Question

Find three consecutive even integers such that the sum of twice the first and three times the second is52 more than twice the third.AnswerKeypadKeyboard Shortcu

Answer 1

18, 20, 22

Step-by-step explanation

Let n be the smallest number. If they all are consecutive integers and all even, then the other two are (n + 2) and (n + 4).

We know that twice the first integer, 2n, plus three times the second, 3(n+2) is twice the third plus 52, 2(n+4) + 52. We have the equation,

`2n+3(n+2)=2(n+4)+52`

To solve for n, apply the distributive property,

`\begin{gathered} 2n+3\cdot n+3\cdot2=2\cdot n+2\cdot4+52 \\ 2n+3n+6=2n+8+52 \end{gathered}`

Add like terms,

`\begin{gathered} (2n+3n)+6=2n+(8+52) \\ 5n+6=2n+60 \end{gathered}`

Subtract 2n from both sides,

`\begin{gathered} 5n-2n+6=2n-2n+60 \\ 3n+6=60 \end{gathered}`

Subtract 6 from both sides,

`\begin{gathered} 3n+6-6=60-6 \\ 3n=54 \end{gathered}`

And divide both sides by 3,

`\begin{gathered} (3n)/(3)=(54)/(3) \\ n=18 \end{gathered}`

If the first even integer is 18, then the next two - remember that they are consecutive even integers, are 20 and 22.

Hence, the three numbers are 18, 20, 22.