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March 1st, 2015, 09:01 AM  #1 
Member Joined: Aug 2014 From: Lithuania Posts: 60 Thanks: 3  System of equations
I need to solve for a and b: $\displaystyle 10a+b7=3a+3b$ $\displaystyle a^2+b^2ab=10a+b$ I can't find a way how to do it. Please help! 
March 1st, 2015, 09:17 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, 09:21 AM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,163 Thanks: 472 Math Focus: Calculus/ODEs 
I would begin by simplifying the first equation by combining like terms so that we have the system: $\displaystyle 7a2b=7\tag{1}$ $\displaystyle a^2+b^2ab=10a+b\tag{2}$ Now, solve (1) for $a$, and then substitute into (2)...you will get a quadratic in $b$... 
March 1st, 2015, 02:27 PM  #4 
Senior Member Joined: Sep 2013 From: Earth Posts: 827 Thanks: 36 
$\displaystyle 10a+b7=3a+3b$ $\displaystyle b3b=3a10a+7$ $\displaystyle 2b=7a+7$ $\displaystyle 2b=7a7$ $\displaystyle b=\frac{7a7}{2}$ $\displaystyle a^2+(\frac{7a7}{2})^2a(\frac{7a7}{2})=10a+(\frac{7a7}{2})$ $\displaystyle a^2+(\frac{49a^298a+49}{4})(\frac{7a^27a}{2})=10a+(\frac{7a7}{2})$ $\displaystyle 4a^2+49a^298a+4914a^2+14a=40a+14a14$ $\displaystyle 39a^2138a+63=0$ $\displaystyle 13a^246a+21=0$ $\displaystyle (13a7)(a3)=0$ $\displaystyle a=3, a=\frac{7}{13}$ $\displaystyle b=\frac{7(3)7}{2}=7$ $\displaystyle b=\frac{7(\frac{7}{13})7}{2}=\frac{21}{13}$ $\displaystyle \therefore a=3,b=7 ; a=\frac{7}{13},b=\frac{21}{13}$ 

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