Help On Algebra I Questions!

Question

Write an expression that describes the area of the rectangle if the length is 2x+3 and the width x+5.

A. (2x+3)(x+5)

B. 2((2x+3)(x+5)

C. 2(2x+3) + 2(x+5)

D. 2(2x+3) + (x+5)

If the area of the rectangle is 72 squared inches determine the value of x.

Answer 1

Answer :

Option A

x = 3

Explanation :

Given Length = 2x + 3 , and Width = x + 5

Area = Length x Width

= ( 2x+ 3 )(x + 5).

Given Area 72 squared inches.

Determine the value of x.

Area = ( 2x + 3)(x + 5)

72 = (2x + 3)( x+ 5 )

72 = 2x² + 10x + 3x + 15

72 = 2x² + 13x + 15

2x² + 13x + 15 - 72 = 0

2x² + 13x - 57 = 0

2x² - 6x + 19x - 57 = 0

2x( x - 3) + 19 ( x - 3) = 0

(2x + 19 ) ( x - 3) = 0

Therefore,

2x + 19 = 0 , x - 3 = 0

2x = - 19 , x = 3

x = -19 / 2 , x = 3

[ Substitute both the values of x in Length and width expression ,and check which value of x is suitable for the equation ]

When x = -19/2

Length = (2x + 3 ) = 2 ( -19/2) + 3 = - 19 + 3 = -16

Since measure of a length cannot be negative, x can't be -19 / 2.

Therefore, x = 3

Answer 2

a) is correct

area = length times width

A=LW = (2x+3)(x+5) = 2x^2 +13x +15

72= 2x^2 + 13x +15

2x^2 +13x - 57 = 0

factor or use the quadratic formula if you can't figure out the factors. try integer factors of -57,

(2x+19)(x-3) =0

x=3 or -19/2

2x+3 = 9

x+5 = 8

or

2x+3 = -19+3 = -16

x+5 = -19/2 +5 = -29/2

ignore the negative solutions, unless you have a twilight zone rectangle