How do you solve (y+2) / y = 1 / (y-5)y+2y=1y5 and find any extraneous solutions?

1 Answer
May 12, 2018

There are two solutions:

y = 2 + sqrt( 14 ) y=2+14

y = 2 - sqrt( 14 ) y=214

Explanation:

( y + 2 ) / y = 1 / ( y - 5 ) y+2y=1y5

Then:

1 * y = ( y + 2 ) * ( y - 5 ) 1y=(y+2)(y5)

So:

y = y^2 - 5y + 2y - 10 = y^2 - 3y - 10 y=y25y+2y10=y23y10

Unifying:

y^2 - 4y - 10 = 0 y24y10=0

As we know in:

a*y^2 + b*y + c = 0 ay2+by+c=0 ,
y = ( -b +- sqrt( b^2 -4*a*c ) ) / ( 2 * a ) y=b±b24ac2a

So:

y = ( 4 +- sqrt( (-4)^2 -4*1*(-10) ) ) / ( 2 * 1 ) y=4±(4)241(10)21

y = ( 4 +- sqrt( 16 + 40 ) ) / 2 = ( 4 +- sqrt( 56 ) ) / 2 = 2 +- sqrt( 56 ) / 2 = y=4±16+402=4±562=2±562=

y = 2 +- sqrt( 14 * 4 ) / 2 = 2 +- 2 / 2 * sqrt( 14 ) = 2 +- sqrt( 14 ) y=2±1442=2±2214=2±14