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Solve the simultaneous equation:

  • #1

Homework Statement


x+2y=3
x^2-4y^2=-33


Homework Equations





The Attempt at a Solution


x-2y=root-33?
 

Answers and Replies

  • #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.
 
  • #3
49
0
Try this:
x+2y=3
=> x=3-2y

<< detailed solution hints deleted by berkeman >>
 
Last edited by a moderator:
  • #4
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.
 
  • #5
HallsofIvy
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Oh, that's cute.
 
  • #6
Mentallic
Homework Helper
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Spidey I wasn't able to follow your reasoning :confused:

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
x+2y=3
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
81
0
I know, I wouldn't have found it until I was halfway through the problem. It was a nice catch.
 

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