**KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6.

## Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6

Question 1.

Solve the following pairs of equations by reducing them to a pair of linear equations :

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

\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\)

\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-1\)

(iii) \(\frac{4}{x}+3 y=14\)

\(\frac{3}{x}-4 y=23\)

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

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

(v) \(\frac{7 x-2 y}{x y}=5\)

\(\frac{8 x+7 y}{x y}=15\)

(vi) 6x + 3y = 6xy

2x + 4y = 5xy

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

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

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

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

Solution:

(i) \(\frac{1}{2 x}+\frac{1}{3 y}=2\) ……………. (i)

\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\) ……………… (ii)

From eqn. (i),

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

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

3y + 2x = 12xy

2x + 3y – 12xy = 0

2x + 3y = 12xy …………….. (i)

From eqn. (ii),

\(\frac{1}{3 x}+\frac{1}{2 y}=\frac{13}{6}\)

\(\frac{2 y+3 x}{6 x y}=\frac{13}{6}\)

6(2y + 3x) = 13 × 6xy

12y + 18x = 78xy

18x + 12y – 78xy = 0

18x + 12y = 78xy ……………… (ii)

Multiplying eqn. (i) by 4,

4(2x + 3y) = 4 × 12xy

8x + 12y = 78xy …………….. (iii)

From eqn. (ii) – eqn. (iii),

\(x=\frac{30 x y}{10}=3 x y\)

\(\frac{x}{x}=3 y\)

1 = 3y

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

2x + 3y = 12xy

\(2 x+3\left(\frac{1}{3}\right)=12 \times x \times \frac{1}{3}\)

2x + 1 = 4x

2x – 4x = 1

-2x = -1

2x = 1

\(x=\frac{1}{2}\)

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

(ii) \(\frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2\) ………….. (i)

\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-1\) ………….. (ii)

From eqn (i)

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

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

\(3 \sqrt{x}+2 \sqrt{x y}=2 \sqrt{x y}\) …………… (i)

From eqn (ii),

\(\frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-1\)

\(\frac{4 \sqrt{y}-9 \sqrt{x}}{\sqrt{x y}}=-1\)

\(-9 \sqrt{x}+4 \sqrt{y}=-\sqrt{x y}\)

\(9 \sqrt{x}-3 \sqrt{y}=\sqrt{x y}\) ……………….. (ii)

Multiplying eqn (i) with 2

\(2(3 \sqrt{x}+2 \sqrt{y})=2 \times 2 \sqrt{x y}\)

\(6 \sqrt{x}+4 \sqrt{y}=4 \sqrt{x y}\) ……………….. (iii)

By adding eqn (ii) + eqn (iii)

\(\sqrt{x}=\frac{5 \sqrt{x y}}{15}\)

\(\frac{\sqrt{x}}{\sqrt{x}}=\frac{5 \sqrt{y}}{15}\)

\(1=\frac{\sqrt{y}}{3}\)

\(3=\sqrt{y} \quad \sqrt{y}=3\)

∴ y = 9

Substituting the value of ‘y’ in eqn (i)

∴ x = 4, y = 9

(iii) \(\frac{4}{x}+3 y=14\)

\(\frac{3}{x}-4 y=23\)

by putting the value of \(\frac{1}{\mathrm{x}}=\mathrm{p}\)

By multiplying eqn. (i) by 4, multiplying eqn. (ii) by 3, and then adding them,

∴ p = 5

\(\frac{1}{x}\)=p=5 ∴ x=\(\frac{1}{5}\)

Substituting the value of ‘p’ in eqn. (i)

4p + 3y = 14

4 × 5 + 3y = 14

20 + 3y = 14

3y = 14 – 20

3y = -6

∴ y = -2

∴ x =\(\frac{1}{5}, y=-2\)

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

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

Let \(\frac{1}{x-1}=p \text { and } \frac{1}{y-2}=q\)

\(5\left(\frac{1}{x-1}\right)+\frac{1}{y-2}=2\) ……………. (i)

\(6\left(\frac{1}{x-1}\right)-3\left(\frac{1}{y-2}\right)=1\) …………… (ii)

5p + 1 = 2 …………… (iii)

6p – 3q = 1 …………… (iv)

By solving equation \(\mathrm{p}=\frac{1}{3}, \quad \mathrm{q}=\frac{1}{3}\)

Substituting \(\frac{1}{x-1}\) for p,

\(p=\frac{1}{x-1}=\frac{1}{3}\)

x – 1 = 3

∴ x = 4

Substituting \(\frac{1}{y-2}\) for q,

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

y – 2 = 3

∴ y = 5

∴ x = 4, y = 5

(v) \(\frac{7 x-2 y}{x y}=5\) ……………. (i)

\(\frac{8 x+7 y}{x y}=15\) ………………… (ii)

From eqn. (i)

\(\frac{7 x-2 y}{x y}=5\)

