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subtracting equations
Fizban140
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I am having math trouble again and my book will not explain this one at all, and I missed the past two weeks of class so I really am lost.
I have 10/(4x-8) - 15/(10-5x)
From what I can gather I have to build up both the denominators to be the same so I find the LCD of the two. To do that I have to factor them.
4(x-2) and 5(2-x) so that makes my LCD 20(x-2)(2-x) right? I tried that and I got some ridiculous answer that is not the right answer so what am I suppose to do? I know after that step I need to raise the denominator by what it is missing so the first one I would multiply by 5 and add 2-x right?
I have 10/(4x-8) - 15/(10-5x)
From what I can gather I have to build up both the denominators to be the same so I find the LCD of the two. To do that I have to factor them.
4(x-2) and 5(2-x) so that makes my LCD 20(x-2)(2-x) right? I tried that and I got some ridiculous answer that is not the right answer so what am I suppose to do? I know after that step I need to raise the denominator by what it is missing so the first one I would multiply by 5 and add 2-x right?
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Anywho. Think about what you want your LCD to look like. If you were trying to add 4/5 + 3/9, what would that LCD look like? Answer spoilered!
Now what would the next step be? You need to 'compensate' (I'm sure there's a better way to think about this, but its how it is set in my head) the numerators for the factors you just added. So for the 4/5's numerator you need to multiply 4*9 for your new numerator, and then add that to 3*5. So your numerator is now...
Ohhhh kay! So you have compatible fractions!
Huzzah!
Now, applying this to your problem...you need to find some way to easily combine those 2 denominators...factoring isn't working...what other way can you combine them?
If you still can't make it work, let me know where you're having issues.
You want the denominators to be the same; LCD is the smallest denominator you can get to achieve this, but in actuality it doesn't really matter what it is as long as it's the same
So for example, if you had
1/2 + 1/4 + 1/8
The LCD is obviously 8, so you could multiply each fraction by something that equals 1 and gives you a denominator of 8 for each fraction:
4/4*1/2 + 2/2*1/4 + 1/1*1/8 = 4/8 + 2/8 + 1/8 = 7/8
You could also just multiply them all together (the simplest in terms of not thinking ahead)-
(4*8/4*8) * 1/2 + (2*8/2*8) * 1/4 + (2*4 / 2*4)*1/8 = 32/64+16/64+8/64 = 56/64 = 7/8
Regardless of what you make the denominator, as long as you're consistent in multiplying by things that're equal to 1 throughout, you'll always get the same answer in the end
Going back to your problem, one thing that may help you out a little - you can always factor out a -1 as well
(x-2) = -1 * (2-x)
However, if you do it your way you should eventually get the same answer as well
http://www.helpwithfractions.com/least-common-multiple.html has a good strategy. (2nd one. The first one is stupid except for small numbers.) I'll go through it, so you have two sources instead of one.
First, the LCM just deals with the numeric part, so you can ignore x and y for now. Lets say we're dealing with the numbers 2, 3, 4, 5, 6, 7, 8, 9.
Basically, it's just breaking down the numbers into primes.
(Now, first, I would knock out all numbers that have multiples already, namely, 2, 3, 4. (any number that is a multiple of 8 will also be a multiple of 2 and 4. Same thing for 9 and 3.) This leaves behind 5, 6, 7, 8, 9. You don't need to do this, but it's good to know.)
Now, factorization:
2: 2
3: 3
4: 2*2
5: 5
6: 2*3
7: 7
8: 2*2*2
9: 3*3
Now, the number that has the most 2s is 8, which has 3. So, the LCM's factorization will have 2*2*2. For 3s, 9 has 3*3, to create 2*2*2*3*3. The complete factorization would be 2*2*2*3*3*5*7, which is 2,520.
Yes, the LCD is the LCM of the denominators.
