Find four consecutive even integers such that if the sum of the first and the third is multiplied by 5 the result is 12 less than 9 times the fourth

Respuesta :

4 consecutive even integers....x, x + 2, x + 4, x + 6

5(x + x + 4) = 9(x + 6) - 12
5(2x + 4) = 9x + 54 - 12
10x + 20 = 9x + 42
10x - 9x = 42 - 20
x = 22

x + 2 = 22 + 2 = 24
x + 4 = 22 + 4 = 26
x + 6 = 22 + 6 = 28

so ur numbers are : 22,24,26, and 28