How do you solve the following system?: $x + 3 y = - 3, y = - x + 12$ $x + 3y = -3 , y=-x + 12$

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Answer 1 · 2018-03-22T17:23:16

$(\frac{39}{2}), (- \frac{15}{2})$ $(39/2),(-15/2)$

$x + 3 y = - 3$ $x + 3y=-3$ and $y = - x + 12$ $y=-x+12$

we can substitute the value of y of the second equation in the first equation so we get,

$x + 3 (- x + 12) = - 3$ $x + 3(-x+12) = -3$

On simplification

$x - 3 x + 36 = - 3$ $x-3x+36=-3$

We get that $x = \frac{39}{2}$ $x=39/2$

Substituting it in either of the initial equations and solving for y, we get $y = - \frac{15}{2}$ $y=-15/2$

So the solution to the given pair of equations is $x = \frac{39}{2}$ $x=39/2$ and $y = - \frac{15}{2}$ $y=-15/2$

How do you solve the following system?: x+3y=−3,y=−x+12x + 3y = -3 , y=-x + 12