# Math Help - Questions Regarding Solve, Simplify, Domain, Absolute Value

1. ## Questions Regarding Solve, Simplify, Domain, Absolute Value

Greetings,

Well, this is my first post here on this forum... I hope this is the right place to put this, I'm really not sure what type of math this is classified as. I am in an algebra class, MTH-95 if it helps any.

Anyhow, I'm having some problems with some of my math work. I've got a lot of my worksheet done so far, but these last few are the ones that I'm really struggling with.
There are eight problems I'm having trouble with and apologize for posting so many. By no means do I expect someone to help me with all of them, I just thought I would post what I'm having problems with and perhaps I'd be able to get help with even a few of them. Any help at all is appreciated.

Simplify:
These ones I'm not exactly sure how to type, so please bare with me... I'll try typing these out and hopefully it'll make sense...
These are fractions within a fraction, which are really throwing me off. As if fractions weren't confusing enough to me as it is...

1.
$\frac{x^2-x-12}{x^2-2x-15}$
_______________
$\frac{x^2+8x+12}{x^2-5x-14}$

2.
$\frac{1}{x-2} + \frac{3}{x-1}$
_______________
$\frac{2}{x-1} + \frac{5}{x-2}$
For these, if I remember right I have to find a common denominator and multiply the problem by that... or something like that. Though I'm not sure how to find the common denominator for factions like these, or what to do with it from there...
And than I've got the whole fraction-within-a-fraction thing to worry about. Completely, utterly lost.

Solve:
3. $y - \frac{14}{y} = 5$

4. $\frac{4}{a-7} = \frac{-2a}{a+3}$
Again, for these I'd have to find a common denominator, correct? And than multiply everything by it? So number 3 the denominator would by Y and for number 4 the denominator would be... A? Maybe I'm no where close...

Simplify, use absolute value notation when necessary. If it cannot be simplified, state this:
5. $\sqrt{9x^2y } \sqrt{3x^5y^2}$

6. $^3\sqrt{ \frac{8m^7}{n^2} } ^3\sqrt{ \frac{n^9}{m^2} }$
Just to clarify, those 3's in front of the square root are supposed to be cubes, I believe, not just the number 3.
The whole square root thing I'm having a problem with. I can find the square root of simple numbers, but toss in powers and multiple numbers and I start getting completely lost.

Determine the domain for each of the following:
7. $g(x) = \sqrt {5x+2 }$

8. $d(x) = -\sqrt {7x-5 }$
These ones don't look too complicated... and yet I have no clue where to go with them. Especially the "domain" thing... What's a "domain"? I don't remember going over that in class yet...
And for number 8, I didn't think you could get the square root of a negative number? So would that simply be marked "no solution"?

And that's it for now.
Again, I apologize for posting so many. I do feel bad about posting eight problems... I feel like I'm asking for a lot. But if I can even got help with a few of them than that's better than being stuck on them.
I am going to a math tutor but unfortunately it's a public tutor at my college so there's always a crowd and only three or four tutors. More often than not I end up sitting and waiting forty-five minutes just to have one question quickly answered before the tutor rushes off again to someone else...

Any help at all would be appreciated.

1.
$\frac{x^2-x-12}{x^2-2x-15}$
_______________
$\frac{x^2+8x+12}{x^2-5x-14}$
1. $\frac{x^2 - x - 12}{x^2 - 2x - 15} = \frac{(x - 4)(x + 3)}{(x - 5)(x + 3)}$

$= \frac{x - 4}{x - 5}$.

$\frac{x^2 + 8x + 12}{x^2 - 5x - 14} = \frac{(x + 2)(x + 6)}{(x - 7)(x + 2)}$

$= \frac{x + 6}{x - 7}$.

So $\frac{\frac{x^2 - x - 12}{x^2 - 2x - 15}}{\frac{x^2 + 8x + 12}{x^2 - 5x - 14}} = \frac{\frac{x - 4}{x - 5}}{\frac{x + 6}{x - 7}}$

$= \frac{x - 4}{x - 5} \cdot \frac{x - 7}{x + 6}$

$= \frac{(x - 4)(x - 7)}{(x - 5)(x + 6)}$

$= \frac{x^2 - 7x - 4x + 28}{x^2 + 6x - 5x - 30}$

$= \frac{x^2 - 13x + 28}{x^2 + x - 30}$.

2.
$\frac{1}{x-2} + \frac{3}{x-1}$
_______________
$\frac{2}{x-1} + \frac{5}{x-2}$
For these, if I remember right I have to find a common denominator and multiply the problem by that... or something like that. Though I'm not sure how to find the common denominator for factions like these, or what to do with it from there...
And than I've got the whole fraction-within-a-fraction thing to worry about. Completely, utterly lost.
Yes, you need a common denominator:

$\frac{1}{x - 2} + \frac{3}{x - 1} = \frac{1}{x - 2} \cdot \frac{x - 1}{x - 1} + \frac{3}{x - 1} \cdot \frac{x - 2}{x - 2}$

$= \frac{x - 1}{(x - 1)(x - 2)} + \frac{3(x - 2)}{(x - 1)(x - 2)}$

$= \frac{x - 1 + 3(x - 2)}{(x - 1)(x - 2)}$

$= \frac{x - 1 + 3x - 6}{(x - 1)(x - 2)}$

$= \frac{4x - 7}{(x - 1)(x - 2)}$.

