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Help On Quadratic Equation!!
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^^^
\[4\pm \sqrt{6}\]/2 is the answer
how did you get that
first solve 3/y-2 -2 by taking lcm
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u will get 7-2y/y-2 on left side of equation
you have \[ \large \frac{3}{y-2}-2=\frac{1}{y-1}\] if you multiply the equation by (y-2)(y-1) you get \[ \large 3(y-1)-2(y-2)(y-1)=y-2 \] this is the equation! can you go on by yourself?
\[\frac{3}{y-2} - 2 = \frac{1}{y-1}\] \[=> \frac{3 - 2(y-2)}{y-2} = \frac1{y-1}\] \[=>\frac{ 7-2y}{y-2} = \frac1{y-1}\] \[=>(7-2y)(y-1) = 1(y-2)\] Now, solve this equation to get a quadratic equation .. then use the quadratic formula .. i.e \[\frac{b \pm (b^2 - 4ac)}{2a}\] to get your answer...
okay thanks, Ill try doing it myself now
:)
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