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

Question

1607 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-12T02:03:44

There are two solutions:

# y = 2 + sqrt( 14 ) #

# y = 2 - sqrt( 14 ) #

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

Then:

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

So:

# y = y^2 - 5y + 2y - 10 = y^2 - 3y - 10 #

Unifying:

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

As we know in:

# a*y^2 + b*y + c = 0 # ,
# y = ( -b +- sqrt( b^2 -4*a*c ) ) / ( 2 * a ) #

So:

# y = ( 4 +- sqrt( (-4)^2 -4*1*(-10) ) ) / ( 2 * 1 ) #

# y = ( 4 +- sqrt( 16 + 40 ) ) / 2 = ( 4 +- sqrt( 56 ) ) / 2 = 2 +- sqrt( 56 ) / 2 = #

# y = 2 +- sqrt( 14 * 4 ) / 2 = 2 +- 2 / 2 * sqrt( 14 ) = 2 +- sqrt( 14 ) #