Answer:
7. LCD = 6v², v = -6
8. LCD = 2n, n = 16/3
Explanation:
You want to solve the rational equations ...
7. 4/(3v) -1/v = (v +2)/(2v²)
8. 3/n -1/2 = (n -5)/n
Least common multiple
The least common multiple of any set of numbers is the product of the unique factors to their highest powers.
7.
The denominators are 3v, v, 2v². The unique factors are 2, 3, v, and the highest powers of these are 1, 1, and 2. Then the LCM is ...
2¹·3¹·v² = 6v² . . . . . least common denominator
We find it convenient to solve rational equations by writing them in the form f(variable) = 0. Multiplying the equation by 6v² and subtracting the right side gives ...
(2v)(4) -(6v) -3(v+2) = 0
-v -6 = 0 . . . . . simplify
v = -6
8.
The denominators are 2 and n. These are both unique factors, and their highest powers are both 1. The LCM is ...
2·n = 2n
Multiplying the equation by this product gives ...
6 -n = 2(n -5)
0 = 2n -10 +n -6 = 3n -16 . . . . . add n-6 to both sides
n = 16/3 . . . . . . . add 16, divide by 3
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Additional comment
Here, nothing was really gained by using the forms f(v) = 0 and f(n) = 0. In some cases, a numerator and denominator factor will cancel when the fractions are combined into a single fraction, so using this form can avoid extraneous solutions. (If you simply multiply the equation by the LCM, as we did here, then you may not realize when you are introducing extraneous solutions.)
One way to find the LCM of denominators is to consider the most complicated denominator expression, then consider what additional factors it must have to be a multiple of the other denominators.
In problem 7, you notice 2v² is already a multiple of v, but must have a factor of 3 included for it to be a multiple of 3v. That makes the LCM be 3·2v² = 6v², as we found above.