Find three consecutive integers such that twice the greatest integer is 2 less than 3 times the least integer.

Givens

Let the first integer = x

Let the second integer = x + 1

Let the third integer = x + 2

Equations

2(x + 2) = 3x - 2 Remove the brackets

Solve

2x + 4 = 3x - 2     Subtract 2x from both sides.

4 = 3x - 2x - 2  

4 = x - 2                Add 2 to both sides

4 + 2 = x

x = 6

Answer

The smallest integer is 6

The middle one is 7

The largest one is 8

Check

2*8 = 16

3*6 - 2 = 16 They check.

16

step-by-step explanation:

2 x 40 = 80

80/5

= 16

squares, rhombus, rectangles