# How do you find two consecutive integers whose product is 5 less than 5 times two their sum?

Nov 27, 2016

There are two pairs of numbers that work:
1st pair: $9$ and $10$
2nd pair: $0$ and $1$

#### Explanation:

Let the first number $= n$
Let the second number $= n + 1$

"Product" refers to multiplying, so we can multiply the two numbers on one side of the equation. On the other side of the equation, "less than" refers to subtraction. "5 times two their sum" is somewhat confusing, but I interpret it as $5 \left(n + n + 1\right)$, ignoring the word "two". Please let me know if I interpreted this incorrectly, as getting the equation wrong makes my answer wrong.

The equation I came up with is:
$\left(n\right) \left(n + 1\right) = 5 \left(n + n + 1\right) - 5$

Multiply the left side, and simplify the right:
${n}^{2} + n = 5 \left(2 n + 1\right) - 5$

${n}^{2} + n = 10 n + 5 - 5$

${n}^{2} + n = 10 n$

Make one side equal to zero so that it becomes a quadratic equation:
${n}^{2} + n - 10 n = 0$
${n}^{2} - 9 n = 0$

Factor to solve:
$\left(n\right) \left(n - 9\right) = 0$

$n = 0 , n = 9$