NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

Explore the NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables (English Medium), with detailed explanations for Exercise 3.6. This resource is designed to simplify complex Pair of Linear Equations in Two Variables concepts, helping Class 10 Maths students grasp the subject effortlessly. For any queries regarding NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6, feel free to leave a comment, and we’ll respond as soon as possible. || Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions Ex 3.6 || NCERT Solutions for Class 10 Maths Chapter 3: Pair of Linear Equations in Two Variables (English Medium)

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution
Exercise-3.6

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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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} $

Answer

Let $ \frac{1}{x}=p $ and $ \frac{1}{y}=q $, then the equations becomes.
$\frac{p}{2}+\frac{q}{3}=2 \Rightarrow 3 p+2 q-12=0\ldots(i)$
$\frac{p}{2}+\frac{q}{2}=\frac{13}{6} \Rightarrow 2 p+3 q-13=0\ldots(ii)$
Using cross-multiplication method, we obtain,
$\frac{p}{-26-(-36)}=\frac{q}{-24-(-39)}=\frac{1}{9-4}$
$\frac{p}{10}=\frac{q}{15}=\frac{1}{5}$
$\frac{p}{10}=\frac{1}{5} \text { and } \frac{q}{15}=\frac{1}{5}$
$p=2 \text { and } q=3$
$ \frac{1}{x}=2 $ and $ \frac{1}{y}=3 $
$ x=\frac{1}{2} $ and $ y=\frac{1}{3} $

(ii) $ \frac{2}{\sqrt{x}}+\frac{3}{\sqrt{y}}=2, \frac{4}{\sqrt{x}}+\frac{9}{\sqrt{y}}=-9 $

Answer

Putting $ \frac{1}{\sqrt{x}}=p $ and $ \frac{1}{\sqrt{y}}=q $ in the given equations, we obtain
$2 p+3 q=2 \ldots(i)$
$4 p-9 q=-1\ldots(ii)$
Multiplying equation (1) by 3 ,
we obtain $ 6 p+9 q=6 \ldots(iii)$
Adding equation (ii) and (iii), we obtain
$10 \mathrm{p}=5$
$p=\frac{1}{2}\ldots(iv)$
$2 \times \frac{1}{2}+3 q=2$
$=3 \mathrm{q}=1$
$=q=\frac{1}{3}$
$p=\frac{1}{\sqrt{x}}=\frac{1}{2}$
$\sqrt{x}=2=x=4$
$p=\frac{1}{\sqrt{y}}=\frac{1}{3}$
$\sqrt{y}=3=y=9$
Hence, $ x=4 $ and $ y=9 $.

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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(iii) $ \frac{4}{x}+3 y=14, \frac{3}{x}-4 y=23 $

Answer

Putting $ \frac{1}{x}=p $ in given equations, we get
$=4 p+3 y=14$
$=4 p+3 y-14=0\ldots(i)$
And, $ 3 p-4 y=23 $
$
=3 p-4 y-23=0\ldots(ii)$
By cross- multiplication, we get,
$=\frac{p}{-69-56}=\frac{y}{-42-(-92)}=\frac{1}{-16-9}$
$=\frac{p}{-125}=\frac{y}{50}=-\frac{1}{25}$
Now,
$=\frac{p}{-125}=-\frac{1}{25}, \text { so, } \mathrm{p}=5$
$=\frac{y}{50}=-\frac{1}{25}, \text { So }\mathrm{y}=-2$

(iv) $ \frac{5}{x-1}+\frac{1}{y-2}=2, \frac{6}{x-1}-\frac{3}{y-2}=1 $

Answer

Putting $ \frac{1}{x-1}=p $ and $ \frac{1}{y-1}=1 $, we get
$=5 p+q=2\ldots(i)$
$=6 p-3 q=1\ldots(ii)$
Now, multiplying equation (i) by 3 we get,
$
=15 p+3 q=6\ldots(iii)$
Adding equations (ii) and (iii)
$
21 p=7
$
$
=p=\frac{7}{21}=\frac{1}{3}
$
Putting value of p in equation (iii) we get,
$=6 \times \frac{1}{3}-3 q=1$
$=-3 q=-1$
$=q=\frac{1}{3}$
we know that,
$p=\frac{1}{x-1}=\frac{1}{3}$
$=3=x-1$
$=x=4$
and, $ q=\frac{1}{y-2}=\frac{1}{3} $
$=3=y-2$
$=y=5$
Hence $ x=4, y=5 $

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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(v) $ \frac{7 x-2 y}{x y}=5, \frac{8 x+7 y}{x y}=15 $

