What is #(5x-1)/(2x-1) - (3x-3)/(1-2x)#?

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Answer 1 · 2018-05-01T21:33:40

#2#

In order to subtract the denominators need to be the same.

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

#=(5x-1)/(2x-1) + (3x-3)/(2x-1)" "larr div -1# out of the bracket

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

#=(8x-4)/(2x-1)#

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

#=2#

Answer 2 · 2018-05-01T21:41:13

4

Put them over a common denominator

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

#[[5x-10x^2-1+2x]-[6x^2-3x-6x+3]]/[2x-4x^2-1+2x]#

#[7x-10x^2-1-6x^2+9x-3]/(4x-4x^2-1)#

#[-16x^2+16x-4]/(-4x^2+4x-1)#

#[4(-4x^2+4x-1)]/(-4x^2+4x-1)#

4