \). First: \( \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \), second: \( \frac{1}{4} – \frac{1}{8} = \frac{1}{8} \), so \( \frac{1}{8} = -\left(-\frac{1}{8}\right) \), true.
Solution: \( (x, y) = (1, -1) \)
2. Formulate the following problems as a pair of equations and then find their solutions.
(i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Let the speed of the boat in still water be \( x \) km/h, speed of the stream be \( y \) km/h.
Upstream speed: \( x – y \), downstream speed: \( x + y \).
First condition: \( \frac{30}{x-y} + \frac{44}{x+y} = 10 \).
Second condition: \( \frac{40}{x-y} + \frac{55}{x+y} = 13 \).
Substitute \( u = \frac{1}{x-y} \), \( v = \frac{1}{x+y} \).
Rewrite: \( 30u + 44v = 10 \), \( 40u + 55v = 13 \).
Multiply the first by 5 and the second by 4: \( 150u + 220v = 50 \), \( 160u + 220v = 52 \).
Subtract: \( 160u – 150u = 52 – 50 \implies 10u = 2 \implies u = \frac{1}{5} \).
Substitute \( u = \frac{1}{5} \) into \( 30u + 44v = 10 \): \( 30 \left(\frac{1}{5}\right) + 44v = 10 \implies 6 + 44v = 10 \implies 44v = 4 \implies v = \frac{1}{11} \).
Solve: \( u = \frac{1}{x-y} = \frac{1}{5} \implies x – y = 5 \), \( v = \frac{1}{x+y} = \frac{1}{11} \implies x + y = 11 \).
Add: \( 2x = 16 \implies x = 8 \), \( y = 11 – 8 = 3 \).
Check: First: \( \frac{30}{5} + \frac{44}{11} = 6 + 4 = 10 \), second: \( \frac{40}{5} + \frac{55}{11} = 8 + 5 = 13 \), both true.
Speed of boat: 8 km/h, Speed of stream: 3 km/h
(ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.
Let the speed of the train be \( x \) km/h, speed of the car be \( y \) km/h.
First condition: 120 km by train, 480 km by car, time = 8 hours: \( \frac{120}{x} + \frac{480}{y} = 8 \).
Second condition: 200 km by train, 400 km by car, time = 8 hours 20 minutes = \( 8 + \frac{20}{60} = \frac{25}{3} \) hours: \( \frac{200}{x} + \frac{400}{y} = \frac{25}{3} \).
Substitute \( u = \frac{1}{x} \), \( v = \frac{1}{y} \).
Rewrite: \( 120u + 480v = 8 \implies 15u + 60v = 1 \), \( 200u + 400v = \frac{25}{3} \implies 24u + 48v = 1 \).
Multiply the first by 4 and subtract the second: \( (60u + 240v) – (24u + 48v) = 4 – 1 \implies 36u + 192v = 3 \).
Simplify: \( 3u + 16v = \frac{1}{4} \). Multiply the second original by 3: \( 72u + 144v = 3 \implies u + 2v = \frac{1}{24} \).
Solve: \( 3u + 16v = \frac{1}{4} \), \( u + 2v = \frac{1}{24} \). Multiply the second by 3: \( 3u + 6v = \frac{1}{8} \).
Subtract: \( (3u + 16v) – (3u + 6v) = \frac{1}{4} – \frac{1}{8} \implies 10v = \frac{1}{8} \implies v = \frac{1}{80} \).
Substitute \( v = \frac{1}{80} \) into \( u + 2v = \frac{1}{24} \): \( u + 2 \left(\frac{1}{80}\right) = \frac{1}{24} \implies u + \frac{1}{40} = \frac{1}{24} \implies u = \frac{1}{24} – \frac{1}{40} = \frac{5-3}{120} = \frac{1}{60} \).
Solve: \( u = \frac{1}{x} = \frac{1}{60} \implies x = 60 \), \( v = \frac{1}{y} = \frac{1}{80} \implies y = 80 \).
Check: First: \( \frac{120}{60} + \frac{480}{80} = 2 + 6 = 8 \), second: \( \frac{200}{60} + \frac{400}{80} = \frac{10}{3} + 5 = \frac{25}{3} \), both true.
Speed of train: 60 km/h, Speed of car: 80 km/h
(iii) 2 women and 5 men can together finish an embroidery work in 4 days while 3 women and 6 men can finish it in 3 days. Find the time to be taken by 1 woman alone and 1 man alone to finish the work.
Let 1 woman finish the work in \( w \) days, 1 man in \( m \) days.
Work rate of 1 woman = \( \frac{1}{w} \), 1 man = \( \frac{1}{m} \).
First condition: 2 women and 5 men in 4 days: \( 4 \left( \frac{2}{w} + \frac{5}{m} \right) = 1 \implies \frac{2}{w} + \frac{5}{m} = \frac{1}{4} \).
Second condition: 3 women and 6 men in 3 days: \( 3 \left( \frac{3}{w} + \frac{6}{m} \right) = 1 \implies \frac{3}{w} + \frac{6}{m} = \frac{1}{3} \).
Substitute \( u = \frac{1}{w} \), \( v = \frac{1}{m} \).
Rewrite: \( 2u + 5v = \frac{1}{4} \), \( 3u + 6v = \frac{1}{3} \).
Multiply the first by 3 and the second by 2: \( 6u + 15v = \frac{3}{4} \), \( 6u + 12v = \frac{2}{3} \).
Subtract: \( 15v – 12v = \frac{3}{4} – \frac{2}{3} \implies 3v = \frac{9-8}{12} = \frac{1}{12} \implies v = \frac{1}{36} \).
Substitute \( v = \frac{1}{36} \) into \( 2u + 5v = \frac{1}{4} \): \( 2u + 5 \left(\frac{1}{36}\right) = \frac{1}{4} \implies 2u + \frac{5}{36} = \frac{1}{4} \implies 2u = \frac{1}{4} – \frac{5}{36} = \frac{9-5}{36} = \frac{1}{9} \implies u = \frac{1}{18} \).
Solve: \( u = \frac{1}{w} = \frac{1}{18} \implies w = 18 \), \( v = \frac{1}{m} = \frac{1}{36} \implies m = 36 \).
Check: First: \( \frac{2}{18} + \frac{5}{36} = \frac{1}{9} + \frac{5}{36} = \frac{4+5}{36} = \frac{1}{4} \), second: \( \frac{3}{18} + \frac{6}{36} = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \), both true.
1 woman alone: 18 days, 1 man alone: 36 days
