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Kipish [7]
3 years ago
10

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 f

ive times the fourth
Mathematics
1 answer:
WITCHER [35]3 years ago
7 0

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

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