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# Applied Linear Equations: Geometry Problem - Problem 1

###### 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!

Using linear equations to solve a geometric problem. In this case we are dealing with a rectangle. We’re given some information about the sides and some information about the perimeter and asked to solve it. In general it’s easier to first interpret what a word question is asking, if you draw a picture. It’s always sort of good to sort of visualize what’s happening.

Here we’re given a rectangle, we have a width and we have a length and we know that these two add up, perimeter of the entire thing is going to add up to 70cm. The question that we actually need figure out is the length is 5 more than twice its width. This is the statement relating length and width, so we need to actually figure out how we can relate that into an equation.

So length is 5 more than twice its width. What I first see is that the length is actually longer than the width. We know that we have length and width and they’re somehow related. And we know that the length is, first is see twice its width, so it’s related to twice its width. If we have width of 5 that means this length is 10, multiply the width by 2 in order to get length. So twice width, this tells us that twice the width is equal to length, but our equation says 5cm more than twice its width. We need to add 5 to one of these sides. If it's 5 more than twice, I mean this is going to be over here with the length and you can always check your work, once you get in the middle of an equation make sure it makes sense.

So say width is 10, we plug in 10 for width, 10 times 20 our length would be 25. Is our length in fact 5cm more than twice its width? Twice its width would be 20 plus 5, 25. We now have a relationship that actually checks out. Our length we know is 2w plus 5. That’s our key relationship.

Perimeter, so we have this relationship we also we’re looking for the perimeter of a rectangle. Perimeter of a rectangle is, we have two widths and two lengths. But we also know that length is equal to this guy over here, plug that in right there, perimeter is equal to 2(2w plus 5) plus 2w. And our last little piece that we need to use is our perimeter is 70. Plugging 70 in we now have a linear equation with w that we can solve for.

Again the hardest part of these problems is taking a word problem, getting that equation. Once we’re here we just use our rules of math. Distribute out this 2, 70 is equal to 4w plus 10 plus 2w. Combine like terms, 70 is equal to 6w plus 10, solving for w, subtract 10 over, 60 is equal to 6w. Finish it up divide by 6, w is equal to 10.

Again the question is asking for the length and the width, so we found our width, plug that into our relationship here plug 10 in, 2 times 10 is 20 plus 5, 25. Somehow I actually ended up choosing the right number in our relationship. That probably isn’t going to work all the time but we managed to figure it out, creating our equation, solving it out for length and width.

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###### Carl Horowitz

B.S. in Mathematics University of Michigan

He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his step-by-step explanations are easy to follow.

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