AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.3

### 10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers

Question 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

Answer:

Given

\(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2

\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1

Put \(\frac{1}{x-1}\) = a and \(\frac{1}{y-2}\) = b,

then the given equations reduce to

5a + b = 2 ……… (1)

6a – 3b = 1 ………. (2)

⇒ b = \(\frac{7}{21}\) = \(\frac{1}{3}\)

Substituting b = \(\frac{1}{3}\) in equation (1) we get

⇒ (x – 1) . 1 = 3 × 1

⇒ x – 1 = 3

⇒ x = 3 + 1 = 4

b = \(\frac{1}{y-2}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{y-2}\)

⇒ (y – 2) . 1 = 3 × 1

⇒ y – 2 = 3

⇒ y = 3 + 2 = 5

∴ Solution (x, y) = (4, 5)

ii) \(\frac{x+y}{xy}\) = 2;

\(\frac{x-y}{xy}\) = 6

Answer:

Given

\(\frac{x+y}{xy}\) = 2

⇒ \(\frac{x}{xy}\) + \(\frac{y}{xy}\) = 2

⇒ \(\frac{1}{y}\) + \(\frac{1}{x}\) = 2

\(\frac{x-y}{xy}\) = 6

⇒ \(\frac{x}{xy}\) – \(\frac{y}{xy}\) = 6

⇒ \(\frac{1}{y}\) – \(\frac{1}{x}\) = 6

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,

then the given equations reduces to

⇒ b = \(\frac{8}{2}\) = 4

Substituting b = 4 in equation (1) we get

a + 4 = 2 ⇒ a = 2 – 4 = -2

but a = \(\frac{1}{x}\) = -2 ⇒ x = \(\frac{-1}{2}\)

b = \(\frac{1}{y}\) = 4 ⇒ y = \(\frac{1}{4}\)

∴ 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

Answer:

Given

\(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2 and \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1

Take \(\frac{1}{\sqrt{x}}\) = a and \(\frac{1}{\sqrt{y}}\) = b,

then the given equations reduces to

2a + 3b = 2 …….. (1)

4a – 9b = – 1 …….. (2)

⇒ b = \(\frac{5}{15}\) = \(\frac{1}{3}\)

Substituting b = \(\frac{1}{3}\) in equation (1) we get

2a + 3\(\left(\frac{1}{3}\right)\) = 2

⇒ 2a + 1 = 2 ⇒ 2a = 2 – 1 ⇒ a = \(\frac{1}{2}\)

∴ Solution (x, y) = (4, 9)

iv) 6x + 3y = 6xy

2x + 4y = 5xy

Answer:

Given

6x + 3y = 6xy

⇒ \(\frac{6x+3y}{xy}\) = 6

⇒ \(\frac{6x}{xy}\) + \(\frac{3y}{xy}\) = 6

⇒ \(\frac{6}{y}\) + \(\frac{3}{x}\) = 6

2x + 4y = 5xy

⇒ \(\frac{2x+4y}{xy}\) = 5

⇒ \(\frac{2x}{xy}\) + \(\frac{4y}{xy}\) = 6

⇒ \(\frac{2}{y}\) + \(\frac{4}{x}\) = 6

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,

then the given equations reduces to

3a + 6b = 6 ……. (1)

4a + 2b = 5 ……. (2)

⇒ b = \(\frac{9}{18}\) = \(\frac{1}{2}\)

Substituting b = \(\frac{1}{2}\) in equation (1) we get

3a +6\(\left(\frac{1}{2}\right)\) = 6

⇒ 3a = 6 – 3

⇒ a = \(\frac{3}{3}\) = 1

but a = \(\frac{1}{x}\) = 1 ⇒ x = 1

b = \(\frac{1}{y}\) = \(\frac{1}{2}\) ⇒ y = 2

∴ Solution (x, y) = (1, 2)

v) \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1

\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10

where x ≠ 0, y ≠ 0

Answer:

Given

\(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1 and

\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10

Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then

the given equations reduce to

5a – 2b = – 1 ……… (1)

15a + 7b = 10 ……… (2)

⇒ b = \(\frac{-13}{-13}\) = 1

Substituting b = 1 in equation (1) we get

5a – 2(1) = -1

⇒ 5a = -1 + 2

⇒ 5a = 1

⇒ a = \(\frac{1}{5}\)

but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5

b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1

⇒ x = \(\frac{6}{2}\) = 3

Solving the above equations

Substituting x = 3 in x + y = 5 we get

3 + y = 5 ⇒ y = 5 – 3 = 2

∴ Solution (x, y) = (3, 2)

vi) \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13

\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2

where x ≠ 0, y ≠ 0

Answer:

