find four consecutive even intepgers such that if the sum if the first and third is multiplied by 3 the result is 4 less than 5 times the fourth

Question

find four consecutive even intepgers such that if the sum if the first and third is multiplied by 3 the result is 4 less than 5 times the fourth

Answer 1

Let x = 1st even integer

Then x+2 = 2nd, x+4 = 3rd, x+6 = 4th. This is because even integers are two apart (like counting by 2's).

Now translate the words into algebra:

Sum of 1st and 3rd: x + x+4 is multiplied by 3: 3(x + x+4).

The result is is four less than five times the 4th: 5(x+6) - 4

Those results are equal (the result "is");

3(x + x+4) = 5(x+6) - 4

3(2x+4) = 5x+30 - 4

6x+12 = 5x+26

x = 14 (which represents the 1st even integer; thankfully 14 is even, so there appears to be a solution).

x+2 = 16

x+4 = 18

x+6 = 20

Check:

Three times sum of 1st and 3rd: 3(14+18) = 96

Five times the fourth: 5(20) = 100, and 4 less than that: 100 - 4 = 96.