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## Algebra 1

### Course: Algebra 1>Unit 2

Lesson 6: Compound inequalities

# A compound inequality with no solution

Sal solves the compound inequality 5x-3<12 AND 4x+1>25, only to realize there's no x-value that makes both inequalities true. Created by Sal Khan and Monterey Institute for Technology and Education.

## Video transcript

Solve for x, 5x - 3 is less than 12 "and" 4x plus 1 is greater than 25. So let's just solve for X in each of these constraints and keep in mind that any x has to satisfy both of them because it's an "and" over here so first we have this 5 x minus 3 is less than 12 so if we want to isolate the x we can get rid of this negative 3 here by adding 3 to both sides so let's add 3 to both sides of this inequality. The left-hand side, we're just left with a 5x, the minus 3 and the plus 3 cancel out. 5x is less than 12 plus 3 is 15. Now we can divide both sides by positive 5, that won't swap the inequality since 5 is positive. So we divide both sides by positive 5 and we are left with just from this constraint that x is less than 15 over 5, which is 3. So that constraint over here. But we have the second constraint as well. We have this one, we have 4x plus 1 is greater than 25. So very similarly we can subtract one from both sides to get rid of that one on the left-hand side. And we get 4x, the ones cancel out. is greater than 25 minus one is 24. Divide both sides by positive 4 Don't have to do anything to the inequality since it's a positive number. And we get x is greater than 24 over 4 is 6. And remember there was that "and" over here. We have this "and". So x has to be less than 3 "and" x has to be greater than 6. So already your brain might be realizing that this is a little bit strange. This first constraint says that x needs to be less than 3 so this is 3 on the number line. We're saying x has to be less than 3 so it has to be in this shaded area right over there. This second constraint says that x has to be greater than 6. So if this is 6 over here, it says that x has to greater than 6. It can't even include 6. And since we have this "and" here. The only x-es that are a solution for this compound inequality are the ones that satisfy both. The ones that are in the overlap of their solution set. But when you look at it right over here it's clear that there is no overlap. There is no x that is both greater than 6 "and" less than 3. So in this situation we have no solution.