Given

\(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 and

\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then

the given equations reduce to

2a + 3b = 13 ……… (1)

5a – 4b = -2 ……… (2)

⇒ b = \(\frac{69}{23}\) = 3

Substituting b = 3 in equation (1) we get

2a + 3 (3) = 13

⇒ 2a = 13 – 9

⇒ a = \(\frac{4}{2}\) = 2

but a = \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)

b = \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)

∴ Solution (x, y) = (\(\frac{1}{2}\), \(\frac{1}{3}\))

vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4

\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2

Answer:

Given

\(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4 and

\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2

Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then

the given equations reduce to

10a + 2b = 4 ……… (1)

15a – 5b = – 2 ……… (2)

⇒ b = \(\frac{16}{16}\) = 1

Substituting b = 1 in equation (1) we get

10a + 2(1) = 4

⇒ 10a = 4 – 2

⇒ a = \(\frac{2}{10}\) = \(\frac{1}{5}\)

but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5 ……. (3)

b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1 …….. (4)

Adding (3) and (4)

⇒ x = \(\frac{6}{2}\) = 3

Substituting x = 3 in x + y = 5 we get

3 + y = 5 ⇒ y = 5 – 3 = 2

∴ Solution (x, y) = (3, 2)

viii) \(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\)

\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)

Answer:

Given

\(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\) and

\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)

Take \(\frac{1}{3x+y}\) = a and \(\frac{1}{3x-y}\) = b, then

the given equations reduce to

⇒ a = \(\frac{2}{8}\) = \(\frac{1}{4}\)

Substituting a = \(\frac{1}{4}\) in equation (1) we get

Solving (3) and (4)

⇒ x = \(\frac{6}{6}\) = 1

Substituting x = 1 in 3x + y = 4

⇒ 3(1) + y = 4

⇒ y = 4 – 3 = 1

∴ The solution (x, y) = (1, 1)

Question 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.

Answer:

Let the speed of the boat in still water = x kmph

and the speed of the stream = y kmph

then speed in downstream = x + y

Speed in upstream = x – y

and time = \(\frac{\text { distance }}{\text { speed }}\)

By problem,

\(\frac{30}{x-y}\) + \(\frac{44}{x+y}\) = 10

\(\frac{40}{x-y}\) + \(\frac{55}{x+y}\) = 13

Take \(\frac{1}{x-y}\) = a and \(\frac{1}{x+y}\) = b, then

the given equations reduce to

30a + 44b = 10 ……… (1)

40a + 55b = 13 ……… (2)

⇒ b = \(\frac{1}{11}\)

Substituting b = \(\frac{1}{11}\) in equation (1) we get

⇒ x = 8

Substituting x = 8 in x – y = 5 we get

8 – y = 5

⇒ y = 8 – 5 = 3

∴ The solution (x, y) = (8, 3)

Speed of the boat in still water = 8 kmph

Speed of the stream = 3 kmph.

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.

Answer:

Let the speed of the train be x kmph

and the speed of the car = y kmph

By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then

the given equations reduce to

15a + 60b = 1 ……… (1)

8a + 16b = \(\frac{1}{3}\) ⇒ 24a + 48b = 1 ……… (2)

⇒ a = \(\frac{-1}{-60}\) = \(\frac{1}{60}\)

Substituting a = \(\frac{1}{60}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{60}\) ⇒ x = 60 kmph

b = \(\frac{1}{y}\) = \(\frac{1}{80}\) ⇒ y = 80 kmph

Speed of the train = 60 kmph and

speed of the car = 80 kmph

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 taken by 1 woman alone and 1 man alone to finish the work.

Answer:

Let the time taken by 1 woman to complete the work = x days

and time taken by 1 man to complete the work = y days

∴ Work done by 1 woman in 1 day = \(\frac{1}{x}\)

Work done by 1 man in 1 day = \(\frac{1}{y}\)

By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,

then the above equations reduce to

2a + 5b = \(\frac{1}{4}\) and 3a + 6b = \(\frac{1}{3}\)

⇒ 8a + 20b = 1 …….. (1) and

9a + 18b = 1 ……… (2)

⇒ b = \(\frac{1}{36}\)

Substituting b = \(\frac{1}{36}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{18}\) ⇒ x = 18 and

b = \(\frac{1}{y}\) = \(\frac{1}{36}\) ⇒ y = 36

∴ Time taken by 1 woman = 18 days

1 man = 36 days