MATH SOLVE

2 months ago

Q:
# solve the following system of equations2x β 3y = 64x+2y=4β

Accepted Solution

A:

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]