# At a Glance - Solving Inequalities

When we solved equations, we found all values of the variable that made the equation true. Similarly, we can solve inequalities to find all values of the variable that make the inequality true. The idea is to get the variable all by itself on one side of the inequality sign so we can easily see the solutions to the inequality.

If you need to, you can always lure it across with hot wings. For some reason, they love those.

To isolate the variable, we do pretty much the same things we did to solve equations. For starters, if we add or subtract the same constant from both sides of an inequality, we get an equivalent inequality. We can also add or subtract the same number of copies of the variable from both sides of the inequality. All this stuff feels eerily familiar, right? Welcome to Punxsutawney

### Sample Problem

What are the solutions to the the inequality x + 2 < 2?

To solve this bad boy, we subtract 2 from both sides of the inequality.

x + 2 < 2
x < 2 – 2
x < 0

This gives us x < 0 as the solutions to the inequality. Solutions are plural, remember, because x can be anything negative. Like your constantly grouchy Uncle Melvin. What's his deal?

We can also represent the solutions by drawing the number line picture that represents the inequality x < 0.

### Sample Problem

What are the solutions to 3x < 9?

Fun fact: we can also multiply or divide both sides of an inequality by the same positive number. You know us—we'll do anything because we can. In this case, let's divide both sides by 3 to get x by itself.

3x ÷ 3 < 9 ÷ 3
x < 3

So our set o' solutions is x < 3: anything smaller than 3 will do the trick.

Be Careful: We can multiply or divide both sides of an inequality by the same positive number. However, if we multiply both sides of an inequality by a negative number, weird things happen, which we'll demonstrate to you now. Some paranormal activity-type business is about to go down in here.

Here's an example with numbers instead of variables, which will more clearly illustrate what's going on. Pick your favorite inequality between two consecutive whole numbers. We'll use 5 < 6.

We know 5 is less than 6 because 5 is to the left of 6 on the number line. You also know this information because it's now only 5 months until your birthday instead of 6. Almost here. Huzzah!

Notice that 5 is closer to 0 than 6 is. Because both 5 and 6 are on the positive side of the number line, whichever number is closer to 0 is smaller. If we multiply each of the numbers 5 and 6 by -1, we're reflecting them across 0 on the number line, so 5 gets sent to -5 and 6 gets sent to -6. Airfare included and all expenses paid. Good thing, too. They really needed to get away.

Notice that -5 is still closer to 0 than -6. Since we're now looking at two numbers on the negative side of the number line, whichever number is closer to 0 is larger. When we compare -5 and -6, we see that -5 > -6. Apparently, -5 ate more than -6 did on this little getaway of theirs.

The moral of the story is that if we multiply both sides of an inequality by a negative number, we also need to switch the direction of the inequality sign. That's important enough to say again: if we multiply or divide both sides of an inequality by a negative number, we also need to switch the direction of the inequality sign. It's actually important enough to say a third time, but we sense we're losing you, so we'll stop.

It's even easier to see what's going on if one number is negative and one is positive. We know -2 < 1. If we multiply both sides of this inequality by -2, we're sending -2 to 4, and 1 to -2. Multiplying a value by a negative number reflects that value across 0 on the number line. Therefore, we also need to switch the direction of the inequality. Here we go.

-2 < 1
-2(-2) < 1(-2)
4 > -2

See how we swapped that "<" symbol out for a ">"? The resulting inequality is 4 > -2.

Now that we have an idea of what's going on, we'll put the variables back in and give it a try.

### Sample Problem

Solve the inequality -x < 3.

The Rock Star Way:

Multiply both sides of the inequality by -1 and switch the direction of the inequality.

x > -3

The Groupie Way (aka the Wrong Way):

If we multiply both sides of the inequality by -1 but forget to switch the direction of the inequality, here's what we'll get:

x < -3

We should be able to tell by looking that there's something wrong here. If all values less than -3 are solutions, then -4 should be a solution. Because -4 is not a solution to the original inequality, we must have missed something along the way.

Since inequalities tend to have more solutions than equations, it's not quite as easy to check our answers to inequalities as it is to check our answers to equations. However, we can still try a few of the numbers we found and make sure they work in the original inequality. We can also try a couple of numbers that we didn't find as solutions and make sure those don't work. With all this plugging in, we'll probably need a surge protector.

#### Example 1

 Solve the inequality x – 4 < 7 and show the solutions using a number line.

#### Example 2

 Solve the inequality 4x + 2 ≤ 3x + 10.

#### Example 3

 Solve the inequality 3(y – 1) > 2(4 + y).

#### Example 4

 Solve the inequality -4(x – 1) < 2x + 3.

#### Exercise 1

Solve the inequality z + 5 ≥ 3 and represent its solutions using a number line.

#### Exercise 2

Solve the inequality 4(x – 1) < 3(x + 1) and represent its solutions using a number line.

#### Exercise 3

Solve the inequality 2x + 3 ≤ x and represent its solutions using a number line.

#### Exercise 4

Solve the inequality 5(x + 2) – 3x ≤ 4 + 2x + 3(x – 1). Write the solution as an inequality, and show the solution on a number line.

#### Exercise 5

Solve the inequality -12x < 144. Write the solution as an inequality, and show the solution on a number line.

#### Exercise 6

Solve the inequality 5[x + 3(1 – x) + 2] ≥ 7. Write the solution as an inequality, and show the solution on a number line.

#### Exercise 7

Solve the inequality 7y + 2 ≤ 65. Write the solution as an inequality, and show the solution on a number line.

#### Exercise 8

Solve the inequality z + z[3 – 2(1 + z)] < 5[1 + 2(z – 3)] – 2z2. Write the solution as an inequality, and show the solution on a number line.