How do you solve #(x-2)/(x^2-7x+10)=1/(5x-10)-1/(x-5)#?

3 Answers
Dec 29, 2015

Answer:

Simplify by getting everything on the same common denominator and then rearrange to find x

Explanation:

#(x-2)/((x-2)(x-5)) = ((x-5) - (5x - 10))/((5x - 10)(x-5))#
#(x-2)/((x-2)(x-5)) = (-4x + 5)/(5(x-2)(x-5)#
#5(x-2) = (5 - 4x)#
#5x + 4x = 5 + 10#
#x = 15/9 = 5/3#

Dec 31, 2015

Answer:

You can simplify the left side.

Explanation:

Obtaining the roots for #x^2-7x+10=0#
#Delta = 49 -40= 9#
#x=(7+-sqrt(9))/2# => #x_1 = 2# and #x_2=5#
Then #x^2-7x+10 = (x-x_1)(x-x_2)=(x-2)(x-5)#

In the expression:
#cancel(x-2)/(cancel((x-2))(x-5))=1/(5x-10)-1/(x-5)#
#2/(x-5)=1/(5x-10)#
#10x-20=x-5#
#9x=15# => #x=5/3#

Jan 9, 2016

Answer:

#x=5/3#

Explanation:

Factor each denominator.

#(x-2)/((x-5)(x-2))=1/(5(x-2))-1/(x-5)#

Simplify the first fraction by dividing #(x-2)/(x-2)=1#.

#1/(x-5)=1/(5(x-2))-1/(x-5)#

Add #1/(x-5)# to both sides.

#2/(x-5)=1/(5(x-2)#

Cross multiply.

#10(x-2)=x-5#

Simplify and solve.

#10x-20=x-5#

#9x-20=-5#

#9x=15#

#x=15/9#

#x=5/3#

When solving rational equations such as these, it's always a good idea to go back and check that this answer isn't an invalid answer. Invalidity would occur if #x=5/3# caused a denominator to equal #0#, but the only two values for #x# that would cause that here are #2# and #5#, so this is a valid answer.