x = -6/11 and y = 13/33
1) Let's solve this by the Substitution Method:
3x =6y -4
4x +3y =-1
Rewriting the 1st equation we have:
[tex]\begin{gathered} x=\frac{6y}{3}-\frac{4}{3} \\ x=2y-\frac{4}{3} \end{gathered}[/tex]2) Let's now plug into the 2nd equation the value of x:
[tex]\begin{gathered} 4x+3y=-1 \\ 4(\frac{6y}{3}-\frac{4}{3})+3y=-1 \\ 8y-\frac{16}{3}+3y=-1 \\ 11y=-1+\frac{16}{3} \\ 11y=\frac{13}{3} \\ y=\frac{13}{3}\cdot\frac{1}{11} \\ y=\frac{13}{33} \end{gathered}[/tex]2.2) Now, let's plug the quantity of y we've just found into the 1st Equation:
[tex]\begin{gathered} 3x=6(\frac{13}{33})-4 \\ 3x=\frac{78}{33}-4 \\ 3x=-\frac{18}{11}\text{ }\times\frac{1}{3} \\ x=-\frac{6}{11} \end{gathered}[/tex]Note that we've simplified the final result.
3) Hence, the answer is:
x = -6/11 and y = 13/33
