# Algebra Word Problem

• Jan 17th 2008, 09:12 AM
quartersworth
Algebra Word Problem
A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle?

I don't quite understand the correct process to go through when doing this, so could you show me how you get the answer? Thanks again guys!
• Jan 17th 2008, 09:15 AM
janvdl
Quote:

Originally Posted by quartersworth
A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle?

I don't quite understand the correct process to go through when doing this, so could you show me how you get the answer? Thanks again guys!

Rectangle 1:
• L = 4x

• W = x

Rectangle 2:
• L = 4x + 5

• W = x + 2

If the area = 270 then that means:

$(4x + 5)(x + 2) = 270$

Can you take it from here?
• Jan 21st 2008, 07:37 PM
quartersworth
Quote:

Originally Posted by janvdl
Rectangle 1:
• L = 4x

• W = x

Rectangle 2:
• L = 4x + 5

• W = x + 2

If the area = 270 then that means:

$(4x + 5)(x + 2) = 270$

Can you take it from here?

How would you finish it?
• Jan 21st 2008, 07:45 PM
Jhevon
Quote:

Originally Posted by quartersworth
How would you finish it?

expand the brackets on the left. (take each term in the first set of brackets and multiply every term in the second set)

by the way, janvdl made a slight error here, a minor oversight, i make them all the time. the question said "The area of the second rectangle is 270 square centimeters greater than the first"

thus, the right side of the equation should be: $4x^2 + 270$
• Jan 21st 2008, 08:42 PM
quartersworth
Quote:

Originally Posted by Jhevon
expand the brackets on the left. (take each term in the first set of brackets and multiply every term in the second set)

by the way, janvdl made a slight error here, a minor oversight, i make them all the time. the question said "The area of the second rectangle is 270 square centimeters greater than the first"

thus, the right side of the equation should be: $4x^2 + 270$

I guess the problem I am having is figuring out the end of the problem itself in my mind.
• Jan 21st 2008, 08:46 PM
Jhevon
Quote:

Originally Posted by quartersworth
I guess the problem I am having is figuring out the end of the problem itself in my mind.

show me what you've tried
• Jan 21st 2008, 09:06 PM
quartersworth
Quote:

Originally Posted by Jhevon
show me what you've tried

I don't even know where to begin.
• Jan 21st 2008, 11:48 PM
janvdl
Quote:

Originally Posted by janvdl
Rectangle 1:
• L = 4x

• W = x

Rectangle 2:
• L = 4x + 5

• W = x + 2

If the area = 270 then that means:

$(4x + 5)(x + 2) = 4x^2 + 270$

Can you take it from here?

Thanks for that spot Jhevon. Heck the things i miss sometimes.

$(4x + 5)(x + 2) = 4x^2 + 270$

$4x^2 + 8x + 5x + 10 = 4x^2 + 270$

$13x + 10 = 270$

$13x = 260$

$x = 20$