How do you solve the system of equations: $y=3x -10$ and $x= 12- 4y$ ?

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Answer 1 · 2015-03-30T12:55:39

Re-write the equations in the form
$ay + bx = c$
then solve as normal.

$y = 3x -10$ becomes
$y-3x = -10$ (equation 1)

$x= 12 -4y$ becomes
$4y + x = 12$ (equation 2)

Multiply equation 1 by 4, giving
$4y-12x = -40$ and subtract from equation 2

$13x = 52$
$rarr x = 4$

Substituting back in any of the earlier equations gives
$y = 2$

How do you solve the system of equations: y=3x -10y=3x -10 and x= 12- 4yx= 12- 4y?