What's wrong here?

[Alkatran]

I took a test in math to determine where I was etc etc...
anyways, the following question came up:

How many answers are there to the following system of equations?
x^2 + 5y = 30
x^2 + (y-3)^2 = 9

a) 0
b) 1
c) 2 (my answer)
d) 3 (correct answer)


The first thing I noticed was that x was the same thing in both cases, the answer would not be changed because of it. As such, x could be almost any value (infinite answers). So obviously, they wanted how many solutions for Y.

My work (in my head) was:
x^2 + 5y = 30
x^2 + (y-3)^2 = 9

*subtract equations from each other
5y - (y-3)^2 = 21

*break (y-3)^2
5y - (y^2 - 6y + 9) = 21
5y - y^2 + 6y - 9 = 21
11y - y^2 = 30

*place ^2 in positive and use ax^2 + bx + c format
y^2 - 11y + 30 = 0

*check for 2 or 1 answers by checking if the sqr() portion of quadratic formula is non zero
sqr(121 - 4*30*1) = sqr(1) = 1

since 1 <> 0, and the answer will be +- 1, there are two answers.

Where is my mistake? :frown:

[Leong]

\begin{align*}\\y=\frac{30-x^2}{5}\\
x^2+[\frac{30-x^2-15}{5}]^2=9\\
x^2(x^2-5)=0\\
x=0\ or \x=\pm\sqrt{5}\end{align*}

When you substitute y=5 which you have found into the first equation, you find 2 solution for x and for y=6, you find x=0.

[JasonRox]

I was thinking...
- 2nd Degree Equation
- 1st Degree Equation

...3 solutions.

I can be wrong with the 1st degree equation.

[Alkatran]

\begin{align*}\\y=\frac{30-x^2}{5}\\
x^2+[\frac{30-x^2-15}{5}]^2=9\\
x^2(x^2-5)=0\\
x=0\ or \x=\pm\sqrt{5}\end{align*}

When you substitute y=5 which you have found into the first equation, you find 2 solution for x and for y=6, you find x=0.

Oh, I get it. I see where I made the mistake... tricky little son of a...

[Tide]

One is a parabola and the other is a circle. A quick sketch of the graphs will tell you how many (real) solutions there are! :-)

[Alkatran]

I see, x is limited by y's limitation... it's a weird concept at first.

[JasonRox]

I can the same answers as the guy above.

I did the following:

Take x^2 + (y-3)^2 = 9, and you get x^2 = y^2 + 6y.

Take the x^2 and put it in the other equation, and you get...

y^2 - 6y + 5y = 30 to... (y - 6)(y - 5) = 0

Now that you know y... plug in the numbers for x, and there you go.

[JasonRox]

I was wrong about the 1st degree. :(

I was doubting it completely because the answer was 3, and I just took the easy way out.

Look at the bright side, you warmed up your brain today.