Find five consecutive integers such that:
“the sum of the first and 3 times the fourth is equal to 39 less than twice the sum of the second, third, and fifth.”
make sure to list all five integers in your answer and show all of your

  17, 18, 19, 20, 21

Step-by-step explanation:

Let x represent the third (middle) integer. Then the first is (x-2) and the given relation is ...

  (x-2) + 3(x+1) = -39 +2((x-1)+x+(x+2))

  4x +1 = 6x -37 . . . . simplify

  38 = 2x . . . . . . . . . add 37-4x

  x = 19 . . . . . . . . . . divide by 2

The integers are ...

  17, 18, 19, 20, 21

Check

  17 + 3·20 = -39 +2(18 +19 +21)

  17 +60 = -39 +2(58)

  77 = -39 +116 . . . . . yes

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.