Answer:
[tex]\boxed{(\frac{3}{2} ,-1)}[/tex]
Step-by-step explanation:
[tex]\left \{ {{2x-3y=6} \atop {4x+2y=4}} \right.[/tex]
It seems this system of equations would be solved easier using the elimination method (the x and y values are lined up).
Multiply everything in the first equation by -2 (we want the 4x to be able to cancel out with a -4x).
[tex]2x-3y=6 \rightarrow -4x+6y=-12[/tex]
Now line up the equations (they are already lined up - convenient) and add them from top to bottom.
[tex]\left \{ {{-4x+6y=-12} \atop {4x+2y=4}} \right.[/tex]
The -4x and 4x are opposites, so they cancel out.
Adding 6y and 2y gives you 8y, and adding -12 and 4 gives you -8.
[tex]8y=-8[/tex]
Divide both sides by 8.
[tex]y=-1[/tex]
Since you have the y-value you can substitute this in to the second (or first equation, it doesn't necessarily matter) equation.
[tex]4x +2(-1)=4[/tex]
Simplify.
[tex]4x -2=4[/tex]
Add 2 to both sides.
[tex]4x=6[/tex]
Divide both sides by 4.
[tex]x=\frac{6}{4} \rightarrow\frac{3}{2}[/tex]
The final answer is [tex]x=\frac{3}{2} ,~y=-1[/tex].
[tex](\frac{3}{2} ,-1)[/tex]
