# SBP-Program

## Solving linear equations with fractions

Example.

Solve the equation

2x 3 = 5 4

Solution.

To solve the equation, we have to find the value of the variable x.

We want to get the x by itself. We isolate the variable x by dividing both sides of the equation by 2/5

2x 2 3 2 x ÷ = ÷ 5 5 4 5 * ÷ = ÷ 1 5 5 4 5

It is obvious that

2 2 ÷ = 1 5 5

Hence

x 3 2 * 1 = ÷ 1 4 5
3 2 x = ÷ 4 5

How to divide fractions? See Rules for dividing fractions.

We take the right-hand side of the equation

3 2 ÷ 4 5

1. Invert the second fraction (the divisor) to get its reciprocal.

The reciprocal for 2/5 is 5/2.

2. Multiply the first fraction (the dividend) by the reciprocal

3 5 15 * = 4 2 8

So, we have the solution of the linear equation

15 x = 8

We should check the solution by substitution 15/8 into the original linear equation

2x 3 2 = 5 4 * = 5 8 4 = 40 4
3 *  10  3 = 4 *  10 4
3 3 = 4 4

The left side equals the right side.

This means that the solution

 15 7 x = = 1 8 8

is correct.