Will you help me figure out the solution to this rational equation? (x/2x - 1) + 1/4 = 2/2x - 1
\[\frac{ x }{ 2x-1 }+\frac{ 1 }{ 4 }=\frac{ 2 }{ 2x-1 }\]
Answer Choices A. x = \[\frac{ 9 }{ 2 }\]B. x = \[\frac{ 7 }{ 2 }\] C. x = \[\frac{ 3 }{ 2 }\] D. x = \[\frac{ 7 }{ 6 }\]
@tom982 Any ideas?
I know I have to have a common denominator...
help anyone!?
Hi sorry, I left this tab open. Firstly, thanks for taking the time to draw out this equation using the editor, it's so much easier to read and it's nice to see people making an effort. Looking at our fractions, we can't merge them as the denominators aren't the same but there are other things we can do, like multiplying through by (2x-1) giving: \[\frac{ x(2x-1) }{ 2x-1 }+\frac{ (2x-1) }{ 4 }=\frac{ 2(2x-1) }{ 2x-1 }\]We're able to do this as we've done the same to both sides. Now we can cancel out some of the terms to get:\[x+\frac{ (2x-1) }{ 4 }=2\]Now we can times everything by 4 to get rid of that last fraction:\[4x+\frac{ 4(2x-1) }{ 4 }=4\times2\]Giving\[4x+(2x-1)=8\]Can you do the last few steps to find x?
1. Add 1 to both sides, canceling out the 1 on the left side of the equal sign. I'm left with \[4x + 2x = 9\]. 2. Combine like terms, so I'm left with 6x = 9. 3. Divide 6 from both sides, so I end up with 9/6, which simplifies to 3/2. The answer is C. x = 3/2. Is this correct @tom982 ?
Fantastic explanation by the way. I appreciate it so much!
Spot on, excellent work. You can check this answer by substituting x=3/2 into your original equation and seeing that both sides are equal. Glad you liked the explanation, I was a bit unsure which way to show you but looks like I picked a good one. Here's another way to do it just for future reference: \[\frac{ x }{ 2x-1 }+\frac{ 1 }{ 4 }=\frac{ 2 }{ 2x-1 }\]\[\frac{ 4x }{ 4(2x-1) }+\frac{ (2x-1) }{ 4(2x-1) }=\frac{ 8 }{ 4(2x-1) }\]Now we have the same denominator for all 3, multiplying through by 4(2x-1) only leaves the numerators as an equation: 4x+(2x-1)=8 which you solved perfectly before.
Wow. I appreciate your time and effort. Gracias por ayudarme! Thank you for helping me! @tom982
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