# The product of two consecutive even integers is 34 less than 7 times their sum, how do you find the two integers?

Jul 5, 2015

Let's call the numbers $2 n \mathmr{and} 2 n + 2$ with $n$ an integer, and the $2$ to make sure it's even.

#### Explanation:

Then we translate the rest of the conditions:
Product: $P = 2 n \cdot \left(2 n + 2\right) = 4 {n}^{2} + 4 n$
Sum: $S = 2 n + \left(2 n + 2\right) = 4 n + 2$

Now $P = 7 \cdot S - 34$

Substituting:
$4 {n}^{2} + 4 n = 7 \cdot \left(4 n + 2\right) - 34 \to$
$4 {n}^{2} + 4 n = 28 n + 14 - 34 \to$

Everything to one side:
$4 {n}^{2} - 24 n + 20 = 0 \to$

Divide everything by $4$ to make life easier:
${n}^{2} - 6 n + 5 = 0 \to$ factorise:
$\left(n - 1\right) \left(n - 5\right) = 0 \to n = 1 \mathmr{and} n = 5$

So the numbers are $2 \mathmr{and} 4$ or $10 \mathmr{and} 12$

Checking:
$2 \cdot 4 = 7 \cdot \left(2 + 4\right) - 34 \to 8 = 42 - 34$ check!
$10 \cdot 12 = 7 \cdot \left(10 + 12\right) - 34 \to 120 = 154 - 34$ check!