7x – 2y = 5xy ………….. (i)

From eqn. (ii)

\(\frac{8 x+7 y}{x y}=15\)

8x + 7y = 15xy ……………… (ii)

Eqm (i) × 7 and Eqn. (ii) × 2, we have

49x – 14y = 35xy ………….. (iii)

16x + 14y = 30xy ……………. (iv)

From eqn. (iii) + eqn. (iv)

∴ \( \frac{65 \mathrm{x}}{65 \mathrm{x}}=\mathrm{y}\)

1 = y

∴ y = 1

Substituting the value of ‘y’ in eqn. (i)

7x – 2y = 5xy

7x – 2 (1) = 5 × x × 1

7x – 2 = 5x

7x – 5x = 2

2x = 2

∴ x = 1

∴ x = 1, y = 1

(vi) 6x + 3y = 6xy ………………. (i)

2x + 4y = 5xy ………………… (ii)

Multiplying eqn. (ii) by 3

3(2x + 4y = 5xy)

6x + 12y = 18xy ………………. (iii)

From eqn. (i) – eqn. (ii)

9y = 9xy

\(\frac{9 y}{9 y}=x\)

1 = x

∴ x = 1

Substituting the value of ‘x’ in eqn (i),

6x + 3y = 6xy

6 × 1 + 3y = 6 × 1 × y

6 + 3y = 6y

6y – 3y = 6

3y = 6

∴ y=\(\frac{6}{3}\)

∴ y = 2

∴ x = 1, y = 2

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

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

Let \(\mathrm{u}=\frac{1}{\mathrm{x}+\mathrm{y}}, \mathrm{v}=\frac{1}{\mathrm{x}-\mathrm{y}}\) then,

10u + 2v = 4 …………. (i)

15u – 5v = -2 …………. (ii)

Multiplying eqn. (i) by 5 and multiplying eqn. (ii) by 2 and then by adding

∴ \(\quad \mathbf{u}=\frac{1}{5}\)

Substituting the value of ‘u’ in eqn. (i)

\(10 \times \frac{1}{5}+2 v=4\)

2 + 2v = 4

2v = 4 – 2

2v = 2

∴ v = 1

∴ u = 1/5 , v =1

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

∴ x + y = 5 x – y = 1

x + y = 5 …………….. (iii)

x – y = 1 …………….. (iv)

From eqn. (iii) + eqn. (iv)

x = 3

∴ Substituting the value of ‘x’ in (iii)

x + y = 5

3 + y = 5

y = 5 – 3

y = 2

∴ x = 3, y = 2

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

\(\frac{1}{2(3 x+y)}-\frac{1}{2(3 x-y)}=\frac{-1}{8}\) ……………. (ii)

Substituting the value of ‘p’ in eqn. (iii)

\(p+q=\frac{3}{4}\)

\(\frac{1}{4}+q=\frac{3}{4}\)

∴ q=\(\frac{3}{4}-\frac{1}{4}\)

∴ q=\(\frac{1}{2}\)

∴ p=\(\frac{1}{3 x+y}=\frac{1}{4}\)

3x + y = 4 ……………. (v)

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

3x – y = 2 ……………… (vi)

From eqn. (v) + eqn. (vi)

3x + y = 4

+3x – y = 2

6x = 6

∴ x=\(\frac{6}{6}=1\)

Substituting the value of ‘x’ in eqn. (v)

3x + y = 4

3 × 1 + y = 4

3 + y = 4

∴ y = 4 – 3

∴ y = 1

∴ x = 1, y = 1

Question 2.

Formulate the following problems as a pair of equations, and hence find their solutions:

(i) Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

(ii) 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 to finish the work, and also that taken by 1 man alone.

(iii) Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and the remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution:

(i) Let the speed of Ritu in still water = x km/hr

and the speed of the current = y km/hr

∴ Downstream speed = (x + y) km/hr

Upstream speed = (x – y)km/hr

Substituting the value of x in equation (1), we get

6 + y = 10 ⇒ y = 10 – 6 = 4

Thus, speed of rowing in still water = 6 km/hr, speed of current = 4 km/hr

(ii) Let the time taken to finish the task by one woman alone = x days

and the time taken to finish the task by one man alone = y days

∴ 1 man can finish the work in 36 days and 1 woman can finish the work in 18 days,

(iii) Let the speed of the train = x km/hr

and the speed of the bus = y km/hr

Case I: Total journey = 300 km

Distance travelled by train = 60 km

Distance travelled by bus = (300 – 60)km

= 240 km

∵ Total time taken = 4 hours

Case II: Distance travelled by train = 100 km

Distance travelled by bus = (300 – 100)km

= 200 km

Total time = 4 hrs 10 mins

Thus, speed of the train = 60 km/hr

and speed of the bus = 80 km/hr

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.6, drop a comment below and we will get back to you at the earliest.