Equations in which the unknown only occurs to the first power are called linear equations. We are familiar with linear equations in one variable and linear equations in two variables and their general forms. Usually students feel equation with fractions are complicated and difficult to solve. In this section we will explain step by step instructions to solve these type of equations. We help you to improve the way you learn and you get exam ready with NCERT Solutions and CBSE Notes on current syllabus with smart tricks. After this chapter you will be able to solve equations with fractions at your own pace within few seconds.

When we come across linear equations with fractions, which are not in the form a/b = c/d, we can follow the following steps to solve the problems:

$\frac{1}{x+1}+\frac{1}{x+2}=\frac{2}{x+10}$

LCM of the denominators is ( x+1) ( x+2) (x+10)

Multiplying both sides by the LCM of the denominators, we get,

$\frac{1(x+1)(x+2)(x+10)}{x+1}+\frac{1(x+1)(x+2)(x+10)}{x+2}=\frac{2(x+1)(x+2)(x+10)}{x+10}$

=> (x+2)(x+10) + (x+1)(x+10) = 2 (x+1)(x+2)

=> x$^2$ + 2x + 10 x+ 20 + x$^2$ + 10x + x + 10 = 2 ( x$^2$ + x + 2x + 2 )

=> 2 x$^2$ + 23 x + 30 = 2 ( x$^2$ + 3x + 2 )

=> 2 x$^2$ + 23x + 30 = 2 x$^2$

=> 17 x = - 26

= > x = -26/17.

We have $\frac{(3x - 5)}{(2x + 3)}=\frac{1}{7}$

=> 7 ( 3x - 5) = 1 ( 2x + 3)

=> 21 x - 35 = 2x + 3

=> 21 x - 2x - 35 -3 = 0

=> 19 x = 38

or x = 2

Substituting x =2 in

$\frac{(3x - 5)} {(2x + 3)}=\frac{1}{7}$, we get,

$\frac{3(2) - 5}{2(2) + 3}=\frac{1}{7}$

$\frac{(6 - 5)}{(4 + 3)}=\frac{1}{7}$

$\frac{1}{7}=\frac{1}{7}$ (True)

$\frac{15}{x+y} + \frac{7}{x-y}\: =\: 10$, where (x + y) and (x - y) are not zero.

The given equation is an example of **linear equations in two variables with fractions**:

Let $\frac{1}{x+y}$ = u and $\frac{1}{x-y}$ = v

Let $\frac{1}{x+y}$ = u and $\frac{1}{x-y}$ = v

Then the given system of equations becomes,

5 u - 2 v = -1 --------(1)

15u + 7 v = 10 ------- (2)

Multiplying Equation (1) by 3, we get

3 ( 5u - 2 v ) = 3 ( -1)

=> 15 u - 6 v = -3 -------(3)

Subtract equation (2) from equation (3) , we have

=> - 13 v = - 13

or v = 1

Substituting v = 1, in Equation (2), we get,

15 u + 7 v = 10

15 u + 7(1) = 10

=> 15 u = 3

or u = $\frac{1}{5}$

So we have, u = $\frac{1}{5}$ and v = 1

Now, u = $\frac{1}{x+y}$ = 1/5

=> x+ y = 5 ---------(4)

=> x+ y = 5 ---------(4)

v = $\frac{1}{x-y}$ = 1

=> x - y = 1 ----------(5)

=> x - y = 1 ----------(5)

Adding Equations (4) and (5), 2x = 6

=> x = 6/2 = 3

Substituting x = 2 in (4), we get,

x+ y = 5

x+ y = 5

=> 3 + y = 5

or y = 2

Therefore the solution to the above pair of equations is ( x, y ) = ( 3, 2 )

Solution: Let x be the number. According to the given statement, we have

$\frac{10+x}{13+x}$ = 3/2

To find: The value of x

By using cross multiplication,

2(10+x) = 3(13+x)

20+2x = 39+3x

x = -19

**Question 2**: One number is 5 times another. The difference of their reciprocals is 11. What are the numbers?

**Solution**: Pick a variable to represent the quantity to be found, and express the other quantities based on the statement.

Let x and y are two numbers.

If x is one number then 5x is another

1/x - 1/5x = 11

$\frac{5-1}{5x}$ = 11

4/5x = 11

or x = 4/55

And, y = 5x = 5 $\times$ 4/55 = 4/11

So 4/55 and 4/11 are required numbers.

Let x and y are two numbers.

If x is one number then 5x is another

1/x - 1/5x = 11

$\frac{5-1}{5x}$ = 11

4/5x = 11

or x = 4/55

And, y = 5x = 5 $\times$ 4/55 = 4/11

So 4/55 and 4/11 are required numbers.