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

2 Answers

Answer:

#x=3/2" "# the root

#x=1" " "# the extraneous root

Explanation:

From the give equation, we multiply both sides by the
LCD#=(2x-1)(x-1)#

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

#(2x-1)(x-1)*(x+1)/(x-1)=(2x-1)(x-1)[2/(2x-1)+2/(x-1)]#

We simplify to obtain

#(2x-1)cancel(x-1)*(x+1)/cancel(x-1)=(2x-1)(x-1)[2/(2x-1)+2/(x-1)]#

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

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

#(2x-1)(x+1)=(x-1)(2)+(2x-1)(2)#

We expand by multiplication

#2x^2-x+2x-1=2x-2+4x-2#

Transpose all terms to the left side of the equation

#2x^2-x+2x-1-2x+2-4x+2=0#

#2x^2-5x+3=0#

We can solve this now by Factoring method

#(2x-3)(x-1)=0#

Equate each factor to 0 and solve for the values of x

First factor
#2x-3=0#

#x=3/2#

Second Factor
#x-1=0#
#x=1#

By checking these values in the original equation, we will find that #x=1# is an extraneous root because it will have a division by zero

The value #x=3/2# is a Root.

God bless....I hope the explanation is useful.

Mar 26, 2016

Answer:

#x=3/2#

Explanation:

Before starting, it is important to note the restrictions in this equation. When the denominator of each fraction is set to not equal #0#, the restrictions are:

#x-1!=0color(white)(XXXXXXXX)2x-1!=0#

#x!=1color(white)(XXXXXXXXXX)2x!=1#

#color(white)(XXXXXXXXXXXXX)x!=1/2#

Thus, the restrictions are #color(red)(|bar(ul(color(white)(a/a)x!=1,1/2color(white)(a/a)|)))#.

Solving the Equation
#1#. Start by finding the L.C.D. (lowest common denominator) for the right side of the equation. Then rewrite the fractions on the right side with the inclusion of the L.C.D.

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

#(x+1)/(x-1)=(2color(darkorange)((x-1)))/((2x-1)color(darkorange)((x-1)))+(2color(blue)((2x-1)))/((x-1)color(blue)((2x-1)))#

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

#2#. Simplify the right side of the equation.

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

#(x+1)/(x-1)=(6x-4)/((2x-1)(x-1))#

#3#. Multiply both sides of the equation by #color(teal)((x-1))# and #color(teal)((2x-1))# to get rid of the denominator.

#(x+1)/color(red)cancelcolor(black)((x-1))xxcolor(teal)((2x-1)color(red)cancelcolor(teal)((x-1)))=(6x-4)/(color(red)cancelcolor(black)((2x-1)(x-1)))xxcolor(red)cancelcolor(teal)((2x-1)(x-1))#

#(x+1)(2x-1)=6x-4#

#4#. Expand the brackets on the left side of the equation.

#2x^2+x-1=6x-4#

#5#. Bring all the terms to the left side of the equation.

#color(violet)2x^2# #color(brown)(-5)x# #color(turquoise)(+3)=0#

#6#. Use the quadratic formula to solve for the values of #x#.

#color(violet)(a=2)color(white)(XXXXXX)color(brown)(b=-5)color(white)(XXXXXX)color(turquoise)(c=3)#

#x=(-b+-sqrt(b^2-4ac))/(2a)#

#x=(-(color(brown)(-5))+-sqrt((color(brown)(-5))^2-4(color(violet)2)(color(turquoise)3)))/(2(color(violet)2))#

#x=(5+-sqrt(1))/4#

#x=6/4color(white)(X),color(white)(X)4/4#

#x=3/2color(white)(X),color(white)(X)1#

However, looking back at the restrictions, ( #color(red)(x!=1,1/2)# ), #x=1# is not a valid solution. Therefore:

#color(green)(|bar(ul(color(white)(a/a)x=3/2color(white)(a/a)|)))#