What are the possible lengths and widths of a rectangle if its length is 8 centimeters longer than the width, and the area is less than or equal to 33 square centimeters?

Length = 11

Width = 3

Step-by-step explanation:

Conditions:

L = W + 8

A ≤ 33

We know both the L and W have to be less than or equal to 33, as the maximum area would be 33 * 1.

First, find the factors of 33.

1, 3, 11, 33.

Put them in pairs.

1,33

3,11

Find the pair with the difference of 8.

33 - 1 = 32

11 - 3 = 8

Assign them to the Length or Width and test them against the conditions.

L = 11

W = 3

11 = 3 + 8

3 * 11 = 33

answer: two five pound bags one two pound

step-by-step explanation:

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .