How do you solve rational equations #(6x + 4 )/(x+4)=(2x+2)/(x-1)#?

Question

↳Redirected from "If an electode reaction has dssolved oxygen as a reactant, is the electrode an anode or a cathode?"

2313 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-13T17:05:53

Put it on a common denominator.

The LCD (Least Common Denominator) would be (x + 4)(x - 1)

#((6x + 4)(x- 1))/((x + 4)(x - 1)) = ((2x + 2)(x + 4))/((x + 4)(x - 1))#

We can now eliminate the denominators, because both fractions are now equivalent.

#6x^2 + 4x - 6x - 4 = 2x^2 + 2x + 8x + 8#

Since this is a quadratic equation, we must put everything to one side of the equation, so that the other side is 0.

#4x^2 - 12x - 12 = 0#

#4(x^2 - 3x - 3x) = 0#

As you can see, this cannot be factored. I will solve by completing the square, but you could also use the quadratic formula.

#4(x^2 - 3x + n)= 12#

#n = (b / 2)^2#

#n = (-3/2)^2#

#n = 9/4#

#4(x^2 - 3x + 9/4 - 9/4) = 12#

#4(x^2 - 3x + 9/4) - 4(9/4) = 12#

#4(x - 3/2)^2 - 9 = 12#

#(x - 3/2)^2 = 21/4#

#(x - 3/2) = +-sqrt(21/4)#

#x = (+-sqrt(21))/2 + 3/2#

#x =(+- sqrt(21) + 3)/2#