I have a volume questions in regards to a box. pleaseeee help!?

Question

Volume: A box 10 inches high has a square base. The box was made 7 inches longer, and 2 inches narrower (or its width was decreased by 2 inches) while the height remained 10 inches. Give the original length of one side of the box if these changes increased the volume of the box by 30.

Answer 1

Let each side of base = x inches
V1 = 10x²
V2 = (x + 7)(x - 2)(10)
10x² + 30 = (x + 7) (x - 2)(10)
x² + 3 = x² + 5x - 14
5x = 17
x = 17 / 5
x = 3 . 4
Original base is a square of side 3.4 "

Answer 2

let x = original length of box
x+7 = new length of box
let y = original width of box
y - 2 is the new width

x = y since the original box had a square bottom
V1 = 10xy = Original volume of box
V2 = 10(x+7)(y-2) = New box volume
V1 + 30 = V2
10xy + 30 = 10(x+7)(y-2)
divide both sides by 10 to get:
xy + 3 = (x+7)(y-2)
since x=y then
x² + 3 = (x+7)(x-2)
x² + 3 = x² + 5x - 14
17 = 5x
x = 17/5 = 3.4 inches

Answer 3

V=x²10
V+30=(x+7)(x-2)10
10x²+30=10x²+50x-140
170=50x
x=3.4"

Answer 4

Imagine the box originally has the width (and length) is a

V = 10a^2

V' = 10(a+7)(a-2) = 10(a^2+5a-14) = V+30 = 10a^2+30

10a^2+50a-140 = 10a^2+30

50a - 170 = 0

a = 17/5 = 3.40inches

Answer 5

Your problem can be reduced to this equation.
(* means multiplication ** means .. raised to the power of)
Let X be the side of the box.
10 * X**2 +30 = (X+7)*(X-2) * 10
10 * (X**2 + 3) = (X+7)*(X-2) * 10
X**2 + 3 = (X+7)*(X-2)
= X**2 + 5*X - 14
3 + 14 = 5*X
17/5=X

Answer 6

The original area of the base is x²

The volume is 10*base area. So the increase of the base area in 30/10 = 3

(x+7)(x-2) = x²+3
x²+5x-14 = x²+3
5x-14=3
5x=17
x = 17/5=3.4
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Answer 7

Do your own homework.