How do you solve $[(x+3y),(3x+y)]=[(-13),(1)]$ ?

Question

2395 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-11T19:08:31

$x=2$ and $y=-5$

We can get two equations from the given:

(1) $x+3y=-13$
(2) $3x+y=1$

If we multiply the second equation from $-3$ we get:
$-9x-3y=-3$ . Adding this new equation to the first equation gives:

$-8x=-16 rarr x =2$ .

Taking this value of $x$ and substituting into (2) gives

$3(2)+y=1 rarr6+y=1 rarry=-5$ ,

so $x=2$ and $y=-5$

How do you solve [(x+3y),(3x+y)]=[(-13),(1)][(x+3y),(3x+y)]=[(-13),(1)]?