How do you solve the rational equation #(y +2)/ y = 1 /( y - 5)#?

2 Answers
Jun 21, 2018

Answer:

#x=2+sqrt14#

and

#x=2-sqrt14#

Explanation:

Since there are only two terms you can X-multiply:

#(y +2)/ y = 1 /( y - 5)#

X-multiply:

#(y +2)( y - 5) = y*1#

#y^2-3y-10=y#

#y^2-4y-10=0#

Now to solve we have to complete the square:

#ax^2+bx+c#

#a# must equal one (it does in your function).

#c=(b/2)^2#

the completed square is #(x+b/2)^2#

#y^2-4y=10#

your #b= -4#

#c=(-4/2)^2=4# and the square is #(x-2)^2#

So let's complete the square, remember we need to add #c# to both sides so we don't alter the equation:

#y^2-4y +c=10+c#

#y^2-4y +4=10+4#

#(x-2)(x-2)=14#

#(x-2)^2=14#

now solve for #x#:

#sqrt((x-2)^2)=+-sqrt14#

#x-2=+-sqrt14#

#x=2+-sqrt14#

Answer:

#y=2\pm\sqrt{14}#

Explanation:

Given rational equation:
#\frac{y+2}{y}=\frac{1}{y-5}#
will have solution such that #y\ne0, y\ne5 \ or \ y \notin{0,5}#
#(y+2)(y-5)=y#
#y^2+2y-5y-10=y#
#y^2-4y-10=0#
Using quadratic formula to solve above quadratic equation,
#y=\frac{-(-4)\pm\sqrt{(-4)^2-4(1)(-10)}}{2(1)}#
#y=\frac{4\pm2\sqrt{14}}{2)}#
#y=2\pm\sqrt{14}#