# The difference between two multiples of ten is 330. The total of them is 710. What are the two numbers?

## is there a way to do this calculation without using simultaneous equations?

Oct 8, 2017

See a solution process below:

#### Explanation:

We can first name our variables. Because they are multiples of $10$ we can write them as: $10 m$ and $10 n$.

From the information in the problem we can write two equations:

Equation 1: $10 m - 10 n = 330$

Equation 1: $10 m + 10 n = 710$

Step 1: Solve the first equation for $m$:

$10 m - 10 n = 330$

$10 \left(m - n\right) = 330$

$\frac{10 \left(m - n\right)}{\textcolor{red}{10}} = \frac{330}{\textcolor{red}{10}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} \left(m - n\right)}{\cancel{\textcolor{red}{10}}} = 33$

$m - n = 33$

$m - n + \textcolor{red}{n} = 33 + \textcolor{red}{n}$

$m - 0 = 33 + n$

$m = 33 + n$

Step 2: Substitute $\left(33 + n\right)$ for $m$ in the second equation and solve for $n$:

$10 m + 10 n = 710$ becomes:

$10 \left(33 + n\right) + 10 n = 710$

$10 \left[\left(33 + n\right) + n\right] = 710$

$\frac{10 \left[33 + n + n\right]}{\textcolor{red}{10}} = \frac{710}{\textcolor{red}{10}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{10}}} \left[33 + 2 n\right]}{\cancel{\textcolor{red}{10}}} = 71$

$33 + 2 n = 71$

$33 - \textcolor{red}{33} + 2 n = 71 - \textcolor{red}{33}$

$0 + 2 n = 38$

$2 n = 38$

$\frac{2 n}{\textcolor{red}{2}} = \frac{38}{\textcolor{red}{2}}$

$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} n}{\cancel{\textcolor{red}{2}}} = 19$

$n = 19$

Step 3: Substitute $19$ for $n$ in the solution to the first equation at the end of Step 1 and calculate $m$:

$m = 33 + n$ becomes:

$m = 33 + 19$

$m = 52$

Solution:

If $m = 52$ then $10 m = 10 \times 52 = 520$

If $n = 19$ then $10 n = 10 \times 19 = 190$

$520 - 190 = 330$

$520 + 190 = 710$

The numbers are: 520 and 190

Oct 8, 2017

the first number is $520$ and the second one is $190$

#### Explanation:

Let the numbers be $10 x$ and $10 y$

Then the difference is:

$10 x - 10 y = 330$

and their sum is:

$10 x + 10 y = 710$

If we sum the equations side by side, we'll get:

$10 x \cancel{- 10 y} + 10 x \cancel{+ 10 y} = 330 + 710$

that's

$20 x = 1040$

and $x = 52$

We can find y by subtracting the equations side by side:

$\cancel{10 x} - 10 y \cancel{- 10 x} - 10 y = 330 - 710$

$- 20 y = - 380$

that's

$y = 19$

then the first number is $10 x = 10 \cdot 52 = 520$

and the second one is $10 y = 10 \cdot 19 = 190$