Grade eight math. hi, i was trying to study for my math exam and come across this problem, but i have no idea how to solve it. i am not factoring yet. just simplification and basic solving. i know the answer but would like an explanation. 25 points! you!

1. First, let's call lengths AD and BC length y. We know that lengths AB and DC are 4x cm.

Now, let us find the formula for the perimeter of the rectangle given these lengths:

P = 4x + 4x + y + y

P = 8x + 2y

From the question, we know that P = 70 cm, thus if we substitute this into the equation above, we get:

70 = 8x + 2y

We can further simplify this to get:

35 = 4x + y

2. The equation for the area of the rectangle given the lengths defined would be:

A = 4x*y

A = 4xy

From the question, we know that the area is 8x^2 + 20x. Thus, if we substitute this into the equation above, we get:

8x^2 + 20x = 4xy

3. To recap, we now have two equations:

a) 35 = 4x + y

b) 8x^2 + 20x = 4xy

What we need to do is to have only x as the variable in the second equation for the area of the rectangle - this will require substituting y with an expression containing x. We can do this from equation a) by rearranging it:

35 = 4x + y

35 - 4x = y (Subtract 4x from both sides)

Now that we know that y = 35 - 4x, we can substitute this value into equation b):

8x^2 + 20x = 4xy

8x^2 + 20x = 4x(35 - 4x)

8x^2 + 20x = 140x - 16x^2

24x^2 + 20x = 140x (Add 16x^2 to each side)

24x^2 = 120x (Subtract 20x from each side)

24x = 120 (Divide each side by x)

x = 5 (Divide each side by 24)

4. Now that we know length x, we can figure out the area in one of two ways:

1) Find length y by substituting x = 5 into y = 35 - 4x:

y = 35 - 4(5)

y = 35 - 20

y = 15

The area is given by 4xy, therefor the area is:

A = 4(5)(15)

= 300 cm squared

2) Alternatively, we can substitute x = 5 into the equation we were given for area in terms of x:

A = 8x^2 + 20x

A = 8(5)^2 + 20(5)

A = 8*25 + 100

A = 200 + 100

A = 300 cm squared

a negative exponent is translated as follows:

so, we have