Solving Rational Equations

[stronger]

5/8- 1/a= 2/5. I solved this problem by mulitiplying each term by 40a/40a and it yielded -3a/8a. However, I also tried to do to do this problem by mulitiplying the numerators by 40a, then dividing by their denominators...and it yielded 20/3 as my answer. Are either of these end results correct? Thank you so much!!!

 

[tkhunny]

I would suggest very strongly that you not try to move the world all at the same time. This will be far less confusing.

5/8 - 1/a = 2/5

Multiply by 8

5 - 8/a = 16/5

Multiply by 5

25 - 40/a = 16

Multiply by a

25a - 40 = 16a

There is a whole lot less chance for an error when there are obviosu pieces. If you do everything at once, there is no obvious piece.

By the way, I would not have appracoed it quite this way. I would have done thiss:

5/8 - 1/a = 2/5

Add 1/a

5/8 = 2/5 + 1/a

Subtract 2/5

5/8 - 2/5 = 1/a

Add the fractions on the left

(25 - 16)/40 = 1/a

Simplify

9/40 = 1/a

a = 40/9

 

[lookagain]

5/8- 1/a= 2/5. However, I also tried to do to do this problem by mulitiplying the numerators by 40a, then dividing by their denominators...and it yielded 20/3 as my answer. Are either of these end results correct?

User stronger,

on the flip side, I would not have you do so many separate steps,
because each step adds another chance to make errors, and
having to combine fractions is a relative nuisance that I would
encourage students not to do in this type of problem. If you're
sure of the least common denominator, I strongly urge you to
clear the problem of all fractions as soon as possible insofar
as you are comfortable enough at some point in the steps.


Consider the following option, especially as it would be shown moreso vertically
so you can make your choice.

Your second method you asked about:


\(\displaystyle \bigg(\frac{40a}{1}\bigg)\frac{5}{8} \ - \ \frac{1}{a}\bigg(\frac{40a}{1}\bigg) \ = \ \bigg(\frac{40a}{1}\bigg)\frac{2}{5}\)

Do as much cancellations as you can with each product so that the denominators become ones,
and thus you will have changed it from two lines (rows) down to one line (row):


\(\displaystyle (5a)5 \ - \ 40 \ = \ (8a)2\)


Get the variable terms together on one side and have the approprate constant
by itself on to the other side of the equation:


\(\displaystyle > 25a \ - \ 40 \ = \ 16a\)
\(\displaystyle \ \ -16a \ + \ \ 40 \ \ -16a \ + \ 40\)
\(\displaystyle -------------\)
\(\displaystyle 9a \ = \ 40\)


\(\displaystyle \frac{9a}{9} \ = \ \frac{40}{9}\)


\(\displaystyle \boxed{a \ = \ \frac{40}{9}}**\)



\(\displaystyle ** \text{Be advised that you must compare this potential solution to}\)
\(\displaystyle \text{any values of the variable which are restricted in the}\)
\(\displaystyle \text{original equation. Here, the variable, a, cannot equal 0,}\)
\(\displaystyle \text{because it makes a term be undefined. But this candidate}\)
\(\displaystyle \text{is not equal to the restricted value, so we do report it as the answer.}\)