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March 1st, 2015, 10:01 AM   #1
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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!
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March 1st, 2015, 10:17 AM   #2
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From the first equation, can you write b in terms of a?
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March 1st, 2015, 10:21 AM   #3
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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) will get a quadratic in $b$...
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March 1st, 2015, 03:27 PM   #4
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$\displaystyle 10a+b-7=3a+3b$

$\displaystyle b-3b=3a-10a+7$

$\displaystyle -2b=-7a+7$

$\displaystyle 2b=7a-7$

$\displaystyle b=\frac{7a-7}{2}$

$\displaystyle a^2+(\frac{7a-7}{2})^2-a(\frac{7a-7}{2})=10a+(\frac{7a-7}{2})$

$\displaystyle a^2+(\frac{49a^2-98a+49}{4})-(\frac{7a^2-7a}{2})=10a+(\frac{7a-7}{2})$

$\displaystyle 4a^2+49a^2-98a+49-14a^2+14a=40a+14a-14$

$\displaystyle 39a^2-138a+63=0$

$\displaystyle 13a^2-46a+21=0$

$\displaystyle (13a-7)(a-3)=0$

$\displaystyle a=3, a=\frac{7}{13}$

$\displaystyle b=\frac{7(3)-7}{2}=7$

$\displaystyle b=\frac{7(\frac{7}{13})-7}{2}=-\frac{21}{13}$

$\displaystyle \therefore a=3,b=7 ; a=\frac{7}{13},b=-\frac{21}{13}$
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