# Linear Equations with Fractions

We are familiar with linear equations in one variable and linear equations in 2 variables and their general forms. To solve the equations for the variable we follow the rules of solving equations. In this section let us see how to solve linear equations with fractions.

## Linear Equations with fractions solver:

### 1. Cross Multiplication Method:

When the linear equations are of the form, \frac{a}{b}=\frac{c}{d}, we have a x d = b x c.
This is called cross multiplication method.

#### Example: Solve linear equations with fractions. $\frac{(3x - 5)}{(2x + 3)}=\frac{1}{7}$

Solution:
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 + 35 = 2x - 2x + 3 + 35
=>                         19 x = 38
=>                        $\frac{19x }{19} = \frac{38}{19}$
=>                             x = 2

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

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

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

$\frac{1}{7}=\frac{1}{7}$

#### $\frac{15}{x+y} + \frac{7}{x-y}\: =\: 10$, where x+y is not 0, x-y is not 0.

Solution:
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)
15 u + 7 v  = 10      ----------------------(2)
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15 u - 15 u - 6v - 7 v = -3 - 10
=>     - 13 v = - 13
=>   $\frac{-13 v}{-13}$ = $\frac{-13}{-13}$
=>           v = 1
Substituting v = 1, in Equation (2), we get,
15 u + 7 v = 10
15 u + 7(1) = 10
=> 15 u + 7 - 7 = 10 - 7
=>           15 u = 3
=>       $\frac{15 u}{15}$    =   $\frac{3}{15}$
= $\frac{1}{5}$
So we have got       u = $\frac{1}{5}$ and v = 1
Now,     u = $\frac{1}{x+y}$ = 1/5          =>     x+ y = 5  -------------------(4)
v = $\frac{1}{x-y}$  = 1             =>      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
=>         3 + y = 5
=>    3 - 3 + y = 5 - 3
=>              y = 2

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

## Solving linear equations with fractions:

When we come across linear equations with fractions, which are not in the form a/b = c/d, we can follow the following procedure.
Step 1: Find the LCM of the denominators.
Step 2: Multiply each term by the LCM so obtained.
Step 3: Solve the equation for the variable x.

#### Solve : $\frac{1}{x+1}+\frac{1}{x+2}=\frac{2}{x+10}$

Solution:
We have $\frac{1}{x+1}+\frac{1}{x+2}=\frac{2}{x+10}$
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)
=> x2 + 2x + 10 x+ 20 + x2 + 10x + x + 10 = 2 ( x2 + x + 2x + 2 )
=>                                  2 x2 + 23 x + 30 = 2 ( x2 + 3x + 2 )
=>                                   2 x2 + 23x + 30 = 2 x+ 6x + 4
=>              2x2 -2 x2 + 23 x - 6x + 30 - 30 = 2 x2 - 2 x 2 + 6x - 6x + 4 -30
=>                                                   17 x  = - 26
= >                                                        x = $\frac{-26}{17}$
Hence we have learned how to do linear equations with fractions.

## How to do linear equations with fractions

The following procedure should be kept in mind while solving linear equations using calculator.
1. Change the mode to Equations.
2. Select the option if the equation is  of single variable or two variables.
3. Input the equation carefully with appropriate parenthesis.
4. The output will be the solution to the equation.

## Practice Questions:

Solve the linear equations with fractions.
1. $\frac{(3x+5)}{(4x+2)} = \frac{(3x+4)}{(4x+7)}$

2. $\frac{x+6}{4}= \frac{5x-4}{8}-\frac{x-3}{5}$

3. $\frac{2}{x}$ - $\frac{3}{y}$   = -5 ,

$\frac{3}{x}$  +  $\frac{1}{y}$ = 9

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