Lesson Plan - Get It!
Q: How does a math teacher get a compound fracture?
A: She breaks her hAND.
In this Inequalities series, you have tackled a large assortment of inequalities.
Now, it is time to tackle the most difficult of all inequalities: the compound inequality!
If you need a review first, catch up in the right-hand sidebar under Related Lessons.
For a compound inequality, it is not just one inequality that is being solved, but multiple, and we must be extremely careful with the wording in the problem to ensure that we do not miss important information!
For a compound inequality, we must first split the inequality, then compare them together, accounting for the connecting word and or or. We will tackle "and" statements first.
For and statements, look for the place where the two inequalities overlap; that is the only place where we can have a solution.
For example, consider this compound inequality: x > 5 and x < 7. When we compare the two graphs, x > 5 is:
and x < 7 is:
So, the only place where both of the inequalities mathc is from 5 to 7, non-inclusive. We can rewrite that as
5 < x < 7, which means x must be greater than 5 but smaller than 7.
What if we have two equations that don't overlap? For example x > 3 and x < 1. We have no solution since a number can't be both larger than 3 and smaller than 1.
Often, we are given one equation with a hidden and, for example: 4 < x+2 < 8. This can also be rewritten as
4 < x+2 and x+2 < 8, which are much easier to solve.
We get 2 < x and x < 6, which we can put together again to get 2 < x < 6. So, x is between 2 and 6.
A simpler way to solve an "and" statement is to leave it in its original form; for example:
–5 < 2x+3 < 9.
To solve without splitting the inequality apart -- or doing things to both sides like in equations -- we must do it to all three parts. So, when we subtract 3, we must subtract 3 from both –5, 2x+3, and 9 to get –8 < 2x < 6. Now, we divide everything by 2 to get –4 < x < 6.
But wait! What if we have to divide or multiply by a negative, as in
–8 < -2x < 6? Instead of flipping the direction of the signs when we divide, we will flip the two outside numbers after you divide to get the right answer!
We end up with –3 < x < 4, which makes sense versus 4 > x.
Now, that we have an idea of how to solve "and" statements, let's move on to "or" statements.
For or statements, instead of the correct values being those where both inequalities overlap, as in and statements, for or statements, the value only has to work for one inequality. If both work, great, but we only need one.
Let's try –4 + a > 1 or –4 + a < -3.
Just solve for both (a > 5 or a < 1).
There is no way to combine an or statement.
Now that we know how to solve for and and or statements, let's practice solving and graphing them.