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

##### 2 Answers

#### Answer:

#### Explanation:

From the give equation, we multiply both sides by the

LCD

We simplify to obtain

We expand by multiplication

Transpose all terms to the left side of the equation

We can solve this now by Factoring method

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

First factor

Second Factor

By checking these values in the original equation, we will find that

The value

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

#### Answer:

#### Explanation:

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

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

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

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

Thus, the restrictions are

**Solving the Equation**

#(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))#

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

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

#(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#

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

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

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