How do you simplify #((2x-5)/(x^2-9))/((3x-1)/(x+3))#?

2 Answers
Feb 18, 2017

Answer:

#(2x-5)/((x+3)(3x-1))#

Explanation:

Rewrite the problem by moving the bottom fraction in line with the top fraction and adding a division symbol. Rewrite #x^2-9# into #(x-3)(x+3)#:

#(2x-5)/((x-3)(x+3)) -: (3x-1)/(x+3)#

Flip the fraction on the left and rewrite the problem as a multiplication problem:

#(2x-5)/((x-3)(x+3)) xx ((x+3)/(3x-1))#

Cancel out the x+3

#(2x-5)/(x+3) xx (1/(3x-1))#

Consolidate the denominator:

#(2x-5)/((x+3)(3x-1))#

Feb 18, 2017

Answer:

#(2x-5)/(3x^2-10x+3)#

Explanation:

Using an example:

Consider #" "6-:2 = 3#
Although not normally done it is both correct and permissible to write this as: #6/1-:2/1#

The shortcut method is: Turn #2/1# upside down and multiply

#6/1xx1/2 = (6xx1)/(1xx2) = 6/2=3#

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Given#" "(2x-5)/(x^2-9)-:(3x-1)/(x+3)#

Write the given expression as: #" "(2x-5)/(x^2-9)xx(x+3)/(3x-1)#

But #x^2-9# is the same as #x^2-3^2=(x-3)(x+3)# giving:

#(2x-5)/((x+3)(x-3))xx(x+3)/(3x-1)#

#(2x-5)/(cancel((x+3))^1(x-3))xx(cancel((x+3))^1)/(3x-1)#

#(2x-5)/((x-3)(3x-1))#

#(2x-5)/(3x^2-10x+3)#