Home > Algebra > Clearing Denominators in Rational Equations

How do you solve #(2x)/(x-4)=8/(x-4)+3#?

Answer 1

Julia Adams

No solution!

First, observe that 4 (division by zero) is not a solution.

Then, multiply both sides by #(x-4)#, you get

#2x = 8+3*(x-4)# #=> 3x-2x= 3*4-8# #=> x = 4# which is impossible! So there is no solution

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Eva Adamson

The equation is unsolvable.

We have:

#(2x)/(x-4) = 8/(x-4)+3#

Multiply all terms by #x-4#.

#2x=8+3(x-4)#

Enlarge the parentheses.

#2x=8+3x-12=3x-4#

Add #4-2x# to both sides.

#4=x# or #x=4#

Unfortunately this leads to a problem that #x=4# is a singularity (mathematicians don't like infinities).

Reorganise the original equation by subtracting #8/(x-4)# from both sides.

#(2x-8)/(x-4)=3#

This results in:

#(2(x-4))/(x-4)=3#

This gives #2=3# the equation makes no sense unless #x=4# in which case both sides are infinite..

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 3

Savannah Anderson

To solve the equation (2x)/(x-4)=8/(x-4)+3, we can start by multiplying both sides of the equation by (x-4) to eliminate the denominators. This gives us 2x = 8 + 3(x-4). Simplifying further, we have 2x = 8 + 3x - 12. Combining like terms, we get 2x - 3x = -4. Simplifying again, we have -x = -4. Finally, dividing both sides of the equation by -1, we find x = 4.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.