How do you solve #(x - 5) / (x - 8) = (x + 1) / (x - 5)#?

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Answer 1 · 2017-06-14T14:26:28

#x=11#

Multiply both sides by #(x-5)(x-8)#. This will allow you to cancel common factors and get rid of the denominators.

#((x-5)(x-5)cancel(x-8))/cancel(x-8)=((x+1)cancel(x-5)(x-8))/cancel(x-5)#

#(x-5)^2=(x+1)(x-8)#

Next, multiply out all the brackets:

#x^2-10x+25=x^2-7x-8#

Bring all the x-terms to one side and the constants to the other:

#x^2-x^2-10x+7x=-8-25#

Simplify and solve for x:

#-3x=-33#

#x=(-33)/-3=11#

Answer 2 · 2017-06-14T14:30:27

x =11 multiply both sides by the denominators to eliminate the fractions then solve for x

# { ( x-8) xx(x-5) xx (x-5)}/(x-5) = {(x-8) xx (x-5) xx ( x+1)}/(x-8)#

Dividing out the denominators gives

# (x -5)xx(x-5) = (x-8) xx ( x+1) # Use the distributive property

# x^2 -10 x + 25 = x^2 -7x -8 # subtract x^2 from both sides gives

# x^2 - x^2 -10x + 25 = x^2 - x^2 -7x -8 # so

# -10x + 25 = -7x -8# add +10x and 8 to both sides

# -10x + 10x +25 + 8 = -7x + 10 x - 8 + 8 # this gives

# 33 = 3x # divide both sides by 3

# 33/3 =( 3x)/3 # so

# 11 = x