Find three consecutive even integers such that the sum of the smallest number and twice the middle number is 20 more than the largest number.

Step-by-step explanation:

Consecutive even integers increase by 2.

x   ← an integer

2x  ← an even integer (the smallest)

2x+2   ← next even integer consecutive to 2x (the middle one)

2x+2+2=2x+4  ← the last even integer consecutive to 2x+2 (the largest)

2(2x+2)    ← twice the middle number

2x+2(2x+2)   ←  the sum of the smallest number and twice the middle

2x+4+20   ←  20 more than the largest number

2x + 2(2x + 2) = 2x + 4 + 20

2x + 4x + 4 = 2x + 24

            -4           -4

            6x = 2x + 20

         -2x       -2x

            4x = 20

            ÷2     ÷2

            2x = 10

2x+2 = 10+2 = 12

2x+4 = 10+4 = 14

huh? im pretty sure it is in 199

step-by-step explanation:

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.