Q:

What is the solution of the system? Use the elimination method. {4x+2y=66xβˆ’4y=βˆ’12 A.The solution is (32,0) B.The solution is (0, 3) .C.There are an infinite number of solutions.D.There is no s

Accepted Solution

A:
[tex]\begin{bmatrix}4x+2y=6\\ 6x-4y=-12\end{bmatrix}[/tex]

[tex]\mathrm{Multiply\:}4x+2y=6\mathrm{\:by\:}3: 12x+6y=18[/tex]
[tex]\mathrm{Multiply\:}6x-4y=-12\mathrm{\:by\:}2: 12x-8y=-24[/tex]

[tex]\begin{bmatrix}12x+6y=18\\ 12x-8y=-24\end{bmatrix}[/tex]

12x - 8y = -24
-
12x + 6y = 18
/
-14y = -42

[tex]\begin{bmatrix}12x+6y=18\\ -14y=-42\end{bmatrix}[/tex]

[tex]Solve\;-14y=-42 \;for\;y \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}-14 \ \textgreater \ \frac{-14y}{-14}=\frac{-42}{-14}[/tex]
y = 3

[tex]\mathrm{For\:}12x+6y=18\mathrm{\:plug\:in\:} \:y=3[/tex]

[tex]12x+6\cdot \:3=18 \ \textgreater \ \mathrm{Multiply\:the\:numbers:}\:6\cdot \:3=18 \ \textgreater \ 12x+18=18[/tex]

[tex]\mathrm{Subtract\:}18\mathrm{\:from\:both\:sides} \ \textgreater \ 12x+18-18=18-18[/tex]

[tex]\mathrm{Simplify} \ \textgreater \ 12x = 0 \ \textgreater \ \mathrm{Divide\:both\:sides\:by\:}12 \ \textgreater \ \frac{12x}{12}=\frac{0}{12} \ \textgreater \ x = 0[/tex]

[tex]Therefore\;the\:solutions\:are\;y=3,\:x=0[/tex]

Hope this helps!