Answer
The solutions are $\frac{-1}{2}$ and $\frac{1}{2}$.
Work Step by Step
4$y^{2}$-1=0
$(2y)^{2}$-$1^{2}$=0
(2y+1)(2y-1)=0
2y+1 =0 or 2y-1=0
2y+1-1 =0 -1 or 2y-1+1=0+1
2y=-1 or 2y=1
$\frac{2y}{2}$=$\frac{-1}{2}$ or $\frac{2y}{2}$=$\frac{1}{2}$
y=$\frac{-1}{2}$ or y= $\frac{1}{2}$
The solutions are $\frac{-1}{2}$ and $\frac{1}{2}$.
Check
Let x= $\frac{-1}{2}$
4$y^{2}$-1=0
4$(-1/2)^{2}$-1=0
4$\frac{1}{4}$-$\frac{4}{4}$=0
0=0
Let x= $\frac{1}{2}$
4$y^{2}$-1=0
4$(1/2)^{2}$-1=0
4$\frac{1}{4}$-$\frac{4}{4}$=0
0=0
