<p>The Square of x is equal to 4 times the square of y. If x is 1 more than twice y, what is the value of x?
A) -4
B) -1/2
C) -1/4
D) 1/4
E) 1/2</p>
<p>OKay before you start screaming about LOOK AT THE BB EXPLANATIONS, let me tell you what I did.</p>
<p>x^2=4y^2
x=2y+1</p>
<p>x^2= (2y)^2
x=2y</p>
<p>So I plugged it in. with the other equation</p>
<p>2y= 2y+1</p>
<p>NOT A SOLUTION</p>
<p>Help? What did I do wrong? If my setup doesn't make sense, I'll be glad to explain what I did some more</p>
<p>x^2= (2y)^2
x = 2y or x = -2y<br>
(Your algebra teacher probably taught that if x^2 = 1, then x = +1 or -1)</p>
<p>x = 2y does not work as you’ve shown; eliminate this as a solution;</p>
<p>try x = -2y as another possible solution
-2y = 2y + 1
y = -1/4
Answer: C)</p>
<p>You also can get the right answer to this problem by trying each answer (backsolve).</p>
<p>OHHH. NEGATIVE! </p>
<p>I get it now. Thank you!</p>
<p>@knowthestuff watch out, the answer is not C. That is the value of y and the question is asking for the value of x, which is E (1/2).</p>
<p>Oh my gawd.</p>
<p>Why is it 1/2??? :(</p>
<p>Start out with the two equations given in the question:</p>
<p>x^2 = 4y^2
and
x = 2y + 1</p>
<p><em>This is only one way of doing this problem, there are many ways of doing it</em></p>
<p>I manipulated the second equation so that I could solve for y in the first equation. I squared both sides of the equation so that:
(x)^2 = (2y+1)^2
= (2y+1)(2y+1)
x^2 = 4y^2 + 4y + 1</p>
<p>Since x^2 = 4y^2</p>
<p>4y^2 = 4y^2 + 4y + 1
-4y^2 -4y^2</p>
<p>0 = 4y + 1
4y = -1
y = -1/4</p>
<p>Now, we solve for x</p>
<p>x = 2y + 1
x = 2(-1/4) + 1
x = (1/2)
Choice E</p>
<p>Like I said, there is definitely more than one way to do this problem. This is just the way that makes sense to me.</p>