# Solve multi-step proportions

Learn how to solve multi-step proportions easily with a couple of good examples.

Quick review:

Suppose you have the following proportion:

Numerator #1 / Denominator #1 = Numerator #2 / Denominator #2

After doing cross-multiplication, it is equivalent to

Numerator #1 × Denominator #2 = Numerator #2 × Denominator #1

Example #1

Solve the proportion (x - 5) / 4 = (x + 3) / 6

(x - 5) / 4 = (x + 3) / 6

The proportion is equivalent to

(x - 5) × 6 = (x + 3) × 4

(x - 5) × 6 = (x + 3) × 4

Use the distributive property

6x - 30 = 4x + 12

Subtract 4x from each side of the equation

6x - 4x - 30 = 4x - 4x + 12

2x - 30 = 0 + 12

2x - 30 = 12

Add 30 to each side of the equation

2x - 30 + 30 = 12 + 30

2x + 0 = 42

2x = 42

Divide each side by 2

2x/2 = 42/2

x = 21

Example #2

Solve the proportion 3 / (x - 4) = -5 / (4x + 1)

3 / (x - 4) = -5 / (4x + 1)

The proportion is equivalent to

3 × (4x + 1) = -5 × (x - 4)

3 × (4x + 1) = -5 × (x - 4)

Use the distributive property

12x + 3 = -5x + 20

Add 5x to each side of the equation

12x + 5x + 3 = -5x + 5x + 20

17x + 3 = 0 + 20

17x + 3 = 20

Subtract 3 from each side of the equation

17x + 3 - 3 = 20 - 3

17x + 0 = 17

17x = 17

Divide each side by 17

17x/17 = 17/17

x = 1

## Recent Articles

1. ### How To Find The Factors Of 20: A Simple Way

Sep 17, 23 09:46 AM

There are many ways to find the factors of 20. A simple way is to...

2. ### The SAT Math Test: How To Be Prepared To Face It And Survive

Jun 09, 23 12:04 PM

The SAT Math section is known for being difficult. But it doesn’t have to be. Learn how to be prepared and complete the section with confidence here.

100 Tough Algebra Word Problems.

If you can solve these problems with no help, you must be a genius!

Recommended