Question

What is the solution to the rational equation x/2x-1 - 1/4 = 2/2x-1 ?

Answer 1

Possibly my worst topic to date. Someone teach us, lol.

Answer 2

Rewrite it as
\(\large\frac{x}{2x-1}-\frac{2}{2x-1}=\frac{1}{4}\)

Answer 3

Next step: subtract and then cross multiply.

Answer 4

so 2x^2-1 and 4x-1 ?

Answer 5

@freckles

Answer 6

can you finish this?

Answer 7

it looks like mathmate is here to your rescues
i will be a cheerleader for now and root for mathmate

Answer 8

If you noticed the common denominator on the left hand side that would make an easy subtraction:
\(\large\frac{x-2}{2x-1}=\frac{1}{4}\)
All you have left is cross multiply and solve for x.

Answer 9

@freckles
Everybody can join in! Get wet!

Answer 10

Check your solutions
to see if they are within the domain of the equation

Answer 11

would it be 7/2?

Answer 12

ohh. I get it now. and yes it's 7/2 ^^

Answer 13

you could have also chose to multiply your original equation by 2x-1 to get rid of the denominator (while keeping in mind that x cannot be 1/2)
\[x-\frac{1}{4}(2x-1)=2 \]
if you don't like that 4 in the bottom
multiply both sides by 4
you have
4x-(2x-1)=8
now solve for x
4x-2x+1=8
2x+1=8
2x=7
x=7/2

Answer 14

What is the simplified form of x^2 - 5x + 6/15xy^2 all over 2x^2 - 7x + 3 /5x^2y? could you help with this one also?

Answer 15

\[\frac{\frac{x^2-5x+6}{15xy^2}}{\frac{2x^2-7x+3}{5x^2y}} \text{ or read \it this way } \\ \frac{x^2-5x+6}{15xy^2} \cdot \frac{5x^2y}{2x^2-7x+3}\]
do some factoring and see if things cancel