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 Algebra Pre-Algebra and Basic Algebra Math Forum

January 11th, 2015, 12:35 PM   #1
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Joined: Oct 2014
From: Norway

Posts: 19
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Hello :) (simultaneous equations)

Can someone help me with this ? Attached Images Skjermbilde 2015-01-11 kl. 21.00.35.png (10.1 KB, 22 views)

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 = 2y-2 substitute (2y-2) 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 $x=2y-2$ Substitute into the first equation. January 11th, 2015, 09:35 PM #4 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 $\displaystyle (x-1)^2+y^2=9$--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 $\displaystyle y=\frac{12}{5}$ $\displaystyle x=2(\frac{12}{5}-2)$ $\displaystyle x=\frac{14}{5}$ $\displaystyle x=-2, y=0$ $\displaystyle x=\frac{14}{5} , y=\frac{12}{5}$ Thanks from Ole Daniel Tags equations, simultaneous Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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