candygirl200:

The length of a rectangle is 4 meters less than twice its width. The area of the rectangle is 70 m2 . Find the dimensions of the rectangle.

2 weeks ago
sammixboo:

Well, the formula to find the area of a rectangle is $$\color{red}{\sf wl = A}$$, YOu were given a decent amount of information in the problem that we need to plug into the formula, such as the area, length, and width. The area is 70m$$\sf ^2$$, while the length is 2w-4, and the width is w. Let's plug it in. $$\sf \color{red}{w(2w-4) = 70}$$. Do you know what to do next?

2 weeks ago
candygirl200:

distribute the property and you would get 2w^2-4w=70

2 weeks ago
sammixboo:

Right, do you know the next step?

2 weeks ago
candygirl200:

move 70 to make it equal to 0

2 weeks ago
candygirl200:

2w^2-4w-70=0

2 weeks ago
candygirl200:

then u put it in a factor method which you will get 7 and -5 and u cannot have a negative distance so you make it a positive 5 and i think your answer will be 7 and 5 m is the dimensions right?

2 weeks ago
sammixboo:

You are definitely on the right track. You're right that we can't have a negative number for a distance, so therefor we will just toss -5 out the window. Our only option left, which is 7, will be our width. We still need to find the length. Do you know how we do that?

2 weeks ago
candygirl200:

yes make the 5 a positive

2 weeks ago
sammixboo:

No, we have to completely forget about the 5 since it is a negative number. We know that our width is 7, so now we have to find the length. The length is the for less than twice the width, so it's $$\sf \color{red}{2w-4=length}$$. If we plug in 7 for w, then we get $$\sf \color{red}{2(7)-4=length}$$, which is?

2 weeks ago
candygirl200:

10

2 weeks ago
sammixboo:

Right! So, our dimensions would be 7 for the width, and 10 for the length

2 weeks ago
candygirl200:

thanks!

2 weeks ago
sammixboo:

NO problem!

2 weeks ago