Answer
$(2y+4,y)$, where $y$ is any real number
Work Step by Step
We are given the system:
$\begin{cases}
4x-8y=16\\
3x-6y=12
\end{cases}$
We will use the addition method. Multiply Equation 1 by 3, Equation 2 by -4 and add the two equations to eliminate $x$ and determine $y$:
$\begin{cases}
3(4x-8y)=3(16)\\
-4(3x-6y)=-4(12)
\end{cases}$
$\begin{cases}
12x-24y=48\\
-12x+24y=-48
\end{cases}$
$12x-24y-12x+24y=48-48$
$0=0$
We got a true statement, therefore the system has an infinity of solutions $(x,y)$, where $4x-8y=16$.
$4x-8y=16$
$4(x-2y)=16$
$x-2y=4$
$x=2y+4$
The solutions can be written:
$(2y+4,y)$, where $y$ is any real number
