- What is Linear equations
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- Linear equations Solutions
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- Graphical Representation of Linear equation in one and two variable
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- Steps to Draw the Given line on Cartesian plane
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- Simultaneous pair of Linear equation
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- Algebraic Solution of system of Linear equation
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- Simultaneous pair of Linear equation in Three Variable
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- Steps to solve the Linear equations

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- Ncert Solutions Exercise 3.1
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- Ncert Solutions Exercise 3.2
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- Ncert Solutions Exercise 3.3,3.4,3.5
- Ncert Solutions Exercise 3.6
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- Ncert Solutions Exercise 3.7

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- Problem and Solutions
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- Linear equations Problems
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- Linear equation worksheet
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- Linear equation word problems
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- Linear equations graphical problems

In this page we have *NCERT Solutions for Class 10th Maths:Linear Equations* for
Exercise 3.6. Hope you like them and do not forget to like , social_share
and comment at the end of the page.

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

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

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

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

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

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

$\frac {3}{x}- 4y= 23$

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

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

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

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

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

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

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

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

Let 1/x = a and 1/y = b, then the equations changes as below:

a/2 + b/3 = 2

⇒ 3a + 2b =12 ...

a/3 + b/2 = 13/6

⇒ 2a + 3b =13 ...

Multiplying (i) by 2 and (ii) by 3 and then subtracting

4b-9b=24 -39

Or b=3

Substituting b=3 in (i)

We get a =2

Now 1/x = 2 and 1/y = 3

Hence, x = 1/2 and y = 1/3

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

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

Let 1/√x = a and 1/√y = b, then the equations changes as below:

2a + 3b = 2 ...

4a – 9b = -1 ...

Multiplying equation

10a = 5

a = 1/2

Putting in equation

2 × 1/2 + 3b = 2

b = 1/3

Now a = 1/√x = 1/2

√x = 2

x = 4

and

b = 1/√y = 1/3

√y = 3

y = 9

Hence, x = 4, y = 9

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

$\frac {3}{x}- 4y= 23$

Putting 1/x = a in the given equations, we get

4a + 3y = 14 -(i)

3a - 4y = 23 -(ii)

Multiplying (i) by 3 and (ii) by 4 and then subtracting

9y +16y=42-92

Or y =-2

Now putting this value in (i), we get

p=5

Also, p = 1/x = 5

⇒ x = 1/5

So, x = 1/5 and y = -2 is the solution.

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

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

Putting 1/(x-1) = a and 1/(y-2) = b in the given equations, we obtain

5a + b = 2 ...

6a – 3b = 1 ...

Now, by multiplying equation

21a = 7

⇒ a = 1/3

Putting this value in equation

6×1/3 – 3b =1

b = 1/3

Now,

a = 1/(x-1) = 1/3

3 = x - 1

x = 4

Also,

b = 1/(y-2) = 1/3

3 = y-2

y = 5

Hence, x = 4 and y = 5 is the solution.

(v) The give equation is

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

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

Now

(7x-2y)/xy = 5

⇒ 7x/xy - 2y/xy = 5

⇒ 7/y - 2/x = 5 ...

8x+7y/xy = 15

⇒ 8x/xy + 7y/xy = 15

⇒ 8/y + 7/x = 15

Putting 1/x = a and 1/y = b, we get,

7b – 2a = 5 ...

8b + 7a = 15 ...

Multiplying equation

49b + 16b = 35 + 30

65b = 65

b = 1

Putting the value of b in equation

8 + 7a = 15

a = 1

Now,

a = 1/x = 1

x = 1

b = 1 = 1/y

y = 1

Hence, x =1 and y = 1 is the solution.

(vi) The given equation is

6x + 3y = 6xy

2x + 4y = 5xy

Now,

6x + 3y = 6xy

6x/xy + 3y/xy = 6

⇒ 6/y + 3/x = 6

2x + 4y = 5xy

⇒ 2x/xy + 4y/xy = 5

⇒ 2/y + 4/x = 5

Putting 1/x = a and 1/y = b , we get,

6b + 3a = 6 -(i)

2b + 4a =5 -(ii)

Multiplying (ii) by 3 and then subtracting from (i)

3a -12a = 6-15

a=1

Now substituting these values in (i)

6b+3=6

b=1/2

Now, a = 1/x = 1 and b = 1/y = 1/2

Hence, x = 1 and y = 2

(vii) The given equation is

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

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

Putting 1/x+y = a and 1/x-y = b in the given equations, we get:

10a + 2b = 4 -(i)

15a – 5b = -2 -(ii)

Multiplying (i) by 5 and (ii) by 2 and then adding

50a +30 a=20 -4

a=1/5

Substituting the value in (i)

b=1

Now p = 1/(x+y) = 1/5 and q = 1/(x-y) = 1

x + y = 5 ...

and x - y = 1 ...

This is again a pair of simultaneous equation

Adding equation

2x = 6

x = 3

Putting value of x in equation

y = 2

Hence, x = 3 and y = 2

(viii) The given equations are

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

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

Putting 1/(3x+y) = a and 1/(3x-y) = b in the given equations, we get

a + b = 3/4 ...

a/2 - b/2 = -1/8

a - b = -1/4 ...

Adding

2a = 3/4 - 1/4

2a = 1/2

a = 1/4

Substituting the value in equation

1/4 - b = -1/4

b= 1/4 + 1/4 = 1/2

Now

p = 1/3x+y = 1/4

3x + y = 4 ...

q = 1/3x-y = 1/2

3x - y = 2 ...

Adding equations

6x = 6

x = 1 ...

Putting the value in equation

3(1) + y = 4

y = 1

Hence, x = 1 and y = 1

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 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.

i) Let the speed of Ritu in still water and the speed of stream be x km/h

and y km/h respectively.

Speed of Ritu while rowing Upstream = (x - y) km/h

Speed of Ritu while rowing Downstream = (x + y) km/h

According to question,

2(x + y) = 20

⇒ x + y = 10 ...

2(x - y) = 4

⇒ x - y = 2 ...

Adding equation

Putting this equation in

y = 4

Hence, Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.

ii) 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 = 1/x

work done by a man in 1 day = 1/y

According to the question,

4(2/x + 5/y) = 1

Or 2/x + 5/y = 1/4

3(3/x + 6/y) = 1

Or 3/x + 6/y = 1/3

Putting 1/x = a and 1/y = b in these equations, we get

2a + 5b = ¼ -(i)

3a + 6b= 1/3 -(ii)

Multiplying (i) by 3 and (ii) by 2 and then subtracting

15b -12 b= ¾ - 2/3

b=1/36

Substituting the values in (i)

a=1/18

Now, x=1/a and y =1/b

So, x = 18 and y = 36

Therefore, number of days taken by a woman = 18 and number of days taken by a man = 36

iii) Let the speed of train and bus be u km/h and v km/h respectively.

According to the given information,

60/u + 240/v = 4

100/u + 200/v = 25/6

Putting 1/u = a and 1/v = b in the equations, we get

60a + 240b = 4 ...

100a + 200b = 25/6

600a + 1200b = 25 ...

Multiplying equation

1200b = 15

b = 15/200 = 1/80

Putting equation

60a + 3 = 4

60a = 1

a = 1/60

Now, a= 1/u = 1/60 and b = 1/v = 1/80

u = 60 and v = 80

Therefore, speed of train = 60 km/h and speed of bus = 80 km/h.

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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