the perimeter of a rectangle is 38 inches, and the area is 70 square inches. find the length and width using one variable

Answer 1

Since the perimeter is 38 inches, we know that the length of the four sides together equals 38. Since it's a rectangle, we know that 2 of the sides are the same length (shorter) and the other 2 are also the same length (longer). So, two times the short side plus two times the long side is equal to the perimeter: 2x+2y=38

(Don't worry; I know they asked for one variable. I'm getting there.)

The area is just the long side times the short side, so: x*y=70

Now, you can pick which one of these you want to use to solve for one variable. You can use both equations to "solve" for x or y. For example:

x*y=70 2x+2y=38

x*y(1/y)=70(1/y) (1/2)(2x+2y)=(1/2)38

x*(y*1/y)=70/y (1/2)2x+(1/2)2y=38/2

x*1=70/y x+y=19

x=70/y x=19-y

Now pick whichever one you like better. Personally, I prefer x=19-y. Plug it into the other equation (NOT the one you used to find this x). In this case, that would be x*y=70. Since x=19-y, you can plug "19-y" into where "x" is in the equation.

x*y=70

(19-y)*y=70

Now, solve for y.

19y-y^2=70

19y-y^2+y^2=70+y^2

19y=70+y^2

19y-19y=70+y^2-19y

0=70+y^2-19y

0=y^2-19y+70

Now, the way I figured this out was by thinking about what can multiply to make 70. 7*10, right? But 7 and 10 won't add to make 19, so I split the even number (10) in half, which gives us 5, and multiply the other number by 2 (to balance taking half of 10) which gives us 14. This gives us 5 and 14, which do add to make 19.

0=(y-5)(y-14)

y-5=0 y-14=0

y=5 y=14

These are the two lengths, but if you want more proof you can plus them back into your original equations for x. Both formulas work.

x=70/y x=19-y

x=70/5 x=19-5

x=14 x=14

So, the lengths are 5 inches and 14 inches. Check your work:

Area: 5*14=70,

Perimeter: 5+5+14+14=38