<p>May 2008
Math Section 8
16.
x=9a
y=9a^2+1</p>
<p>If a is greater than 0 in the equations above, what is “y” in terms of “x”?</p>
<p>What makes this problem so difficult??</p>
<p>y = ax + 1
LOL, i dunno why</p>
<p>y=x^2 +1</p>
<p>is it that simple?</p>
<p>No…remember Joe Bloggs?
I wish there were more questions like this :D</p>
<p>"y=x^2 +1</p>
<p>is it that simple?"</p>
<p>LOL, now u see why it’s a tough question 
ppl still get it wrong hehe</p>
<p>@BigB14… if it was y = x^2 + 1
then it would be y = 81a^2 + 1</p>
<p>Wouldn’t it be y=(x/3)^2 + 1</p>
<p>x=9a
y=9a^2+1</p>
<p>a= (x/9)
y=9(x/9)^2+1
y=9[(x^2)/(81)]+1
y=((x^2)/9)+1</p>
<p>is it y=9a^2 +1 or y=(9a)^2+1?</p>
<p>x=9a
y=9a^2 + 1</p>
<p>y=9a(a) + 1
= (x)(a) + 1</p>
<p>y=ax+1</p>
<p>Just so you know. y = ax+1 is not in terms of x…</p>
<p>The answer is (x/3)^2 + 1.</p>
<p>Of the 7 people to try to answer the question. Only 3 got it right. That’s less than 50%, so therefore it is a hard question.</p>
<p>wow, I guess it is hard…</p>
<p>all you have to do is solve for a in terms of x, then plug what you get into the second equation…</p>
<p>right?</p>
<p>what are the answer choices? It would make it easier for us :P</p>
<p>i need to know where the parenthesis are…</p>
<p>Actually this is not a hard question, we need answer choices cos it can be written in many ways.</p>
<p>The Answer is y=(x^2/9)+1
I think it’s a fairly simple problem, as long as one notices that you need ton get x in terms of a first, rather than assuming 9a^2 is equal to (9a)^2, which it is not.</p>
<p>i remember back in may this prob only took me like 30 sec, but few days ago i went over it again, i actually got stuck on it! took me like 2 mins lol… so yes this prob is hard because most people dont see y</p>