$\frac{2}{x - 1} + \frac{5}{x - 2} = \frac{2}{x - 1}\cdot \frac{x - 2}{x - 2} + \frac{5}{x - 2} \cdot \frac{x - 1}{x - 1}$

$= \frac{2(x - 2)}{(x - 1)(x - 2)} + \frac{5(x - 1)}{(x - 1)(x - 2)}$

$= \frac{2(x - 2) + 5(x - 1)}{(x - 1)(x - 2)}$

$= \frac{2x - 4 + 5x - 5}{(x - 1)(x - 2)}$

$= \frac{7x - 9}{(x - 1)(x - 2)}$.

So $\frac{\frac{1}{x - 2} + \frac{3}{x - 1}}{\frac{2}{x - 1} + \frac{5}{x - 2}} = \frac{\frac{4x - 7}{(x - 1)(x - 2)}}{\frac{7x - 9}{(x - 1)(x - 2)}}$

$= \frac{4x - 7}{(x - 1)(x - 2)}\cdot \frac{(x - 1)(x - 2)}{7x - 9}$

$= \frac{4x - 7}{7x - 9}$.

Solve:
3. $y - \frac{14}{y} = 5$

4. $\frac{4}{a-7} = \frac{-2a}{a+3}$
Again, for these I'd have to find a common denominator, correct? And than multiply everything by it? So number 3 the denominator would by Y and for number 4 the denominator would be... A? Maybe I'm no where close...
3. $y - \frac{14}{y} = 5$.

Multiply both sides of the equation by $y$.

$y^2 - 14 = 5y$

$y^2 - 5y - 14 = 0$

$(y - 7)(y + 2) = 0$

$y - 7 = 0$ or $y + 2 = 0$

$y = 7$ or $y = 2$.

4. $\frac{4}{a-7} = \frac{-2a}{a+3}$

Cross multiply

$4(a + 3) = -2a(a - 7)$

$4a + 12 = -2a^2 + 14a$

$2a^2 - 10a + 12 = 0$

$2(a^2 - 5a + 6) = 0$

$2(a - 6)(a + 1) = 0$

$a - 6 = 0$ or $a + 1 = 0$

$a = 6$ or $a = -1$.

Simplify, use absolute value notation when necessary. If it cannot be simplified, state this:
5. $\sqrt{9x^2y } \sqrt{3x^5y^2}$

6. $^3\sqrt{ \frac{8m^7}{n^2} } ^3\sqrt{ \frac{n^9}{m^2} }$
Just to clarify, those 3's in front of the square root are supposed to be cubes, I believe, not just the number 3.
The whole square root thing I'm having a problem with. I can find the square root of simple numbers, but toss in powers and multiple numbers and I start getting completely lost.
5. $\sqrt{9x^2y} = \sqrt{(3x)^2y}$

$= \sqrt{(3x^2)^2}\cdot \sqrt{y}$

$= |3x|\sqrt{y}$

$= 3|x|\sqrt{y}$.

6. $\sqrt[3]{\frac{8m^7}{n^2}}\cdot \sqrt[3]{\frac{n^9}{m^2}}$

$= \sqrt[3]{\frac{8m^7}{n^2}\cdot\frac{n^9}{m^2}}$

$= \sqrt[3]{\frac{8m^7n^9}{m^2n^2}}$

$= \sqrt[3]{8m^5n^7}$

$= \sqrt[3]{8}\cdot \sqrt[3]{m^5n^7}$

$= 2\sqrt[3]{m^5n^7}$.

Determine the domain for each of the following:
7. $g(x) = \sqrt {5x+2 }$

8. $d(x) = -\sqrt {7x-5 }$
These ones don't look too complicated... and yet I have no clue where to go with them. Especially the "domain" thing... What's a "domain"? I don't remember going over that in class yet...
And for number 8, I didn't think you could get the square root of a negative number? So would that simply be marked "no solution"?
The domain of a function is the set of values that $x$ can take.

7. $g(x) = \sqrt{5x + 2}$.

Since you can not have the square root of a negative number (I assume you're dealing with real numbers and not complex...), this means that the stuff under the square root must be $\geq 0$.

So $5x + 2 \geq 0$

$5x \geq -2$

$x \geq -\frac{2}{5}$.

So the domain of $g(x)$ is $x \geq -\frac{2}{5}$, which can also be written as $x \in \left[-\frac{2}{5}, \infty\right)$.

8. $d(x) = -\sqrt {7x-5 }$

Same thing again, the stuff under the square root can't be negative.

So $7x - 5 \geq 0$

$7x \geq 5$

$x \geq \frac{5}{7}$.

So the domain of $d(x)$ is $x \geq \frac{5}{7}$, which can also be written as $x \in \left[\frac{5}{7}, \infty\right)$.

7. Oh, wow, thank you very much. Your help is very much appreciated.