How To Solve Linear Inequalities

Say you have to go grocery shopping and you're on a budget. You want to buy pasta and bread for a large group, but you can't spend more than twenty dollars. In theory, you could buy only bread and no pasta, or lots of bread and only one box of pasta. How many different combinations of pasta boxes and loaves of bread could you buy? And how can you get the most of each for your money?

Problems like these are called ​linear inequalities​: equations whose graph is a line, but instead of using the equals sign, they use inequality symbols like > or <.

​To​ ​solve a linear inequality​ (or any equation), you have to find all the combinations of ​x​ and ​y​ that make that equation true.

You can solve linear inequalities algebraically or you can represent the solutions on a graph (or both!). Let's walk through some example problems together.

This process is ​almost​ the same as solving a linear equation, but with a key exception. Take a look at the problem below.

\(-4x – 6 > 12 – x\)

First, get all the ​x​-es on the same side of the "greater than" sign. Add ​x​ to both sides to cancel out the ​x​ on the right-hand side and only have ​x​ on the left.

\(- 4x (+ x) – 6 > 12 – x (+ x)\)
\(-3x – 6 > 12\)

Now add six to both sides:

\(-3x – 6 (+ 6) > 12 (+ 6)\)
\(- 3x > 18\)

So far this has been exactly like any linear equation. But now things are about to change up! ​When you divide both sides of an inequality by a negative number, you have to switch the direction of the inequality symbol​.

So for −3​x​ > 18, we're going to divide both sides by −3, and then we're going to flip the > sign to a < sign.

\(x < -6\)

How about graphing? Once again, the process is really similar to linear equations, but there's an important difference. Since you have to indicate ​all​ of the combinations of ​x​ and ​y​ that make an inequality true, you're going to graph the line like usual and then you're going to shade in the section of the graph that gives you the rest of the possible solutions.

For example, how would you graph the inequality ​y​ < 3​x​ + 6?

First, you would notice that the inequality is in ​slope-intercept form​, which means we can use the ​y​-intercept and the slope to quickly graph the line.

The ​y​-intercept is 6, so draw a point at (0, 6), then use the fact that the slope is 3 to go up three units and one unit to the right, then draw a point. Your point should be at (1, 9). To make a line neat and pretty, it's nice to get three points, so draw one more point by starting at (1, 9) and going up three, over one again. You'll get a point at (2, 12). Now draw a line by connecting the points.

Great! You just graphed the equality ​y​ = 3​x​ + 6, but remember the original equation is ​y​ < 3​x​ + 6. Use this simple trick to shade the correct portion of the graph: ​**when the inequality is in slope-intercept form, if you have ​y​ < , then shade in everything underneath the line. If you have ​y​ >, then shade in everything above the line.**​

But do double-check to make sure! When you shade in a whole section of the graph, it means that any of those points should make the equation true. Grab a random point that you've shaded in and plug ​x​ and ​y​ into the original inequality. If it works, you're good to go. If it doesn't, you need to double-check your graphing and/or your algebra.

One last thing: ​when you have > or <, the line on the graph needs to be dotted! When the inequality uses ≥ or​ ≤, ​the line must be solid.​ This shows whether or not the points on the line itself are included in the solution.

Solving a system of linear inequalities is very similar to solving systems of equations. ​Graphing​ is the easiest way to solve linear inequalities.

To graph a system of linear inequalities, graph your first inequality like you did above and shade in the areas above or below your line. Then graph the second inequality. Once again, you're going to shade in all sections of the graph that make the inequality true. Most of the time, there will be one area on the graph that you've shaded over twice! This is the ​solution​ to the system of inequalities, because it's ​the section of the graph where both of the inequalities are true​.