$$\frac { 2 y + 3 } { 4 } + \frac { y - 5 } { 6 } = 0$$
$y=\frac{1}{8}=0.125$
$$3\left(2y+3\right)+2\left(y-5\right)=0$$
$$6y+9+2\left(y-5\right)=0$$
$$6y+9+2y-10=0$$
$$8y+9-10=0$$
$$8y-1=0$$
$$8y=1$$
$$y=\frac{1}{8}$$
Show Solution
Hide Solution
