# Frustrating word problems

• Apr 1st 2008, 02:43 AM
Fails_at_Math
Frustrating word problems
Hello, I am new to these forums. After hitting a wall on my take-home quiz, I am on the hunt for some assistance. Our professor told us we can use any means necessary to get help, so I am doing just that! Always been terrible at word problems, algebra in general for that matter. Have two problems here that left me with nothing but a blank paper and a nasty migraine!

1. The area of a rectangle is 80 meters^2 (squared). The length is six more than twice its width. Find the length and width.

2. The profit P, generated for a video store is given by the equation p = x^2 - 15x - 50, where X indicates the number of videos sold. How many videos must be sold for the profit to be \$200?

• Apr 1st 2008, 03:09 AM
mr fantastic
Quote:

Originally Posted by Fails_at_Math
Hello, I am new to these forums. After hitting a wall on my take-home quiz, I am on the hunt for some assistance. Our professor told us we can use any means necessary to get help, so I am doing just that! Always been terrible at word problems, algebra in general for that matter. Have two problems here that left me with nothing but a blank paper and a nasty migraine!

1. The area of a rectangle is 80 meters^2 (squared). The length is six more than twice its width. Find the length and width.

2. The profit P, generated for a video store is given by the equation p = x^2 - 15x - 50, where X indicates the number of videos sold. How many videos must be sold for the profit to be \$200?

1. Solve 80 = x(2x + 6) => 2(x - 5)(x + 8) = 0 for x, where x is the width. Keep the answer that makes sense.

2. Solve 200 = x^2 - 15x - 50 => (x - 25)(x + 10) = 0 for x. Keep the answer for x that makes sense.
• Apr 1st 2008, 03:21 AM
angel.white
Quote:

Originally Posted by Fails_at_Math
1. The area of a rectangle is 80 meters^2 (squared). The length is six more than twice its width. Find the length and width.

First you need to write a formula that relates length, width, and area. Fortunately, this is a common formula: A=L*W

We know A since it is given, so plug that into the formula:

\$\displaystyle 80 = L*W\$

Now we have two variables, but lets get it down to one so that we can actually solve for something. How do length and width relate to eachother? Well we were told that length is 6 more than twice the width, so L = 6+2W

Now we can substitute this value in for L

\$\displaystyle 80 = (6+2W)*W\$

\$\displaystyle 80 = 6W +2W^2\$

\$\displaystyle 40 = W^2 + 3W\$

\$\displaystyle 0 = W^2 + 3W - 40\$

Solve by factoring (Hint: -40 = -5*8)
This will give you two solutions (use the positive answer since you can't have negative width).
Then plug that answer into the equation again and solve for length.
Quote:

Originally Posted by Fails_at_Math
2. The profit P, generated for a video store is given by the equation p = x^2 - 15x - 50, where X indicates the number of videos sold. How many videos must be sold for the profit to be \$200?

p is profit = 200 so plug that in:

\$\displaystyle 200 = x^2 -15x - 50\$

\$\displaystyle 0 = x^2 -15x - 250\$

Now you just factor it (hint: -250 = -25 * 10)
This will give you two solutions, evaluate which makes the most sense (hint: it doesn't make sense to sell negative videos)
• Apr 1st 2008, 03:31 AM
Fails_at_Math
I see now how it relates back to the rest of the problems we have been doing in class. Just hit a wall trying to put the words into equation form, but I guess that's the real challenge of word problems. Thanks MHF!
• Apr 1st 2008, 03:42 AM
angel.white
Quote:

Originally Posted by Fails_at_Math
Just hit a wall trying to put the words into equation form, but I guess that's the real challenge of word problems. Thanks MHF!

For me the real challenge of word problems is that I can't read :/

I don't even know what the question was, I just copied Mr. Fantastic's answer and changed the variable letters. I don't even know what you just posted, I'm just randomly hitting keys right now, I don't even know what I'm typing! (Surprised)

• Apr 1st 2008, 05:01 AM
mr fantastic
Quote:

Originally Posted by angel.white
For me the real challenge of word problems is that I can't read :/

I don't even know what the question was, I just copied Mr. Fantastic's answer and changed the variable letters. I don't even know what you just posted, I'm just randomly hitting keys right now, I don't even know what I'm typing! (Surprised)