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Solving Linear Inequalities - Problem 2

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 for this particular example we solve it just as we would any equation. Distribute out all of our numbers and then try to isolate our x terms. So distribute this 4 through, 4x plus 12 minus x and then distribute this -3 through -3 times 2 is -6, -3 times -x is plus 3x.

Now we combine like terms, so 4x minus x plus 12 stays the same, and just for ease of use we can rearrange this put our x terms in front 3x minus 6. Now we want to isolate our x’s if we subtract 3x from both sides, what ends up happening? They actually both disappear. So subtract 3x, subtract 3x we end up with 12 is greater than or equal to -6.

We lost our x in the process but what does this mean? We end up with a true statement, 12 is in fact greater than or equal to -6, so what this actually means is that x doesn’t make a difference. This will work for any value of x we plug in.

We look at it right here 3x plus 12 is just a number 3x minus 6 is going to be less than that number. That x doesn’t actually make any difference at all so we can plug in whatever we want this will always be a true statement.

So depending on how your teacher wants you might have to right this in different ways, you can either right all real numbers. This funny little double r basically means x can be anything or you could right your interval notation -3 to infinity. X can be whatever it wants this statement will always hold true.

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