- #1
d3v3st4t10n
- 3
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Homework Statement
x+2y=3
x^2-4y^2=-33
Homework Equations
The Attempt at a Solution
x-2y=root-33?
Solve the simultaneous equation:
- Thread starter d3v3st4t10n
- Start date
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Homework Help Overview
The problem involves solving a system of simultaneous equations: a linear equation and a quadratic equation. The equations presented are x + 2y = 3 and x^2 - 4y^2 = -33.
Discussion Character
- Mixed
Approaches and Questions Raised
- Participants discuss various methods for manipulating the equations, including substituting expressions for x and y. Some suggest using the difference of squares to factor the quadratic equation.
Discussion Status
There is an ongoing exploration of different approaches to solve the equations. Some participants have provided hints and suggestions, while others express confusion regarding the reasoning presented. No consensus has been reached yet.
Contextual Notes
Participants note the importance of correctly applying algebraic identities and the implications of negative values in square roots. There is also mention of homework constraints that may affect the approach taken.
x+2y=3
x^2-4y^2=-33
x-2y=root-33?
- #2
HallsofIvy
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First, that's not a solution, you haven't found a value for x and y! Second, sqrt(a- B) is NOT sqrt(A)- sqrt(B).
Probably it is simplest to solve the first equation for x, replace the "x" in the second equation by that expression.
Probably it is simplest to solve the first equation for x, replace the "x" in the second equation by that expression.
- #3
v0id19
- 47
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Try this:
x+2y=3
=> x=3-2y
<< detailed solution hints deleted by berkeman >>
x+2y=3
=> x=3-2y
<< detailed solution hints deleted by berkeman >>
Last edited by a moderator:
- #4
spideyunlimit
- 61
- 0
LOL dude, use a^2 - b^2 = (a+b)(a-b)
and u have the a+b eqn. with you..
Get the a-b eqn from the above equation, and solve a+b and a-b for a and b.
and u have the a+b eqn. with you..
Get the a-b eqn from the above equation, and solve a+b and a-b for a and b.
- #5
HallsofIvy
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Oh, that's cute.
- #6
Mentallic
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Spidey I wasn't able to follow your reasoning 
It is always easiest (imo) to make x or y the subject in the linear equation: i.e. the first equation becomes [tex]x=3-2y[/tex] or [tex]y=\frac{3-x}{2}[/tex] and then substitute either of these values into the quadratic equation so that becomes: [tex](3-2y)^2-4y^2=-33[/tex] or if you made y the subject: [tex]x^2-4(\frac{3-x}{2})^2=-33[/tex]
Oh and remember you can't take the square root of a negative value.
It is always easiest (imo) to make x or y the subject in the linear equation: i.e. the first equation becomes [tex]x=3-2y[/tex] or [tex]y=\frac{3-x}{2}[/tex] and then substitute either of these values into the quadratic equation so that becomes: [tex](3-2y)^2-4y^2=-33[/tex] or if you made y the subject: [tex]x^2-4(\frac{3-x}{2})^2=-33[/tex]
Oh and remember you can't take the square root of a negative value.
- #7
spideyunlimit
- 61
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x+2y=3
x^2-4y^2=-33
-33 = (x+2y)(x-2y)
got it now?
x^2-4y^2=-33
-33 = (x+2y)(x-2y)
got it now?
- #8
Mentallic
Homework Helper
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Ahh the 4y2 slipped my mind. Didn't think difference of 2 squares would've had a factor equal to the linear equation.
- #9
Chaos2009
- 81
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I know, I wouldn't have found it until I was halfway through the problem. It was a nice catch.
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