Answer

$ \frac{7}{y}-\frac{2}{x}=5 \ldots(i)$
$=\frac{8 x+7 y}{x y}=15$
$=\frac{8}{y}+\frac{7}{x}=15\ldots(ii)$
Putting $ \frac{1}{x}=p $ and $ \frac{1}{y}=q $ in (i) and (ii) to get,
$7 q-2 p=5\ldots(iii)$
$8 q+7 p=15\ldots(iv)$
multiplying equation (iii) by 7 and equation (iv) by 2 . we get,
$
49 q-14 p=35\ldots(v)$
$
16 q+14 p=30\ldots(vi)$
After adding equations (v) and (vi) . we get,
$65 q=65$
$=q=1$
Putting value of q in equation (iv), we get,
$8+7 p=15$
$=7 p=15-8=7$
$=p=1$
Now,
$\mathrm{p}=\frac{1}{x}=\frac{1}{1}=1$
$\mathrm{q}=\frac{1}{y}=\frac{1}{1}=1$
Hence, $ x=1 $ and $ y=1 $

(vi) $ 6 x+3 y=6 x y, 2 x+4 y=5 x y $

Answer

$ 6 x+3 y=6 x y $
dividing equation by $ xy, \frac{6 x}{x y}+\frac{3 y}{x y}=\frac{6 x y}{x y}=\frac{6}{y}+\frac{3}{x}=6 \ldots(i)$
$2 x+4 y=5 x y
$
dividing equation by $ x y, \frac{2 x}{x y}+\frac{4 y}{x y}=\frac{5 x y}{x y}=\frac{2}{y}+\frac{4}{x}=5 \ldots(ii)$
Putting $ \frac{1}{x}=p$ and $ \frac{1}{y}=q $, we get,
$6 q+3 p-6=0$
$2 q+4 p-5=0$
By cross multiplication method, we get,
$
=\frac{p}{-30-(-12)}=\frac{q}{-24-(-15)}=\frac{1}{6-24}
$
$
=\frac{p}{-18}=\frac{q}{-9}=\frac{1}{-18}
$
After comparing we get,
$ p=1 $ and $ q=\frac{1}{2} $
Now ,
$ p=\frac{1}{x}=1 $ and $ q=\frac{1}{y}=\frac{1}{2} $
Hence, $ x=1 $ and $ y=2 $

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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(vii) $ \frac{10}{x+y}+\frac{2}{x-y}=4, \frac{15}{x+y}-\frac{5}{x-y}=-2 $

Answer

Putting $ \frac{1}{x+y}=p $ and $ \frac{1}{x-y}=q $, we get
$ 10 p+2 q-4=0 \ldots(i)$
$ 15 p-5 q+2=0 \ldots(ii)$
By applying cross multiplication method, we get,
$=\frac{p}{4-20}=\frac{q}{-60-(-20)}=\frac{1}{-50-30}$
$=\frac{p}{-16}=\frac{q}{-80}=\frac{1}{-80}$
After comparing we get,
$ p=\frac{1}{5} $ and $ q=1 $
Now,
$p=\frac{1}{x+y}=\frac{1}{5} \text { So, } x+y=5 \ldots(iii)$
$q=\frac{1}{x-y}=1=\text { So, } x-y=1\ldots(iv)$
Adding equations (iii) and (iv) we get,
$2 x=6$
$=x=\frac{6}{2}=3$
Putting value of equation (iii) we get,
$ y=2 $
Hence, $ x=3 $ and $ 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}
$

Answer

Putting $ \frac{1}{3 x+y}=p $ and $ \frac{1}{3 x-y}=q $ we get,
$p+q=\frac{3}{4} \ldots (i)$
$\frac{p}{2}-\frac{q}{2}=-\frac{1}{8}$
$p-q=-\frac{1}{4} \ldots (ii)$
Addomh (i) and (ii) we get,
$2 p=\frac{3}{4}-\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$
$=p=\frac{1}{4}$
Putting value of p in (ii) we get,
$ =\frac{1}{4}-q=-\frac{1}{4} $
$
=q=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}
$
Now, $ p=\frac{1}{3 x+y} $ so, $ 3 x+y=4 \ldots(iii)$
$
q=\frac{1}{3 x-y}=3 x-y=2
$
Adding equations (iii) and (iv) we get,
$ 6 x=6 $
$
=x=1
$
Putting value of $ x $ in equation (iii) we get,
$3(1)+y=4$
$=y=1$
Hence, $ x=1 $ and $ y=1 $

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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.

