y-1/x+3=-3/4 how do you solve for y
\(\frac{ y-1}{x+3} =- \frac{3}{4}\\ \\x+3 eq 0\\ \\x eq -3 \\ \\D=R\setminus \left \{ -3 \right \}\)\(4(y-1)=3(x+3)\\ \\4y-4=3x+9\\ \\4y=3x+9+4\\ \\4y=3x+13\ \/:4\\ \\y=\frac{3}{4}x +\frac{13}{4}\)
