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OpenStudy (anonymous):
Solve 2-y/4+3y/5=y+2/3 please!
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OpenStudy (anonymous):
Is it like this..\[\frac{2-y}{4+3y}*\frac{1}{5}=\frac{y+2}{3}\]
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
Do it like this...\[\frac{3(2-y)}{(4+3y)}*\frac{1}{5}*5=\frac{5(y+2)}{3}*3\]\[\frac{(6-3y)}{(4+3y)}*(4+3y)=(5y+10)(4+3y)\]\[6-3y=15y^2+20y+30y+40\]\[6-3y+3y-6=15y^2+50y+40+3y-6\]So you get a quadratic equation..\[15y^2+53y+34=0\]Now solve for y..
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