How do you solve the system of equations $3x + 9y = 1$ and $- 12x + 3y = 16$ ?

Question

1886 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-28T04:06:49

$x=-47/39$ and $y=20/39$

Lets call $3x+9y=1$ the first equation and $-12x+3y=16$ the second equation.

First multiply the first equation by $4$ so that it has $12$ as a coefficient of the $x$ term

$4xx(3x+9y)=4xx(1)$

$12x+36y=4$

Comparing to the second equation, we get

$" "12x+36y=4$
$-12x+3y=16$

Adding the two equations to each other gives

$0x+39y=20$

$39y=20$

$y=20/39$

Finally, take $y=20/39$ and plug it into either of the first two original equations and solve for $x$

$3x+9color(red)(y)=1$

$3x+9(color(red)(20/39))=1$

$3x+180/39=1$

$3x=1-180/39$

$3x=39/39-180/39$

$3x=-141/39$

Dividing both sides by $3$ gives

$x=-47/39$

ANSWER: $x=-47/39$ and $y=20/39$

How do you solve the system of equations 3x + 9y = 13x + 9y = 1 and - 12x + 3y = 16- 12x + 3y = 16?