solve for x and y (x+y-8)/2 = (x-2y-14)/3 = (3x+y-12)/11 That's the question, no questions asked!
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Originally Posted by ice_syncer solve for x and y (x+y-8)/2 = (x-2y-14)/3 = (3x+y-12)/11 That's the question, no questions asked! $\displaystyle \frac{x+y-8}{2}=\frac{x-2y-14}{3}$ . Simplify : $\displaystyle x+7y+4=0$ ---- 1 $\displaystyle \frac{3x+y-12}{11}=\frac{x+y-8}{2}$ . SImplify : $\displaystyle 5x+9y-64=0$ ---- 2 1x5 $\displaystyle 5x+35y+20=0$ ---- 3 3-2 .. carry on here .
Originally Posted by mathaddict $\displaystyle \frac{x+y-8}{2}=\frac{x-2y-14}{3}$ . Simplify : $\displaystyle x+7y+4=0$ ---- 1 $\displaystyle \frac{3x+y-12}{11}=\frac{x+y-8}{2}$ . SImplify : $\displaystyle 5x+9y-64=0$ ---- 2 1x5 $\displaystyle 5x+35y+20=0$ ---- 3 3-2 .. carry on here . I did the same thing, but the book here said that the answer is x=2, y=6
Originally Posted by ice_syncer I did the same thing, but the book here said that the answer is x=2, y=6 Lets give it a test then : $\displaystyle \frac{x+y-8}{2}=\frac{x-2y-14}{3}$ , sub x=2 and y=6 $\displaystyle 0\neq-8$ so the answers are wrong .
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