1. Separating constants

Hi guys, I need an idea of exactly what I'm supposed to do here. I'm working on a series of problems where the only instructions given are to 'separate the constants from the terms'

The terms are: a. -2a^3/7

b. (-3x)^2/8

c. -1/5x

Am I supposed to set them equal to 0 and get the variable on the left side?

2. Originally Posted by Lewong
Hi guys, I need an idea of exactly what I'm supposed to do here. I'm working on a series of problems where the only instructions given are to 'separate the constants from the terms'

The terms are: a. -2a^3/7

b. (-3x)^2/8

c. -1/5x

Am I supposed to set them equal to 0 and get the variable on the left side?
No, you are not to "set them equal" to anything. You are only asked to "separate" the constant from the variable.

However, I need to know if "-2a^3/7" is "-2a^(3/7)" or "-2a^(3)/7= (-2/7)a^3". In the first case, the "constant" is -2 while the "term" is a^(3/7). In the second, the "constant" is (-2/7) and the "term" is a^3.

The same ambiguity applies to (b): is it (-3x)^(2/8) or ((-3x)^2)/8. Since the first results in (-3)^(1/4), which is not a real number, I suspect you meant ((-3a)^2)/8= ((-3)^2 a^2)/8= (9/8)a^2.

Finally, in (c) do you mean -1/(5x) or (-1/5)x?

3. I'll try to type these a little less crpytically here, heh

a. -2a^3/7

In this one 2a^3 is the numerator and 7 is the denominator, the negative sign is in the middle.

b. (-3x)^2/8

(-3x)^2 is the numerator, 8 is the denominator.

c.-1/5x

-1 is the numerator, 5x is the denominator.

4. Originally Posted by Lewong
I'll try to type these a little less crpytically here, heh

a. -2a^3/7
This is a weird kind of question, but I would do something like $-\frac{2a^3}{7}=(-\frac{2}{7})(a^3)$

5. Originally Posted by Lewong
I'll try to type these a little less crpytically here, heh

a. -2a^3/7

In this one 2a^3 is the numerator and 7 is the denominator, the negative sign is in the middle.

b. (-3x)^2/8

(-3x)^2 is the numerator, 8 is the denominator.

c.-1/5x

-1 is the numerator, 5x is the denominator.
Cover up the variable and that will leave a constant. Remember that $\frac{ax}{b} = \frac{a}{b}x$ and the constant is $\frac{a}{b}$

b. Note that $(ab)^2 = a^2b^2$

c. Note that $\frac{a}{bc} = \frac{a}{b} \times \frac{1}{c}$

6. Ok, so for (b) I would put: (-3^2/8)(x^2)

Honestly, this is would I thought I was supposed to do originally, but it just seemed too simple in the context of a test containing much more advanced math.

7. Originally Posted by Lewong
Ok, so for (b) I would put: (-3^2/8)(x^2)

Honestly, this is would I thought I was supposed to do originally, but it just seemed too simple in the context of a test containing much more advanced math.
You could but it would be more proper to square the -3 to give $\frac{9}{8}$

8. And be careful to distinguish between " $-3^2$", which is equal to -9, and $(-3)^2$, which is equal to 9. Use parentheses!