Question

Three consecutive integers are as such when they are taken in increasing order and multiplied by 2, 3, and 4 respectively, they add up to 56. Find these numbers.
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Answer 1

Answer:

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Answer 2

First, let's name the three consecutive integers.

Let's call the first integer: n

Then the next two integers will be (n+1) and (n+2)

If we then multiply them as described in the problem and sum these products to 56 we can write an equation as:

2n+3(n+1)+4(n+2)=56

We can now solve this equation for n :

2n+(3×n)+(3×1)+(4×n)+(4×2)= 56

2n+3n+3+4n+8=56

2n+3n+4n+3+8=56

(2+3+4)n+(3+8)=56

9n+11=56

9n= 56−11

9n=45

n=45/9

n=5

Therefore:

n+1=5+1=6

n+2=5+2=7

The three consecutive integers are: 5, 6, 7