You have to solve the equation system by using the elimination method.
The first step is to multiply each equation by a number, so that the first term have the same coefficient.
The leading coefficients of both equations are "3" and "2", a common factor for these numbers is "6", so you have to multiply the first equation by "2" and the second equation by "3" so that both end up having the same coefficient "6"
[tex]\begin{cases}3x-3y=-6 \\ 2x-2y=-16\end{cases}[/tex]
Multiply the first equation by 2
[tex]\begin{gathered} 2(3x-3y=-6) \\ 2\cdot3x-2\cdot3y=2\cdot(-6) \\ 6x-6y=-12 \end{gathered}[/tex]
Multiply the second equation by 3
[tex]\begin{gathered} 3(2x-2y=-16) \\ 3\cdot2x-3\cdot2y=3\cdot(-16) \\ 6x-6y=-48 \end{gathered}[/tex]
The second step is to subtract both equations:
As you can see, when you subtract both equations, the x and y terms are eliminated. This means that this equation system has no solution.
