Solving Multi-step Inequalities

Contributor: Mason Smith. Lesson ID: 11241

How do you eat an elephant? One bite at a time. Same thing with multi-step inequalities: you solve them one step at a time. Learn the method with online practice, graphs, and real-world word problems!


Algebra I, Algebra II, Pre-Algebra

learning style
personality style
Grade Level
High School (9-12)
Lesson Type
Quick Query

Lesson Plan - Get It!

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Q: What did the doctor say to the multi-step inequality?

A: I can solve your problem with a few operations.

Solving one-step inequalities is a great first step, but what happens when inequalities begin to get more difficult?

(If you need a review of the earlier Inequalities series lessons, catch up in the right-hand sidear under Related Lessons.)

  • After all, a simple inequality can cover a simple problem, but in the real world, we rarely encounter a problem that can be so easily tackled, so how do we take on multi-step inequalities?

The answer to that is simple; one at a time!

Just remember, we are trying to get the variable (x) by itself, and you can work through an inequality exactly as you would work through an equation, following the same steps.

The trick for both of them is to get the variable by itself, then solve for anything attached to it.

Also remember: whenever we multiply or divide by a negative number, we need to flip the direction of the inequality.

Let's try some practice! Solve the inequality with me and then we will graph it:

160 + 4f ≤ 500

First, let's get the f by itself by subtracting 160 from both sides; that leaves:

4f ≤ 340

Now we can divide by 4 to get our final answer:

f ≤ 85

When we graph it, it looks like this:

graph 1

Let's try another one that is more difficult:

7 - 2t ≤ 21

First, let's bring the 7 over by subtracting to get:

-2t ≤ 14

Now the tricky bit: When we divide by –2, then we have to switch the direction of the inequality to get:

t ≥ -7,

which we can easily graph. Use a closed circle at 7, since the inequality is ≥, then shade all numbers greater than 7, since t must be larger than 7.

graph 2

Let's work through a word problem before we move on to the practice section:

To win the ribbon for heaviest pumpkin, the average weight of Susan's two pumpkins must be more than 419 kg. One of her pumpkins weighs 487 kg, so what is the least number of kilograms that Susan's pumpkin can weigh?

The average of two things would be:  a + b 
We know one pumpkin weighs 487 kg, so we have:  487 + p 

They must be greater than 419, which means we will use (>).

So our inequality will look like this:  487 + p  > 419


We solve this by multiplying by 2 on both sides to get 487 + p > 838 which, when we subtract 487 from both sides, leaves us with the inequality p > 351. The pumpkin must weigh more than 351 kg for Susan to win.

Using your knowledge of how to solve inequalities, taking your time, and being careful about traps, you can tackle any multi-step inequality you may encounter, so let's practice this in the next section.

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