The length of a rectangle exceeds its width by 7 inches and the area is 30 square inches. what are the length and width of the rectangle?

Let's assume w represent the width of the rectangle.

Given that, the length of a rectangle exceeds its width by 7 inches. So,

length = w + 7.

Given, area is 30 square inches.

Since Area = l* w

So, we can set up an equation as following:

(w + 7 )* w = 30

w² + 7w = 30 By distribution property.

w² + 7w - 30 = 0 Subtract 30 from each sides.

Next step is to solve the above equation by factoring to get the value of w.

First step is to breakdown the constant - 30 into two multiples so that their addition will result the coefficient of w =7.

So, -30 = -3 * 10.

Addition of -3 and 10 will give 7.

So, next step is to replace 7w with -3w + 10w. Therefore,

w² - 3w + 10w - 30 = 0

(w² - 3w) + (10w - 30)= 0 Make the group of terms.

w ( w - 3) + 10 (w - 3) = 0 Take out the common factor from each group.

(w - 3) (w + 10 ) = 0 Take out the common factor (w - 3).

w - 3 =0 and w + 10 = 0 Set up each factor equal to 0.

So, w = 3 and - 10.

Width cannot be negative 10.

So, w = 3.

Now length = w + 7 = 3 + 7 = 10

Hence length of the rectangle is 10 inches and width is 3 inches.

no, the equation would be invalid since they are both not equal.

step-by-step explanation:

here you got

4y-6+4y 2-9

first, you would simplify the equation on the form.

4y-6+4y 2-9

then, move all the terms that contains a y to the left and all the other terms to the right.

+4y-4y+-9-6

finally, you will simplify the left and the right side of the equation.

y=-3