SOLUTION: Solve algebraically using one variable: The perimeter (distance around) of a rectangle is 38 inches and its area (length times width) is 70 square inches. Find the length an

Algebra ->  Equations -> SOLUTION: Solve algebraically using one variable: The perimeter (distance around) of a rectangle is 38 inches and its area (length times width) is 70 square inches. Find the length an      Log On


   

Question 87128: Solve algebraically using one variable: The
perimeter (distance around) of a rectangle is 38
inches and its area (length times width) is 70
square inches. Find the length and the width of
this rectangle.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
The formula for the perimeter is
P=2W%2B2L <---------where W is the width and L is the length
Since we know the perimeter is 38 in, we can say this:
38=2W%2B2L
The formula for the area is:
A=W%2AL
Since we know the area is 70 in, we can say this:
70=W%2AL

So we have 2 equations:

38=2W%2B2L
70=W%2AL


Now lets solve for one variable:
70=W%2AL+ Start with any given equation (I'm choosing the 2nd equation)
70%2FW=L+ Divide both sides by W to solve for L (you can choose to solve for any variable)

Since L=70/W, we can replace every L with (70/W) for the first equation like this:

38=2%2870%2FW%29%2B2W Plug in L=70%2FW

38=140%2FW%2B2W Multiply

0=140%2FW%2B2W-38 Subtract 38 from both sides

W%2A0=W%28140%2FW%2B2W-38%29 Multiply both sides by W

0=140%2B2W%5E2-38W Distribute

0=2W%5E2-38W%2B140 Rearrange the terms

Now lets use the quadratic formula to solve for W

Starting with the general quadratic

aW%5E2%2BbW%2Bc

the general form of the quadratic equation is:

W+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29

So lets solve 2%2AW%5E2-38%2AW%2B140

W+=+%2838+%2B-+sqrt%28+%28-38%29%5E2-4%2A2%2A140+%29%29%2F%282%2A2%29 Plug in a=2, b=-38, and c=140



W+=+%2838+%2B-+sqrt%28+1444-4%2A2%2A140+%29%29%2F%282%2A2%29 Square -38 to get 1444



W+=+%2838+%2B-+sqrt%28+1444%2B-1120+%29%29%2F%282%2A2%29 Multiply -4%2A140%2A2 to get -1120



W+=+%2838+%2B-+sqrt%28+324+%29%29%2F%282%2A2%29 Combine like terms in the radicand (everything under the square root)



W+=+%2838+%2B-+18%29%2F%282%2A2%29 Simplify the square root



W+=+%2838+%2B-+18%29%2F4 Multiply 2 and 2 to get 4

So now the expression breaks down into two parts

W+=+%2838+%2B+18%29%2F4 or W+=+%2838+-+18%29%2F4

Lets look at the first part:

W=56%2F4 Add the terms in the numerator
W=14 Divide

So one answer is
W=14
Now lets look at the second part:

W=20%2F4 Subtract the terms in the numerator
W=5 Divide

So another answer is
W=5

So our solutions are:
W=14 or W=5

Notice when we graph 2%2AW%5E2-38%2AW%2B140 we get:

+graph%28+500%2C+500%2C+-5%2C+24%2C+-5%2C+24%2C2%2Ax%5E2%2B-38%2Ax%2B140%29+

and we can see that the roots are W=14 and W=5. This verifies our answer

So this means the width can be either 5 or 14. Lets solve for each length:

Lets find the length using W=5
70=5%2AL Plug in W=5
14=L Divide both sides by 5
So the length is 14 when the width is 5


Lets find the length using W=14
70=14%2AL Plug in W=14
5=L Divide both sides by 14
So the length is 5 when the width is 14 (notice the length and width switched places)

Check:
Lets check the area formula:
70=5%2A14 Plug in W=5 and L=14
70=70 works

70=14%2A5 Plug in W=14 and L=5
70=70 works


Lets check the perimeter formula:
38=2%2A5%2B2%2A14 Plug in W=5 and L=14
38=38 works

38=2%2A14%2B2%2A5 Plug in W=14 and L=5
38=38 works

So the solutions work in every equation.
So to recap, the widths are
W=5 or W=14

and the lengths are
L=5 or L=14

note: the length and the width cannot be the same number