__Example 1__

Solve the equation:
__Answer__

When the equation is an equation of two fractions like above, then we can eliminate the denominators by multiplying the nominator of each fraction with the denominator of the other.

<=> 3 (2x +1)=4 ( x - 1) <=> 6x + 3=4x - 4 <=> 6x - 4x = -3 - 4 <=> 2x = -7 <=>

<=> x = -7/2 |

__Example 2__

Solve the equation:

__Answer__

Here we follow the algorithm that we described in the previous paragraph.

<=>
<=> 3(5x+1) - x = 2(7x+4) <=>

<=> 15x + 3 - x = 14x + 8 <=> 15x
- x - 14x = -3 + 8 <=> (15 - 1 - 14)x = 5 <=>

<=> 0x = 5 which has no solutions in R |

__Example 3__

Solve the equation:

__Answer__

<=> <=>

2(2x - 3) - (3x + 1) = x - 3 - 4 <=> 4x - 6 -3x - 1 = x - 3 - 4 <=>

(4 - 3 -1)x = 0 <=>

<=> 0x = 0 |