Is there a way to simplify an expression like the following where x is only present in the denominators?
1/x - 2/(4x - 2)?
Exactly same here - the LCD is x * (4x - 2)OK, so normally to add/subtract 2 fractions I would use the LCD.
I tried that but it only made the whole expression more complex.
Like the 1/5 - 2/13 suggestion, I would simply and calculate using a common denominator of 5 x 13.
I can’t see how to do this in this example.
And that exactly is my point. What is simpler is sometimes in the eye of the beholder.OK, so normally to add/subtract 2 fractions I would use the LCD.
I tried that but it only made the whole expression more complex.
Like the 1/5 - 2/13 suggestion, I would simply and calculate using a common denominator of 5 x 13.
I can’t see how to do this in this example.
Better is that the least common divisor is x(2x- 1) since, as Jomo suggested, \(\displaystyle \frac{2}{4x-2}= \frac{1}{2x-1}\)Exactly same here - the LCD is x * (4x - 2)
and continue.....
Exactly same here - the LCD is x * (4x - 2)
and continue.....
I think when rational expressions (fractions) are involved, "simplify" means to express with only one vinculum and no common factors.What do you mean by "simplify"?
[MATH]\dfrac{1}{x} - \dfrac{2}{4x - 2} = \dfrac{1}{x} - \dfrac{2}{2(2x - 1)} = \dfrac{1}{x} - \dfrac{1}{2x - 1}.[/MATH]
That is the only transformation of the expression that is unambiguously simpler. We could turn the expression into a single fraction with a quadratic denominator and a linear numerator, but why is that simpler than fractions with linear denominators?
Almost certainly you are correct, but it is a usage that makes no sense to a student. who likely does not know (and does not need to know) the word "vinculum." The problem would be clearer if it said "For each expression below, find a single fraction that is equivalent in value and simplify that fraction if possible.I think when rational expressions (fractions) are involved, "simplify" means to express with only one vinculum and no common factors.