SOLUTION: A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square cent

Question 285614: A rectangle is 4 times as long as it is wide. A second rectangle is 5 centimeters longer and 2 centimeters wider than the first. The area of the second rectangle is 270 square centimeters greater than the first. What are the dimensions of the original rectangle?
I would like to see how to solve the problem in steps, if possible. Thank you!
Found 2 solutions by JBarnum, MathTherapy:
Answer by JBarnum(2146) (Show Source): You can put this solution on YOUR website!
, because w=4l. A=LW is the first rectangle
"A rectangle is 4 times as long as it is wide."
"A second rectangle is 5 centimeters longer and 2 centimeters wider than the first."
"The area of the second rectangle is 270 square centimeters greater than the first" this is the second rectangle.
so now wee need to substitute to be able to solve for

use the quadratic formula to solve for L, below is the solver to show steps, but for now or but the lengths cant be negative so
now we can find all the other lengths of the 2 rectangles
first rectangle:

___________

___________

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=4481 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 6.24253398558978, -10.4925339855898. Here's your graph:

Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!

Let the width of the 1st rectangle be W. Then its length = 4W, since its length is 4 times its width

The width and length of the 2nd rectangle are W + 2, and 4W + 5, respectively, since the second rectangle is 2 centimeters wider and 5 centimeters longer than the 1st.

The area of the 1st rectangle is then: W*4W, or , and the area of the 2nd rectangle is (W + 2)(4W + 5 ), or .

Since the area of the 2nd rectangle is 270 square centimeters greater than the area of the 1st rectangle, then we�ll have:

- 13W = - 260

Now, since W, or width = 20, then the dimensions of the 1st rectangle are: W, or width = cm, and L, or length = cm ------> 4W, or 4(20)
Check:
Width of 1st rectangle = 20 cm
Length of 1st rectangle = 4(20) = 80 cm

Area of 1st rectangle = 20 * 80 = 1,600 sq centimeters

Width of 2nd rectangle = 22 cm (20 + 2)
Length of 2nd rectangle = 85 cm (4*20 + 5)

Area of 2nd rectangle = 22 * 85 = 1,870 sq centimeters

Area of 2nd rectangle (1,870 sq centimeters) is 270 sq centimeters greater than area of 1st rectangle (1,600 sq centimeters).

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