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Solving Linear Inequalities - Problem 1

Teacher/Instructor Carl Horowitz
Carl Horowitz

University of Michigan
Runs his own tutoring company

Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!

Solving linear inequalities; so how we solve linear inequalities is exactly the same as how we solve equations except we just have to be careful if we ever divide by a negative. So for this particular problem I’m going to do it two ways.

In general I always try to keep my x term to the positive side, so what that would mean for this case if we take this 3x over here our x is just going to be positive. So if we subtract the 3x over, subtract 3x from over here 2x minus 7 is greater than or equal to 9. Getting the x by itself add the 7 over 2x is greater than or equal to 16. And lastly solving this out divide by 2, x is greater than or equal to 8.

So keeping our x term positive we get x is greater than or equal to 8. I’m going to do the same problem but instead of bringing everything to the what side is this? Left side, we are going to bring to over to the right. So 5x minus 7 is greater than or equal to 3x plus 9. So this time we are going to bring everything to the other side. Subtract 5x this time and now we end up with a negative coefficient on our x term. Bringing the 9 over to the other side, -7 minus 9 subtracting is adding a negative, -16 is greater than or equal to -2x, lastly dividing by negative 2 we again end up with 8 and x. But the problem is is that we divide by negatives so we have to be careful to flip our sign. So this used to be a less than now turned to greater than or equal to.

So two different ways of doing the same exact problem, just making sure that if you divide by negative you switch your sign.

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