Question 171821


{{{6+(1)/(y+1)=(6)/(y+3)}}} Start with the given equation.



{{{6(y+1)(y+3)+((1)/cross((y+1)))cross((y+1))(y+3)=((6)/cross((y+3)))(y+1)cross((y+3))}}} Multiply <font size="4"><b>EVERY</b></font> term on both sides by the LCD {{{(y+1)(y+3)}}}. Doing this will eliminate all of the fractions.



{{{6(y+1)(y+3)+1(y+3)=6(y+1)}}} Simplify



{{{6(y^2+4y+3)+1(y+3)=6(y+1)}}} FOIL



{{{6y^2+24y+18+y+3=6y+6}}} Distribute



{{{6y^2+24y+18+y+3-6y-6=0}}} Subtract 6y from both sides. Subtract 6 from both sides.



{{{6y^2+19y+15=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ay^2+by+c}}} where {{{a=6}}}, {{{b=19}}}, and {{{c=15}}}



Let's use the quadratic formula to solve for y



{{{y = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{y = (-(19) +- sqrt( (19)^2-4(6)(15) ))/(2(6))}}} Plug in  {{{a=6}}}, {{{b=19}}}, and {{{c=15}}}



{{{y = (-19 +- sqrt( 361-4(6)(15) ))/(2(6))}}} Square {{{19}}} to get {{{361}}}. 



{{{y = (-19 +- sqrt( 361-360 ))/(2(6))}}} Multiply {{{4(6)(15)}}} to get {{{360}}}



{{{y = (-19 +- sqrt( 1 ))/(2(6))}}} Subtract {{{360}}} from {{{361}}} to get {{{1}}}



{{{y = (-19 +- sqrt( 1 ))/(12)}}} Multiply {{{2}}} and {{{6}}} to get {{{12}}}. 



{{{y = (-19 +- 1)/(12)}}} Take the square root of {{{1}}} to get {{{1}}}. 



{{{y = (-19 + 1)/(12)}}} or {{{y = (-19 - 1)/(12)}}} Break up the expression. 



{{{y = (-18)/(12)}}} or {{{y =  (-20)/(12)}}} Combine like terms. 



{{{y = -3/2}}} or {{{y = -5/3}}} Simplify. 



So the answers are {{{y = -3/2}}} or {{{y = -5/3}}}