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solve 2y/y+1=2-8/2y
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\[\frac{ 2y }{ y+1 }=\frac{ 2-8 }{ 2y }\]
\[2y(2y)=(y+1)(2-8)\] thats after cross multiplying
see what happened?
?
ok now we want to keep the 4y^2 on the left so it stays positive and move the - 6y and - 6 over to the left by adding them to both sides setting the equation = to 0
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you are going to have to use the quadratic equation which is \[\frac{ -b \pm \sqrt{b^2 - 4(a)(c)} }{ 2a }\]
in your eqaution, 4 = a, 6 = b and 6 = c
ax^2 + bx + c
\[\frac{ -6\pm \sqrt{6^2 - 4(4)(6)} }{ 2(4) }\]
thats what i wanted you to do
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