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 March 1st, 2015, 10:01 AM #1 Member   Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3 System of equations I need to solve for a and b: $\displaystyle 10a+b-7=3a+3b$ $\displaystyle a^2+b^2-ab=10a+b$ I can't find a way how to do it. Please help!
 March 1st, 2015, 10:17 AM #2 Math Team   Joined: Apr 2010 Posts: 2,778 Thanks: 361 From the first equation, can you write b in terms of a?
 March 1st, 2015, 10:21 AM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 Math Focus: Calculus/ODEs I would begin by simplifying the first equation by combining like terms so that we have the system: $\displaystyle 7a-2b=7\tag{1}$ $\displaystyle a^2+b^2-ab=10a+b\tag{2}$ Now, solve (1) for $a$, and then substitute into (2)...you will get a quadratic in $b$...
 March 1st, 2015, 03:27 PM #4 Senior Member   Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 $\displaystyle 10a+b-7=3a+3b$ $\displaystyle b-3b=3a-10a+7$ $\displaystyle -2b=-7a+7$ $\displaystyle 2b=7a-7$ $\displaystyle b=\frac{7a-7}{2}$ $\displaystyle a^2+(\frac{7a-7}{2})^2-a(\frac{7a-7}{2})=10a+(\frac{7a-7}{2})$ $\displaystyle a^2+(\frac{49a^2-98a+49}{4})-(\frac{7a^2-7a}{2})=10a+(\frac{7a-7}{2})$ $\displaystyle 4a^2+49a^2-98a+49-14a^2+14a=40a+14a-14$ $\displaystyle 39a^2-138a+63=0$ $\displaystyle 13a^2-46a+21=0$ $\displaystyle (13a-7)(a-3)=0$ $\displaystyle a=3, a=\frac{7}{13}$ $\displaystyle b=\frac{7(3)-7}{2}=7$ $\displaystyle b=\frac{7(\frac{7}{13})-7}{2}=-\frac{21}{13}$ $\displaystyle \therefore a=3,b=7 ; a=\frac{7}{13},b=-\frac{21}{13}$

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