How do you solve #x+ \frac { 6 } { x } = 5#?

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Answer 1 · 2016-10-12T02:58:27

#x=2# and #x=3#

I would approach this problem by finding the common denominator on both sides of the equation then simplify the equation.

#x+6/x=5#

The common denominator (LCD) is #x#

#(x^2+6)/(1 cancel x)=(5x)/(1 cancel x)#

Simplify the #x#'s or use another method but you will see that you will end up simplifying the #x#

#x^2+6=5x#

#x^2-5x+6=0#

Here, too you can use any method you are comfortable with, as for me I always find the #Delta#

#Delta=b^2-4ac#, with #a=1#, #b=-5# and #c=6#

#Delta=(-5)^2-4(1)(6)=1=>sqrt Delta=+-1#

#x_1=(-b+sqrt Delta)/(2a)# and #x_2=(-b-sqrt Delta)/(2a)#

#x_1=(5+1)/(2)=3# and #x_2=(5-1)/(2)=2#

So, #x=2# and #x=3# is your solution.