# How do you solve the system of equations 5x - 3y = 0 and - 5x + 12y = 0?

Mar 29, 2018

x=0
y=0

#### Explanation:

Just add the two linear equations together

$5 x - 3 y = 0$
$- 5 x + 12 y = 0$

$0 + 9 y = 0$
$y = 0$

Put the y value into the first equation to figure out x
$5 x - 3 \left(0\right) = 0$
$5 x = 0$
$x = 0$

Mar 29, 2018

$\textcolor{b l u e}{x = 0}$

$\textcolor{b l u e}{y = 0}$

#### Explanation:

$5 x - 3 y = 0 \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \left[1\right]$

$- 5 x + 12 y = 0 \setminus \setminus \setminus \setminus \left[2\right]$

Add $\left[1\right]$ and $\left[2\right]$

$\setminus \setminus \setminus \setminus 5 x - 3 y = 0$
$\setminus - 5 x + 12 y = 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus 0 + 9 y = 0$
$\setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus \setminus y = 0$

Substituting this value of y in $\left[1\right]$

$5 x - 3 \left(0\right) = 0$

$5 x = 0$

$x = 0$

So solutions are:

$\textcolor{b l u e}{x = 0}$

$\textcolor{b l u e}{y = 0}$

This is an example of a homogeneous system. $\left(0 , 0\right)$ is always a solution to these systems and is known as the trivial solution.