# Simultaneous equations-one linear, one quadratic!

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#1
Solve x+y=2 and 3x^2-2c+y^2=2 simultaneously.

Well, i got up to 3y^2-12y+12-2y+4+y^2=2 so far?! Is that right? What next?
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15 years ago
#2
is the -2c meant to be -2c or -2x?
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#3
(Original post by El Stevo)
is the -2c meant to be -2c or -2x?
sorry, -2x!
0
15 years ago
#4
next collect all the terms together to get a quadratic in y. solve for y, and sub into one of the equations to get x. then you will have a pair of solutions for x and y.
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15 years ago
#5
I got y=1, y=3/2 and corres. values of x are x=1 x=1/2. hope that helps
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15 years ago
#6
Solve x+y=2 and 3x^2-2c+y^2=2 simultaneously.

 x + y = 2 -----> change this to y=2 - x

 3x² - 2x + y² = 2

In  put (2-x) instead of y

 3x² - 2x + (2-x)² = 2

...3x² - 2x + 4 - 4x + x² = 2
Rearrange in the form ax² + bx + c = 0
...4x² - 6x + 2 = 0
Take out a factor of 2
...2(2x² - 3x +1) = 0
...2x² - 3x +1=0
Solving this using the quadratic formula gives, x = 0.5 and x = 1
Substituting into either equation to find the corresponding y values gives,
(0.5, 1.5) and (1, 1)
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#7
thank you!!!
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