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January 11th, 2015, 12:35 PM   #1
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Hello :) (simultaneous equations)

Can someone help me with this ?
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Last edited by greg1313; January 11th, 2015 at 01:48 PM.
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January 11th, 2015, 01:13 PM   #2
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solve for x in the second equation ...

x = 2y-2

substitute (2y-2) for x in the first equation, then solve for y; you should get two solutions.
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January 11th, 2015, 01:14 PM   #3
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$x=2y-2$

Substitute into the first equation.
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January 11th, 2015, 09:35 PM   #4
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$\displaystyle (x-1)^2+y^2=9$--1

$\displaystyle 2y-x=2$--2, $\displaystyle x=2y-2$

Substitute $\displaystyle x=2y-2$ into the first equation.

$\displaystyle (2y-2-1)^2+y^2=9$

$\displaystyle (2y-3)^2+y^2=9$

$\displaystyle 4y^2-12y+9+y^2-9=0$

$\displaystyle 5y^2-12y=0$

$\displaystyle y(5y-12)=0$

$\displaystyle y=0$ and $\displaystyle y=\frac{12}{5} $

When $\displaystyle y=0$

$\displaystyle x=2(0)-2$

$\displaystyle x=-2$



When $\displaystyle y=\frac{12}{5} $

$\displaystyle x=2(\frac{12}{5}-2) $

$\displaystyle x=\frac{14}{5} $


$\displaystyle x=-2, y=0$
$\displaystyle x=\frac{14}{5} , y=\frac{12}{5}$
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