How do you solve #-4/(x-1)=7/(2-x)+3/(x+1)#?

2 Answers

Answer:

Follow the steps below.

The rearranged equation is #-13x+5=0#, to which the solution is #x=5/13.

Explanation:

(I'm doing this in more and slower steps than normal to make it super clear. We multiply be each denominator in turn and cancel, then multiply, collect like terms and simplify)

First multiply everything by #(x-1)# and cancel:

#-4=(7(x-1))/(2-x)+(3(x-1))/(x+1)#

Now multiply everything by #(x+1)# and cancel:

#-4(x+1)=(7(x-1)(x+1))/(2-x)+3(x-1)#

Now multiply everything by #(2-x)# and cancel:

#-4(x+1)(2-x)=7(x-1)(x+1)+3(x-1)(2-x)#

First multiply out each set of parentheses:

#-4(2x-x^2+2-x)=7(x^2+x-x-1)+3(2x-x^2-2+x)#

Collect like terms within the parentheses:

#-4(x-x^2+2)=7(x^2-1)+3(3x-x^2-2)#

Multiply the constants through the parentheses:

#-4x+4x^2-8 = 7x^2 - 7 + 9x - 3x^2 - 6#

Collect all like terms on one side and equate the total to zero:

#(0x^2 )-13x+5=0#

We don't write the term with a coefficient of #0#, so the equation is #-13x+5=0#

Therefore #x=(-5)/-13=5/13.#

Jul 27, 2017

Answer:

#color(red)(x=5/13)#

Explanation:

Given:
#color(white)("XXX")-4/(x-1)=7/(2-x)+3/(x+1)#
(I will re-write #(2-x)# as standard form: #(-x+2)# in the work below)

Multiplying all terms by #(x-1)(-x+2)(x+1)# to get rid of the denominators:
#color(white)("XXX")-4(-x+2)(x+1)=7(x-1)(x+1)+3(x-1)(-x+2)#

From here on, it is simply a matter of standard algebraic operations:
#rarrcolor(white)("xx")-4(-x^2+x+2)=7(x^2-1)+3(-x^2+3x-2)#

#rarrcolor(white)("xx")4x^2-4x-8=7x^2-7-3x^2+9x-6#

#rarrcolor(white)("xx")4x^2-4x+8=4x^2+9x-13#

#rarrcolor(white)("xx")-4x+8=9x-13#

#rarrcolor(white)("xx")13x=5#

#rarrcolor(white)("xx")x=5/13#

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Here is the result verification in spreadsheet form:
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