Answer:
Step-by-step explanation:
[tex]\left\{\begin{array}{ccc}y=-2x+12&(1)\\3y-x+6=0&(2)\end{array}\right\qquad\text{substitute (1) to (2):}\\\\3(-2x+12)-x+6=0\qquad\text{use the distributive property}\\(3)(-2x)+(3)(12)-x+6=0\\-6x+36-x+6=0\qquad\text{combine like terms}\\(-6x-x)+(36+6)=0\\-7x+42=0\qquad\text{subtract 42 from both sides}\\-7x=-42\qquad\text{divide both sides by (-7)}\\\boxed{x=6}\qquad\text{put it to (1)}\\\\y=-2(6)+12\\y=-12+12\\\boxed{y=0}[/tex]
Answer:
[tex]\text{\fbox{(6,~0)}}[/tex]
Step-by-step explanation:
[tex]\left \{ {{\text{y~=~-2x~+~12}} \atop {\text{3y~-~x~+~6~=~0}} \right. \\ \\ \text{We~already~have~the ~value~of ~y ~so~ substitute~ this~ value~~ of ~y ~into~ the ~second ~equation.} \\ \\ \text{3(-2x~+~12)~-~x~+~6~=~0} \\ \\ \text{Distribute~ 3 ~inside~ the~ parentheses.} \\ \\ \text{-6x~+~36~-~x~+~6~=~0} \\ \\ \text{Combine~ like~ terms. ~You ~can~ subtract~ -6x ~and ~x ~and ~add ~36 ~and ~6.} \\ \\ \text{-7x~+~42~=~0} \\ \\ \text{Subtract~ 42 ~from~ both~ sides ~of~ the ~equation.} \\ \\ \text{-7x~=~-42} \\ \\ \text{Now ~solve~ for ~x ~by ~dividing~ both~ sides ~by~ -7.} \\ \\ \text{\fbox{x~=~6}} \\ \\ \text{To~ find~ y, ~substitute~ 6 ~for~x~ into~ the~first~ equation.} \\ \\ \text{y~=~-2(6)~+~12} \\ \\ \text{Multiply ~-2~ and~ 6.} \\ \\ \text{y~=~-12~+~12} \\ \\ \text{Combine~ like ~terms~ to ~complete~ solving~ for ~y.} \\ \\ \text{\fbox {y~=~0}} \\ \\ \text{The~ solution~ to ~this ~system ~of ~equations ~is ~\fbox{(6~,~ 0)}~.}[/tex]
