# How do you find two numbers such that the first number added to three times the second is 31, while three times the first less twice the second is 16?

Nov 25, 2016

The first number is 10 and the second number is 7

#### Explanation:

Let's define the two numbers we are looking for as $x$ and $y$.

The first number added to three times the second number is 31 can be written as:

$x + 3 y = 31$

Three time the first less twice the second is 16 can be written as:

$3 x - 2 y = 16$

Solve the first equation for $x$L

$x + 3 y - 3 y = 31 - 3 y$

$x = 31 - 3 y$

Substitute $31 - 3 y$ for $x$ in the second equation and solve for $y$:

$3 \left(31 - 3 y\right) - 2 y = 16$

$93 - 9 y - 2 y = 16$

$93 - 11 y = 16$

$93 - 11 y + 11 y - 16 = 16 + 11 y - 16$

$93 - 16 = 11 y$

$77 = 11 y$

$\frac{77}{11} = \frac{11 y}{11}$

$y = 7$

Now substitute $7$ for $y$ in the solution to the first equation to calculate $x$:

$x = 31 - 3 \cdot 7$

$x = 31 - 21$

$x = 10$