Subtracting Rational Expressions

[Chocobitties]

Hi! First time poster. So I am working furiously toward my math finals for college. I got stuck on this problem in the pre-assessment because it looks like to me the system has the wrong info for the equation. I need to know if I'm crazy or have misunderstood the steps involved, or if I need to notify my school of a bad question . Here's the question with the solution:

Shouldn't the terms being multipled be 5x/5x and 7x/7x instead of 5/5 and 7/7, and shouldn't the denominator be 35x^2, not 35x? The solution I got was 5x^2+25-7xy over 35x^2

Thanks in advance!!

 

[stapel]

The text in the image is quite small. The following is what I think it says:

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Solution 2 of 2

You have been asked to determine the following difference:

. . . . .\(\displaystyle \dfrac{x\, +\, 5}{7x}\, -\, \dfrac{y}{5x}\)

Solution:

. . . . .\(\displaystyle \begin{align} \dfrac{x\, +\, 5}{7x}\, -\, \dfrac{y}{5x} &=\, \dfrac{x\, +\, 5}{7x}\, \cdot\, \dfrac{5}{5}\, -\, \dfrac{y}{5x}\, \cdot\, \dfrac{7}{7}

\\ \\ &=\, \dfrac{5x\, +\, 25}{35x}\, -\, \dfrac{7y}{35x}

\\ \\ &=\, \dfrac{(5x\, +\, 25)\, -\, (7y)}{35x}

\\ \\ &\, \dfrac{5x\, +\, 25\, -\, 7y}{35x} \end{align}\)

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Shouldn't the terms being multipled be 5x/5x and 7x/7x instead of 5/5 and 7/7...

Why? Don't each of the denominators already have one copy of "x"? So isn't that part already "common"? (For further information on finding common denominators, try here.)

The solution I got was 5x^2+25-7xy over 35x^2

How did you multiply (x + 5) by (5x) and get that 5x(x + 5) = 5x(x) + 5x(5) somehow equalled 5x^2 + 25?

 

[Chocobitties]

Thanks for the quick reply (and apologies because I think I double posted this topic and the other post hasn't shown up yet). I think I had a case of the midnight brain not working right. No matter how much we want to study, sometimes the body tells us we need sleep, and it does this in a sinister way by refusing to properly remember how to perform a task!