Answer:
[tex]\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}[/tex]
[tex]x=-4,\:y=3[/tex]
Step-by-step explanation:
Given the system of the equations
[tex]x+3y=5;\:x=-6y+14[/tex]
solving by elimination method
[tex]\begin{bmatrix}x+3y=5\\ x=-6y+14\end{bmatrix}[/tex]
[tex]\mathrm{Arrange\:equation\:variables\:for\:elimination}[/tex]
[tex]\begin{bmatrix}x+3y=5\\ x+6y=14\end{bmatrix}[/tex]
[tex]x+6y=14[/tex]
[tex]-[/tex]
[tex]\underline{x+3y=5}[/tex]
[tex]3y=9[/tex]
[tex]\begin{bmatrix}x+3y=5\\ 3y=9\end{bmatrix}[/tex]
solve [tex]3y=9[/tex] for [tex]y[/tex]:
[tex]3y=9[/tex]
[tex]\frac{3y}{3}=\frac{9}{3}[/tex]
[tex]y=3[/tex]
[tex]\mathrm{For\:}x+3y=5\mathrm{\:plug\:in\:}y=3[/tex]
Solve [tex]x+3\cdot \:3=5[/tex] for x:
[tex]x+3\cdot \:3=5[/tex]
[tex]x+9=5[/tex]
[tex]x=-4[/tex]
Therefore,
[tex]\mathrm{The\:solutions\:to\:the\:system\:of\:equations\:are:}[/tex]
[tex]x=-4,\:y=3[/tex]
