KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6

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),
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 1
\(x=\frac{30 x y}{10}=3 x y\)
\(\frac{x}{x}=3 y\)
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
\(x=\frac{1}{2}\)
\(x=\frac{1}{2}, \quad y=\frac{1}{3}\)

KSEEB Solutions

(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)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 2
\(\sqrt{x}=\frac{5 \sqrt{x y}}{15}\)
\(\frac{\sqrt{x}}{\sqrt{x}}=\frac{5 \sqrt{y}}{15}\)
\(1=\frac{\sqrt{y}}{3}\)
\(3=\sqrt{y} \quad \sqrt{y}=3\)
∴ y = 9
Substituting the value of ‘y’ in eqn (i)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 3
∴ 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,
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 4
∴ 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 =\(\frac{1}{5}, y=-2\)

KSEEB Solutions

(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)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 5
∴ \( \frac{65 \mathrm{x}}{65 \mathrm{x}}=\mathrm{y}\)
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

KSEEB Solutions

(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)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 6
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
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 7
∴ \(\quad \mathbf{u}=\frac{1}{5}\)
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)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 8
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)
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 9
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

KSEEB Solutions

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
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 17
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
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 18
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 1
∴ 1 man can finish the work in 36 days and 1 woman can finish the work in 18 days,

KSEEB Solutions

(iii) Let the speed of the train = x km/hr
and the speed of the bus = y km/hr
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 20
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
MP Board Class 10th Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 21

Case II: Distance travelled by train = 100 km
Distance travelled by bus = (300 – 100)km
= 200 km
Total time = 4 hrs 10 mins
KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 2
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.

Leave a Comment