Question

Jared wanted to find three consecutive even integers whose sum was 4 times the first of those integers. He let k represent the first integer, then wrote and solved this equation: k + k + 1) + (K + 2) = 4k. Did he get the correct answer? Complete the explanation. Jared did not get the correct answer. The solution to his equation is k = giving and as the three integers. However, 3 and 5 are not even integers. He should have used the equation k + (k + ) + (k + ) = 4k, which gives k = and the correct answer

Answer 1

Jared's solution was:

$k+k+1+k+2=4k\Rightarrow3k+3=4k\Rightarrow3=4k-3k\Rightarrow3=k\Rightarrow k=3$

Jared's got the following three integers: 3, 3+1, 3+2, or 3, 4, 5.

The correct solution

First of all. to represent three consecutive even integers whose sum was 4 times the first of those integers is:

$k+k+2+k+4=4k_{}$

We can find k:

$3k+6=4k\Rightarrow6=4k-3k\Rightarrow6=k\Rightarrow k=6_{}$

The even integers are then: 6, 6+2, 6+4 or 6, 8, 10.