Question

Help please! Find three consecutive even integers, such that the sum of the first two even integers is 20 less than 3 times the largest even integers

Answer 1

Let x = 1st even integer

Let x + 2 = 2nd even integer

Let x + 4 = 3rd even integer

Now look at words and translate Three times the sum of first and third is 24 greater than 4 times the second

3( x + x + 4) = 24 + 4(x + 2) now simplify

3(2x + 4) = 24 + 4(x + 2) distribute

6x + 12 = 24 + 4x + 8 combine

6x + 12 = 32 + 4x subtract 4x from each side

-4x -4x

2x +12 = 32 subtract 12 from each side

-12 -12

2x = 20 divide both sides by 2

x = 10

This tells us that the first even number is 10, the second is 12, and the third is 14.

CHECK

3(10 + 14) = 24 + 4(12)

3(24) = 24 + 48

72 = 72√