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Applied Linear Inequalities - Concept 11,877 views

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!

In Algebra II, we are sometimes asked to use linear inequalities to solve word problems. Solving linear inequalities word problems is very similar to solving word problems with linear equations. With word problems, we extract the relevant information and then set up a linear inequality to solve it.

Here we have a word problem talking about a linear inequality okay. Inequalities is very much the same idea as an inequality but instead of dealing with a specific number as your answer you're actually going to deal with a region okay.
So for this problem you score 87, 94, and 83 on your first three Math tests, what must you score on your fourth test to obtain an average of at least 90 okay? So the trick for this one is to remember how you calculate average, okay always average out you add up all the elements that you have together divided by the number that you're dealing with. So to find the average of these tests, we have a score of 87, 94 and 83. If we divide this by three we would get the average of our first three tests, but that's not what we're talking about we are talking about the average of our first four. We don't know what our fourth score is. Anytime we don't know what something is call it x. So our fourth test is the score of x to find the average of this four you divide by 4.
Okay and we want this to be at least 90, so we know we are comparing it to 90. Trick now here is we want we know that this has to be greater, the question is is it going to be greater than or greater than equal to. Look our problem it says at least, so 90 is okay. So that means we can have it equal to, if it's an average greater than 90 we would know that it couldn't be equal to or just be greater than okay.
So solving this out how we solve linear inequalities is pretty much exactly the same as equalities. The first one get rid of our denominator multiply both sides by 4 and these sides our 4 cancels leaving with just our numerator. And then 90 times 4 is 360, combining like terms 87+94+83 some bigger numbers so I'll use my calculator this ends up being 264+x has to be greater than or equal to 360. Subtract that 264 over, x has to greater than or equal to 96. Okay so if we get a 96, 97, 98, 99 or 100 our test average would be above 90.
So by using inequalities and average we're able to calculate our grade that we need to get in the average.

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