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

Question

Hazel Anglin · Accepted Answer

#x=3/2# 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)|)))#

Julia Adams · Answer

#x=3/2" "# the root

#x=1" " "# the extraneous root

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.

Ethan Adkins · Answer

undefined To solve the equation (x+1)/(x-1) = 2/(2x-1) + 2/(x-1), we can follow these steps:

1. Start by finding a common denominator for the fractions on the right side of the equation. The common denominator will be (x-1)(2x-1).

2. Multiply both sides of the equation by the common denominator to eliminate the fractions.

3. Simplify the equation by distributing and combining like terms.

4. Solve for x by isolating the variable on one side of the equation.

5. Check the solution by substituting the value of x back into the original equation to ensure it satisfies the equation.

Here are the steps in detail:

1. The common denominator is (x-1)(2x-1).

2. Multiply both sides of the equation by (x-1)(2x-1):

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

3. Simplify the equation:

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

Expand and simplify both sides:

2x^2 - x + 2x - 1 = 2(x^2 - 2x + 1) + 2(2x^2 - 3x + 1)

4. Continue simplifying:

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

Combine like terms:

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

5. Rearrange the equation to have all terms on one side:

0 = 6x^2 - 2x^2 + 6x - x + 4 + 1

Combine like terms:

0 = 4x^2 + 5x + 5

6. Set the equation equal to zero and attempt to factor or use the quadratic formula to solve for x. However, in this case, the equation cannot be factored easily, so we will use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 4, b = 5, and c = 5.

Substitute these values into the quadratic formula:

x = (-5 ± √(5^2 - 4(4)(5))) / (2(4))

Simplify:

x = (-5 ± √(25 - 80)) / 8

x = (-5 ± √(-55)) / 8

Since we have a negative value under the square root, the equation has no real solutions. Therefore, there are no solutions to the equation (x+1)/(x-1) = 2/(2x-1) + 2/(x-1).