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

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Answer 1 · 2018-03-29T10:33:40

x=0
y=0

Just add the two linear equations together

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

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

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

Answer 2 · 2018-03-29T10:40:18

$color(blue)(x=0)$

$color(blue)(y=0)$

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

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

Add $[1]$ and $[2]$

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

Substituting this value of y in $[1]$

$5x-3(0)=0$

$5x=0$

$x=0$

So solutions are:

$color(blue)(x=0)$

$color(blue)(y=0)$

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

How do you solve the system of equations 5x - 3y = 05x - 3y = 0 and - 5x + 12y = 0- 5x + 12y = 0?