<p>Can someone show me step by step how to do this?</p>
<p>x^2= 4(y^2)
if x= 2y+1
replacement:</p>
<p>(2y+1)^2= 4y^2
(2y+1)(2y+1)
4y^2 + 4y+ 1 = 4y^2
4y+1= 0
4y= -1
y= -1/4</p>
<p>x= 2y+1
x= 1/2</p>
<p>Wow thanks. I realized my dumb mistake at “4y+1= 0”. I didn’t bring 1 to the other side and was stuck.</p>