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January 11th, 2015, 12:35 PM  #1 
Newbie Joined: Oct 2014 From: Norway Posts: 19 Thanks: 0  Hello :) (simultaneous equations)
Can someone help me with this ? Last edited by greg1313; January 11th, 2015 at 01:48 PM. 
January 11th, 2015, 01:13 PM  #2 
Math Team Joined: Jul 2011 From: Texas Posts: 3,031 Thanks: 1620 
solve for x in the second equation ... x = 2y2 substitute (2y2) for x in the first equation, then solve for y; you should get two solutions. 
January 11th, 2015, 01:14 PM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
$x=2y2$ Substitute into the first equation. 
January 11th, 2015, 09:35 PM  #4 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
$\displaystyle (x1)^2+y^2=9$1 $\displaystyle 2yx=2$2, $\displaystyle x=2y2$ Substitute $\displaystyle x=2y2$ into the first equation. $\displaystyle (2y21)^2+y^2=9$ $\displaystyle (2y3)^2+y^2=9$ $\displaystyle 4y^212y+9+y^29=0$ $\displaystyle 5y^212y=0$ $\displaystyle y(5y12)=0$ $\displaystyle y=0$ 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} $ $\displaystyle x=2, y=0$ $\displaystyle x=\frac{14}{5} , y=\frac{12}{5}$ 