How to solve an algebraic equation with fractions with an unknown in the denominator

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brycewaters10

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Can someone help figure out how to solve this equation:

(3 / 5x) - (2 / 6x) = (2 / x) + (7 / 3x)

There must be something I'm missing, whatever I try just doesn't work out... Thank you in advance!
 
brycewaters10 said:
Can someone help figure out how to solve this equation:

(3 / 5x) - (2 / 6x) = (2 / x) + (7 / 3x)

There must be something I'm missing, whatever I try just doesn't work out... Thank you in advance!
Click to expand...
Can you solve for 'u' in the following equation:

3*u/5 - u/3 = 2*u + 7*u/3
 
brycewaters10 said:
Well, I get 4u = 65u, so only valid for u=0... Right?
Click to expand...
Correct.... but can you please review your original post and make sure that you have posted the correct problem?
(3 / 5x) - (2 / 6x) = (2 / x) + (7 / 3x)
As posted it reads:

35x − 26x = 2x + 73x\displaystyle \displaystyle{\dfrac{3}{5x} \ - \ \dfrac{2}{6x} \ = \ \dfrac{2}{x} \ + \ \dfrac{7}{3x}}
 
Yes, that's the equation, I just didn't realize you can put in actual fractions in these posts. I'm helping my child (in 9th grade) with algebra and this is an equation they got to solve. It's the first one since he's been doing algebra that I've had difficulty approaching... I tried finding a common denominator -- 30x, but then I end up with (8 / 30x) = (130 / 30x), and I'm not sure how to solve this for x when x is in the denominator. Just by looking at it I can see that it's only valid for x=0, but I'm not sure how to actually solve this...
 
The general procedure is to multiply by the common denominator. That leads to 8 = 130, which is never true; so this equation has no solution. (The equation is not valid when x=0.)

The usual way to do this is to multiply every term by 30x right at the start, but it can also be done as you have done it, with this final step.

Keep in mind that "no solution" is a valid answer!

One more thing: you don't have to use LaTeX, which SK used to write the equation (fractions are not the easiest thing to write in that form), but when you write in plain text, it's safer to write like this: 3 / (5x) - 2 / (6x) = 2 / (x) + 7 / (3x) , to make it clear what is in the denominators.
 
brycewaters10 said:
Yes, that's the equation, I just didn't realize you can put in actual fractions in these posts. I'm helping my child (in 9th grade) with algebra and this is an equation they got to solve. It's the first one since he's been doing algebra that I've had difficulty approaching... I tried finding a common denominator -- 30x, but then I end up with (8 / 30x) = (130 / 30x), and I'm not sure how to solve this for x when x is in the denominator. Just by looking at it I can see that it's only valid for x=0, but I'm not sure how to actually solve this...
Click to expand...
If two (non zero) quantities are equal then so is their recriprocal. EX: 2/3 = 2/3 and 3/2=3/2. So if (8 / 30x) = (130 / 30x) and you do not want the x's in the denominator simply write 30x/8 = 30x/130. The numerators are the same so you only have equality if the denominators are the same OR the numerators both equal 0. Since the denominators are not equal we must have 30x=0 or x=0. But be careful, each side of (8 / 30x) = (130 / 30x) is undefined when x=0. So we must reject that answer. For completeness, even if only one denominator equaled 0 when x=0, we would reject the solution x=0.
 
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