The length of a rectangle is 5 yd more than double the width, and the area of the of the rectangle is 42 yd to the second power . What is the dimensions of the rectangle

The answer to your question is l = 12 and w = 7/2 or 3.5

Step-by-step explanation:

Data

length = l

width = w

Area = 42 yd²

Process

1) Write equations that help to solve this problem

Area = length x width = l x w

Length = 2width + 5  or   l = 2w + 5

2) Substitution

                       42 = lw                               Equation l

                          l = 2w + 5                        Equation ll

                       42 = (2w + 5)w                   Substitute equation ll in l

3.- Expand

                       42 = 2w² + 5w

                       2w² + 5w - 42 = 0

Solve the equation by factoring

-Multiply 2 by 42

                             2 x 42 = 84

-Find the prime factors of 84

                         84 = 2² 3 7

-Find two numbers that added gives +5 using the prime factors

 These numbers are +12 and -7    

-Substitute the values in the equation

                         2w² -7w + 12w - 42

-Factor by grouping

                        w(2w - 7) + 6(2w - 7)                  

                           (2w - 7)(w + 6)

-Find w

                    2w₁ - 7 = 0             w₂ + 6 = 0

                      w₁ = 7/2               w₂ = -6   This value is discarted because

                                                                  there are no negative values

-Find l

                     l = 2(7/2) + 5

                     l = 7 + 5

                     l = 12

i believe it's a.

i'm not sure on this one, most likely correct.

idk what the heck he answer is, i’m just putting this here so i finish my account