Question 158038
This is not a mixture problem.

This is a geometry word problem.

A painting canvas has a length 6 inches longer than its width. If the area is 112 in^2, what are the dimensions of the canvas?

Let width = x

Let length = x + 6

The area given is 112 inches^(2).

The area of a rectangle (this canvas is a rectangle) is given by the formula
A = L*W

We plug and chug.

112 = (x + 6)(x)

When I multiply the right side, a quadratic will be the result.

112 = x^2 + 6x

Subtract 112 from both sides and equate the entire polynomial to zero.

x^2 + 6x - 112 = 0

We now factor the left side of this quadratic equation.

14 times -8 = -112

So, the factor becomes (x + 14) (x - 8).

Set each factor to equal 0 and solve for x.

x + 14 = 0

x = -14....This answer for x is REJECTED because length CANNOT be negative.

x - 8 = 0

x = 8

We now know the value of x.

Since the width = x and I just found the value of x, the width is 8 inches.

The length = x + 6

The length = 8 + 6

The length = 14 inches

The dimensions of the canvas are:

width = 8 inches 

length = 14 inches