Answer

Let the speed of Ritu in still water and the speed of stream be $ x \mathrm{~km} / \mathrm{h} $ and $ y \mathrm{~km} / \mathrm{h} $ respectively.
While rowing upstream, Ritu's speed slows down and the speed will be her speed minus speed of stream and while rowing downstream her speed will increase and will be equal to sum of her speed and speed of stream. Therefore,
The speed of Ritu while rowing
Upstream $ =(x-y) \mathrm{km} / \mathrm{h} $
Downstream $ =(x+y) \mathrm{km} / \mathrm{h} $
According to the question:
Ritu can row downstream 20 km in 2 hours, and
distance $ = $ speed $ \times $ time
$ \Rightarrow 2(x+y)=20 $
$ \Rightarrow x+y=10 \ldots(1)$
also, Ritu can row upstream 4 km in 2 hours
$ \Rightarrow 2(x-y)=4 $
$\Rightarrow x-y=2\ldots(2)$
Adding equation (1) and (2),
we obtain
$\Rightarrow x+y+x-y=10+2$
$\Rightarrow 2 x=12$
$\Rightarrow x=6$
Putting this in equation (1),
$
6+y=10
$
we obtain $ y=4 $
Hence, Ritu's speed in still water is $ 6 \mathrm{~km} / \mathrm{h} $ and the speed of the current is $ 4 \mathrm{~km} / \mathrm{h} $.

(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.

Answer

Let the number of days taken by a woman and a man be $x$ and y respectively.
Therefore, work done by a woman in 1 day $ =\frac{1}{x} $
and work done by a man in 1 day $ =\frac{1}{y} $
According to the question,
2 women and 5 men take 4 days to complete the work
i.e. they take 4 days to complete one work
$\Rightarrow 4\left(\frac{2}{x}+\frac{5}{y}\right)=1$
$\Rightarrow \frac{2}{x}+\frac{5}{y}=\frac{1}{4}$
Also, 3 women and 6 men take 3 days to complete the work i.e. they take 3 days to complete one work
$
\Rightarrow 3\left(\frac{3}{x}+\frac{6}{y}\right)=1
$
$
\Rightarrow \frac{3}{x}+\frac{6}{y}=\frac{1}{3}
$
Putting $ \frac{1}{x}=p $ and $ \frac{1}{y}=q $ in these equations,
We obtain
$2 p+5 q=\frac{1}{4}$
$\Rightarrow 8 p+20 q=1$
And
$3 p=6 q=\frac{1}{3}$
$\Rightarrow 9 p+18 q=1$
By cross multiplication, we obtain
$\frac{p}{-20-(-18)}=\frac{q}{-9-(-8)}=\frac{1}{144-180}$
$\frac{p}{-2}=\frac{q}{-1}=\frac{1}{-36}$
$\frac{p}{-2}=\frac{-1}{36} \text { and } \frac{q}{-1}=\frac{1}{-36}$
$p=\frac{1}{18} \text { and } q=\frac{1}{36}$
$p=\frac{1}{x}=\frac{1}{18} \text { and } q=\frac{1}{y}=\frac{1}{36}$
$x=18, y=36$
Hence, number of days taken by a woman, $ x=18 $
Number of days taken by a man, $ y=36 $

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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(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 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.

Answer

Let the speed of train and bus be $ x \mathrm{~km} / \mathrm{h} $ and $ y \mathrm{~km} / \mathrm{h} $ respectively. According to the given information,
It takes her 4 hours if she travels 60 km by bus and rest (i.e. 240 km ) by train
As ti $ m e=\frac{\text { distance }}{\text { speed }} $
We have
$
\frac{60}{u}+\frac{240}{v}=4\ldots(1)$
and also, If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer i.e. 4 hours and 10 minutes. Also, 1 hour $ =60 $ minutes.
$ \Rightarrow 10 $ minutes $ =\frac{1}{60} \times 10=\frac{1}{6} $ hour
And $ \Rightarrow 4 $ hours and 10 minutes $ =\left(4+\frac{1}{6}\right)=\frac{25}{6} $ hours
We have
$
\frac{100}{u}+\frac{200}{v}=\frac{25}{6}\ldots(2)$
Putting $ \frac{1}{u}=p $ and $ \frac{1}{v}=q $ is these equations, we obtain
$60 p+240 q=4\ldots(3)$
$600 p+1200 q=25\ldots(4)$
Multiplying equation (3) by 10 , we obtain
$
600 p+2400 q=40\ldots(5)$
Subtracting equation (4) from (5), we obtain
$ 1200 \mathrm{q}=15 $
$
q=\frac{15}{1200}=\frac{1}{80}\ldots(6)$
Substituting in equation (3), we obtain
$
60 p+3=4
$
$ 60 \mathrm{p}=1 $
$ p=\frac{1}{60} $
$ p=\frac{1}{u}=\frac{1}{60} $ and $ q=\frac{1}{v}=\frac{1}{80} $
$ u=60 \mathrm{~km} / \mathrm{h} $ and $ v=80 \mathrm{~km} / \mathrm{h} $
Hence, speed of train $ =60 \mathrm{~km} / \mathrm{h} $
Speed of bus $ =80 \mathrm{~km} / \mathrm{h} $.

NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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NCERT Math Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables solutions (English Medium) || CBSE Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.6 Math Solution

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