# The sum of 3 numbers is 61. The second number is 5 times the first. While the 3rd is 2 less then the first. How do you find the numbers?

Oct 18, 2015

$9$, $45$, and $7$

#### Explanation:

Let's say that your numbers are $x$, $y$, and $z$.

The first thing to write down is that their sum is equal to $61$

$x + y + z = 61 \text{ " " } \textcolor{p u r p \le}{\left(1\right)}$

Next, you know that the second number, $y$, is five times bigger than the first, $x$. This is equivalent to saying that you need to multiply $x$ by $5$ to get $y$

$y = 5 \cdot x \text{ " " } \textcolor{p u r p \le}{\left(2\right)}$

Finally, you know that if you subtract $2$ from $x$ you get the $z$

$z = x - 2 \text{ " " } \textcolor{p u r p \le}{\left(3\right)}$

Use equations $\textcolor{p u r p \le}{\left(2\right)}$ and $\textcolor{p u r p \le}{\left(3\right)}$ to find the value of $x$ in equation $\textcolor{p u r p \le}{\left(1\right)}$

$x + 5 x + x - 2 = 61$

$7 x = 63 \implies x = \frac{63}{7} = \textcolor{g r e e n}{9}$

Now take this value of $x$ and find $y$ and $z$

$y = 5 \cdot \left(9\right) = \textcolor{g r e e n}{45}$

$z = \left(9\right) - 2 = \textcolor{g r e e n}{7}$

The three numbers are $9$, $45$, and $7$.