Solving the System of Equations: x^2 + 5y = 30 & x^2 + (y-3)^2 = 9

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Discussion Overview

The discussion revolves around solving a system of equations involving a quadratic equation and a circle. Participants explore how many solutions exist for the equations: x^2 + 5y = 30 and x^2 + (y-3)^2 = 9. The conversation includes various approaches to finding solutions and interpretations of the equations.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests that since x appears in both equations, it could take on many values, leading to infinite solutions for y.
  • Another participant proposes a substitution method to express y in terms of x and derives a quadratic equation to find potential solutions.
  • Some participants note the nature of the equations, identifying one as a parabola and the other as a circle, which may affect the number of real solutions.
  • There is a mention of finding two solutions for y when substituting back into the first equation, indicating a potential for multiple solutions.
  • One participant expresses uncertainty about the classification of the equations, initially considering one as first degree but later correcting themselves.
  • Another participant describes a method to derive y values and subsequently find corresponding x values, suggesting a systematic approach to the problem.

Areas of Agreement / Disagreement

Participants express differing views on the number of solutions, with some suggesting two or three solutions while others remain uncertain. The discussion does not reach a consensus on the exact number of solutions.

Contextual Notes

Some participants' reasoning involves assumptions about the relationships between x and y, and there are unresolved mathematical steps in deriving the solutions. The discussion reflects varying interpretations of the equations' graphical representations.

Who May Find This Useful

Readers interested in algebra, particularly in solving systems of equations, and those exploring the graphical implications of quadratic and circular equations may find this discussion relevant.

Alkatran
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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:
 
Last edited:
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[tex]\begin{align*}\\y=\frac{30-x^2}{5}\\<br /> x^2+[\frac{30-x^2-15}{5}]^2=9\\<br /> x^2(x^2-5)=0\\<br /> x=0\ or \x=\pm\sqrt{5}\end{align*}[/tex]

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

...3 solutions.

I can be wrong with the 1st degree equation.
 
Leong said:
[tex]\begin{align*}\\y=\frac{30-x^2}{5}\\<br /> x^2+[\frac{30-x^2-15}{5}]^2=9\\<br /> x^2(x^2-5)=0\\<br /> x=0\ or \x=\pm\sqrt{5}\end{align*}[/tex]

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...
 
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! :-)
 
I see, x is limited by y's limitation... it's a weird concept at first.
 
I can the same answers as the guy above.

I did the following:

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

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

[tex]y^2 - 6y + 5y = 30[/tex] to... [tex](y - 6)(y - 5) = 0[/tex]

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

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