Can someone please help?

Question

Solve for x if the area of the following rectangle is 20 and the perimeter is 18.


I dont get it...

Answer 1

the length of the rectangle is 5. the width of the rectangle is 4.

5x4=20

5+5+4+4=18

hope this helps! :)

Answer 2

set up the two equations.
if the rectangle has length = x and width = y
then Area = 20 = xy
and Perimeter = 18 = 2x + 2y

xy = 20
2x + 2y = 18

solving the second equation for y
2y = 18 - 2x
y = 9 - x

substitute into xy = 20
x(9-x) = 20
9x - x^2 = 20 get this = 0 to solve as a quadratic
0 = x^2 - 9x + 20
0 = (x -4)(x -5)
x = 4 or x = 5

If x = 4 then y = 9 - 4 = 5

Rectangle is 4 by 5 in dimensions.

Answer 3

Let x be one side of the rectangle,

then

area of rectangle = x y = 20

perimeter = 2(x + y) = 18

substitute for y

2(x + y) = 2(x + 20/x) = 18

now solve for x

x+20/x = 9
x^2 + 20 = 9x
x^2 - 9x + 20 = 0
(x - 4)(x -5) = 0
x = 4 or 5

Answer 4

rectangle with sides a and b
area = ab = 20
perimeter = 2(a +b) =18

a+b = 18/2 =9
a +b =9 ---> b = 9 -a
area = ab = a(9 -a) =20
-a^2 +9a = 20
-a^2 +9a -20 =0

a^2 -9a +20 =0

Discriminant = (-9)^2 -4(20)(1) = 1

a1 = (9 -1)/2 = 4
a2 = (9 +1)/2 =5

dimensions of the rectangle are

a = 4 and b =5

Answer 5

let w = width, l = length.
set up 2 equations:
20 = l x w
18 = 2l + 2w = 2 (l + w)

isolate l in the second equation and get:
l = 9 - w
plug that into the 1st equation
20 = w (9 - w)
20 = 9w - w^2
w^2 - 9w +20 = 0
(w - 5)(w - 4) = 0
w = 4, 5

therefore the longer one is the length
so length = 5
width = 4

Answer 6

let x be the length of one side
(9-x) is the other side

x(9-x)=20 is the area
x^2 - 9x + 20 = 0
(x-4)(x-5) = 0
x = 4 or 5

Answer 7

You can't solve for "x" when you don't have an equation.

Answer 8

where is x?