Question 318849
I'm assuming this is the problem?
{{{8/y-1/3=5y }}}
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Plug the solution into the original equation to verify.
{{{8/9-1/3=5(9)}}}
{{{24/27-9/27=45}}}
{{{5/9=45}}}
False, so {{{y=9}}} cannot be the solution.
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{{{8/y-1/3=5y}}} 
Multiply both sides by {{{3y}}} to get rid of fractions.
{{{24-y=15y^2}}}
{{{15y^2+y-24=0}}}
Use the quadratic formula,
{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{y = (-1 +- sqrt( 1^2-4*15*(-24) ))/(2*15) }}} 
{{{y = (-1 +- sqrt( 1+1440 ))/(30) }}} 
{{{highlight(y = (-1 +- sqrt(1441))/(30)) }}} 
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{{{20/(a^2+a-6) + 6/(a+3) = 5/(a-2) }}}
{{{20/((a+3)(a-2)) + 6/(a+3) = 5/(a-2) }}}
Multiply both sides by {{{(a+3)(a-2)}}}
{{{20 + 6(a+2) = 5(a+3) }}}
{{{20+6a-12=5a+15}}}
{{{6a+8=5a+15}}}
{{{highlight(a=7)}}}