Question 88959
{{{1/y-1/(y+5)=1/4}}}


{{{4y(y+5)(1/y-1/(y+5))=4y(y+5)(1/4)}}} Multiply both sides by the LCD 4y(y+5)


{{{4(y+5)-4y=y(y+5)}}} Distribute


{{{4y+20-4y=y^2+5y}}} Distribute again


{{{20=y^2+5y}}} Combine like terms


{{{0=y^2+5y-20}}} Subtract 20 from both sides




Now let's use the quadratic formula to solve for y:



Starting with the general quadratic


{{{ay^2+by+c=0}}}


the general solution using the quadratic equation is:


{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}


So lets solve {{{y^2+5*y-20=0}}} ( notice {{{a=1}}}, {{{b=5}}}, and {{{c=-20}}})


{{{y = (-5 +- sqrt( (5)^2-4*1*-20 ))/(2*1)}}} Plug in a=1, b=5, and c=-20




{{{y = (-5 +- sqrt( 25-4*1*-20 ))/(2*1)}}} Square 5 to get 25




{{{y = (-5 +- sqrt( 25+80 ))/(2*1)}}} Multiply {{{-4*-20*1}}} to get {{{80}}}




{{{y = (-5 +- sqrt( 105 ))/(2*1)}}} Combine like terms in the radicand (everything under the square root)




{{{y = (-5 +- sqrt(105))/(2*1)}}} Simplify the square root




{{{y = (-5 +- sqrt(105))/2}}} Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


{{{y = (-5 + sqrt(105))/2}}} or {{{y = (-5 - sqrt(105))/2}}}



which split up to



{{{y=-5/2+sqrt(105)/2}}} or {{{y=-5/2-sqrt(105)/2}}}





Which approximate to


{{{y=2.6234753829798}}} or {{{y=-7.6234753829798}}}



So our solutions are:

{{{y=2.6234753829798}}} or {{{y=-7.6234753829798}}}


Notice when we graph {{{x^2+5*x-20}}} (just replace y with x), we get:


{{{ graph( 500, 500, -17.6234753829798, 12.6234753829798, -17.6234753829798, 12.6234753829798,1*x^2+5*x+-20) }}}


when we use the root finder feature on a calculator, we find that {{{x=2.6234753829798}}} and {{{x=-7.6234753829798}}}.So this verifies our answer