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Question

Find 2 consecutive even integers such that their sum = the difference of 3 times the larger in 2 times the smaller

Answer 1

Two consecutive even integers, n and n + 2

n + n + 2 = 3(n + 2) - 2(n)

2n + 2 = 3n + 6 - 2n

2n + 2 = n + 6

n = 4

n + 2 = 6

Answer 2

Let's call the smaller integer 2n and the larger one 2n+2

The statement (There is some weird wording as to which is subtracted) says I₁+I₂ = 3I₂-2I₁

In tems of n: 2n + 2n+2 = 3(2n+2)-2(2n) or 2n = 4 (which is I₁)

Check 4+6 =? 3(6)-2(4) yes

Answer 3

Let L = the Larger consecutive even Integer

Let S = the Smaller consecutive even Integer

sum of the integers = 3 times the Larger minus 2 times the Smaller

L+S = 3L-2S

2L =3S

L and S are 2 consecutive even integers, so

L=S+2

2L = 3S,

substitute S+2 for L

2(S+2) = 3S

2S +4 = 3S

3S-2S =4

S=4

L=2+S = 2+4 = 6

check the answers

6+4=10

= 3(6)-2(4) = 18-8 = 10

4 and 6 are the two consecutive integers