[LS1toSVT]

Is it ok to ask for homework help here? I am in a Math 100 class that is taught through the computer though we meet once a week. I've been sick and need to catch up by a chapter, and go figure, it is the most difficult. The online tools are not helping and my professor isn't reachable until next week, blahblahblah I am stuck.

If homework help is not allowed here, please go ahead and delete/lock. I am just at such a stand still and am panicing.


Here are some print screen shots of some of the problems since I don't know how else to write them out w/o it making sense. Any help would be appreciated soo much, I'm just not even sure where to start with this...


1.
Question: I know I need to factor out the top and bottom. I know the top is (z-2)(z-3) but the bottom, is that factorable? I am horrible at this an can't recall or figure it out for those numbers

2.

Question: How do I know what to pull apart out of that to get the LCD? What do I do for the second part??

3.

Question: I don't even know where to start.

That's it for now, any help at all would be great, I hate being stuck like this! Oh and by the way there is no book for the class, it's all based on the computer but it's not giving a explanation that's working for me. Thanks so much!

[boa808]

1. denominator is (z+4)(z-3)

2. LCD = xyz

3. start by getting all the denominators to be the same. if you notice, (y^2 -9) = (y+3)(y-3). basically multiply the numerator and denominator of the first term by (y+3). and multiply the numerator and denominator of the last term by (y-3). all three terms should then have the same denominator which is (y^2 -9). now, if you multiply both sides of the equation by (y^2 -9), you can get rid of the denominator. then solve for y from the resulting equation.

[tealove]

for the first problem, you can factor the bottom into (z+4)(z-3). so the expression would be [(z-2)(z-3)]/[(z+4)(z-3)]. if you cancel out the (z-3) from the top and bottom, your final answer is (z-2)/(z+4)

for the second problem, the lowest common denominator is the smallest number that is a multiple of the denominators. in the case of your problem, the LCD would be x^6*y^5*z^3. uhh, i'm not really sure how to explain it really well.... say you have two fractions 1/9 and 1/10. the least common denominator would be 90 because it is the least common multiple of 9 and 10. but for 1/3 and 1/6, the LCD is 12, not 18 (=3*6). okay, that's not really a helpful example. if you're still confused, i can try to explain it better...

anyway, for part two of that problem, you need to change both expressions to have the LCD as the denominator.

3a/(x^6yz) would turn into (3a*y^4*z^2)/(x^6*y^5*z^3). you need to times both the numerator and denominator (top and bottom) by y^4 and z^2. since you multiply the top and bottom by the same exact thing, the expression is equivalent to the original and you turn the denominator into the LCD.
likewise, you need to multiply the top and bottom of the second expression by x^5 to get (3a*x^5)/(x^6*y^5*z^3).

for the third problem, you need to change both expressions that you're adding to have the same denominator. uhh, since (y^2-9) = (y-3)(y+3), you have to multiply (x+3) to both the top and bottom of the first thing you're adding (which is y/(y-3)) and you get:
y(y+3)/[(y-3)(y+3)] + y/[(y-3)(y+3)] = [y + y(y+3)]/[(y-3)(y+3)] = y[1+(y+3)]/[(y-3)(y+3)] = y(y+4)/[(y-3)(y+3)]

.... which does not equal to (y+4)/(y+3), so it's false? hahaha maybe I'M doing it wrong...

[clairdeluned]

I think the first two have been answered, so I'll help with the third

Okay, so you start with y/(y-3) + y/(y^2 - 9) = (y+4)/(y+3)

First multiply both sides of the equation by (y^2)-9, or (y+3)(y-3). In the y/(y-3) term, the (y-3)'s would cancel out on the top and bottom, leaving you with y(y+3). In the second term, the whole denominator would cancel out with the top, leaving you with y. For the right side of the equation, (y+3) would cancel out on the top of bottom, leaving you with (y+4)(y-3).

Thus, it would be:
y(y+3) + y = (y+4)(y-3)

Multiply all the stuff in the parentheses out. This would get you:

y^2 + 3y + y = y^2 -3y + 4y -12

Rearrange by getting all the y terms on the left side, leading you to:
3y = -12; thus, y = -4

Plug -4 back into the original equation, and double check that it works

(-4)/(-4-3) + (-4)/[(-4)^2 -9] = (-4 + 4)/(-4+3)

4/7 - 4/7 = 0/(-7); or 4/7 - 4/7 = 0, which is true I tried showing a total step-by-step process and not leaving out anything.. let me know if something doesn't make sense!

[LS1toSVT]

Thank you guys so much for the much needed help! Giving it another try now! Thanks again!