Perimetr of a rectangle

[liesl123]

The perimeter of a rectangle is 64 m. If the width were doubled and the legnth were increased by 8 m, the perimeter would be 106 m. What are the length and width of the rectangle?

I am at a loss I am not sure how to start this problem.

I am guessing ...

2W + 2L =64
2(2W) + (2L+8) =106

But I am not sure how to complete this problem. I was thinking I should solve for W then solve for L.

 

[srmichael]

The perimeter of a rectangle is 64 m. If the width were doubled and the legnth were increased by 8 m, the perimeter would be 106 m. What are the length and width of the rectangle?

I am at a loss I am not sure how to start this problem.

I am guessing ...

2W + 2L =64
2(2W) + (2L+8) =106

But I am not sure how to complete this problem. I was thinking I should solve for W then solve for L.

Close.

2W + 2L = 64
2(2W) + 2(L+8) = 106

After simplification you get:

2W + 2L = 64
4W + 2L = 90

From here you can either use substitution or, in this case the easier way, elimination of variables by subtracting the two equations which would eliminate L and you can solve for W then solve for L.

 

[liesl123]

Thanks

Here is my work

4W + 2L = 64
2(2W) +2(L+8) = 106

4W + 2L = 64
4W+2L+16=106
4W+2L=90

Now subtract the two statements
4W+2L=90
2W-2L=64
2W=26
W=13

Now solve for L by plugging in @
2(13) 22L=64
26*2L=64
2L=38
L=19

Yeah I understand thanks so much

 

[TchrWill]

The perimeter of a rectangle is 64 m. If the width were doubled and the legnth were increased by 8 m, the perimeter would be 106 m. What are the length and width of the rectangle?

I am at a loss I am not sure how to start this problem.

I am guessing ...

2W + 2L =64
2(2W) + (2L+8) =106

But I am not sure how to complete this problem. I was thinking I should solve for W then solve for L.

Exactly! Now how would you go about it?

First, correct (2L + 8) to (2L + 16)
4W + 2L + 16 = 106
2W + 2L = 64

Subtracting, 2W + 16 = 106 or 2W = 42 making W = 21

I'll let you finish it.