*Contributor: Mason Smith. Lesson ID: 11240*

Whether you notice or not, you use inequalities in different ways. Solving them is possible if you start with one-step linear inequalities. Find out how with online practice and real-world examples!

categories

subject

Math

learning style

Visual

personality style

Lion

Grade Level

Middle School (6-8), High School (9-12)

Lesson Type

Quick Query

Q: Why did the parents think their little variable was sick?

A: The nurse said he had to be isolated!

Now that you have an understanding, from the first lesson in this *Inequalities* series, of what an inequality is and represents; it's time for you to learn to solve inequalities to get your final answer.

If you missed it or need a review, catch up in the right-hand sidebar under **Related Lessons**.

Remember, an inequality is just like an equation, in that anything you do to the left side of the equation, you must also do on the right side, so the statement remains balanced.

If an equation says, for example, x + 4 < 9 , for every value of x where the statement is true, the inequality is considered to be true. The maximum number where the statement would be true is just under five since, 4.9999999 + 4 is 8.9999999 which is < 9.* This means x can be all values less than five *non-inclusive *(without the final number, in this case, 5). (If we were to have the inequality x ≤ 5, then we would say that x is less than or equal to 5 [another way is to say less than 5 *inclusive*]).

*This follows from the idea that with numbers, we can go out as many decimal places (or 9s) as we'd like with our 0.9999, but it still won't be 1. This is a difficult idea in math that takes some time to wrap your head around, so no worries if it doesn't quite make sense. Just know that when we say x < 9, it can mean all the values up to x being 8.99999999... with the 9s repeating forever. It still will never be x = 9, since 8.9999 is not equal to 9.

Let's practice solving some inequalities with addition and subtraction:

x + 9 < 15

We just subtract 9 from both sides and we get our answer of x < 6.

d - 3 > -6

We just add 3 to both sides and get our answer of d > -3

But what if the inequality were slightly more difficult and had multiplication and division?

What values can p have if *2p > 8*?

P can be all numbers greater than 4, which we know by dividing 8 by 2 to get p by itself.

What values can r have in ^{2}/_{3 }r < 6?

Well, by multiplying by the reciprocal (^{3}/_{2}) we can find r < (^{18}/_{2}) or r < 9, simple right?

However there is one major exception to solving inequalities where they differ greatly from equations. With inequalities, if you divide or multiply by a *negative*, then you have to remember to switch the sign of the inequality.

So, if we have –2p > 8, when we divide by –2, we have p < -4 instead of p > -4.

Let's do a few examples to make sure we are clear on this idea:

Let's take the equation -3 ≤ ^{x}/_{-5}

When we multiply both sides by –5 to get x by itself, we end up with -5 * -3 ≥ x, or 15 ≥ x.

Note the direction of the inequality throughout the solving process.

-4x > 8

When we divide the whole equation by the –4, we get x < -2.

Now that we know how to solve simple inequalities using addition, subtraction, multiplication, and division, let's practice a bit.

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