Question

Find four consecutive even integers such that if the sum of the first and third is multiplied by 3, the result is 28 less than five times the fourth

Answer 1

Answer:

$A=-10\\B=-8\\C=-6\\D=-4$

Step-by-step explanation:

We can solve the problem in this way:

- The 1st information that we have is that the four numbers are even and consecutive integer. Let's call these numbers A, B, C and D. This means that:

$D=C+2\\C=B+2\\B=A+2$

(since the distance between two even consecutive numbers is 2)

- The 2nd information that is given is that the sum of the first and third multiplied by 3 equals 28 less than 5 times the fourth, so:

$3(A+C)=5D-28$ (1)

Now we rewrite eq(1) by using:

$D=C+2$

and

$A=C-4$ (obtained by combining the 3 first equations)

So we get:

$3(C-4+C)=5(C+2)-28$

Solving this equation for C,

$6C-12=5C+10-28\\C=-6$

So the other numbers are:

$A=-10\\B=-8\\C=-6\\D=-4$