For what you're taking about they mean the same thing; the least common multiple of a group of values will be the least common denominator for adding them together
The above post covers this fairly well; with variables, the process is the same - you want to factor it into the smallest possible factors, and then figure out what's the minimal overlapping set
So for
6x+12 , 3x + 12 , 5x+10, and 12
First factor them all:
2*3*(x+2) , 3 * (x+4) , 5 * (x+2) , 2 * 2 * 3
So the LCM will be
2 * 2 * 3 * 5 * (x+2) * (x+4)
Well now the book has me confused. They way I understand it is if I am trying to add 1/3 and 1/4 I need to multiply 1/3 by 4 which gives me 4/12 and 14 by 3 which gives me 3/12 but my book is doing it different. For some reason they are only multiplying the numerator or atleast it looks like it. They put a parentheses with the LCD next to the term being multiplied and they do not do anything with it until the end of the problem, once everything is reducded then it is multiplied. This is very confusing to me. At the start of these problems I need to find the LCD and multiply everything by that number correct? I do I raise the denominator by the missing number and then raise the numerator by that?
No. With (eg) 1/(w-1), 3/(w-1), 6/w, and 9/w, the LCD would be (w-1)*w.
EDIT: I'm not sure what your book is doing. Once you have the LCD, you want each term to have the LCD as the denominator; otherwise you can't actually add them together, which is your goal. Multiply both the numerator and denominator by (LCD/denominator). (eg) if the LCD is 40, and the fraction is 4/10, you should end up with 16/40.
Actually I got it, I was way overthinking this. I am horrible at math, for whatever reason the easiest shit confuses the fuck out of me and then I get stuck on a couple problems for an hour.
To expand on my earlier example, if you have 1/(w-1), 3/(w-1), 6/w, and 9/w, remember that you can multiply anything by one. In this case, 1 = w/w (unless w=0) and 1 = (w-1)/(w-1). (unless w=1) (This is ignorable in this case, as 0 and 1 mess everything up in the original equation; however, in general, you need to watch out for divisions by zero.)
In any case, you want everything to have the same denominator, so, (1/(w-1))*(w/w)->w/w(w-1), 3/(w-1)*(w/w)->3w/w(w-1), (6/w)*((w-1)/(w-1))-> 6(w-1)/w(w-1), and (9/w)*((w-1)/(w-1))->9(w-1)/w(w-1).
Now, you can add everything together:
(w+3w+6(w-1)+9(w-1))/w(w-1).
Simplify:
(4w+15(w-1))/w(w-1)
Anyways the problem is Karen can ride her bike from home to school in the same amount of time as she can walk from home to the post office. She rides 10 mph faster than she walks. THe distance from her home to school is 7 miles and the distance from her home to the post office is 2 miles, how fast does karen walk?
I set it up like this 7/x = 2/(x-10) and got the wrong answer so I tried 7/(x+10) = 2/x and got the wrong answer, how am I setting it up wrong?
I have no idea how to set up what goes where, my good only gives one example that I figured out on my own without reading the instructions because it was so simple. And there missing numbers? Two equations? I do not understand and the next one is even worse.
Molly bought $5.28 worth of oranges and 8.80 worth of apples she bought 2 more pounds of oranges than apples. If applies cost twice as much per pound as oranges then how many pounds of each did she buy?
I set that one up like this which I know is wrong. PpP is price per pound, A is amount and Tp is total price
PpP A Tp
O a/2 2a 5.28
A a a 8.80
Dp = Distance to post office in miles = 2
Sw = Karen's speed (in mph) while walking = Sb - 10
Sb = Karen's speed (in mph) while biking = Sw + 10
Speed is distance over time, so we have:
Ts = Time it takes her to bike to school in hours = Ds/Sb = 7/Sb = 7/(Sw + 10)
Tp = Time it takes her to walk to the post office in hours = Dp/Sw = 2/Sw
From the problem description we know that Ts=Tp, so 7/(Sw + 10) = 2/Sw
Those denominators are annoying, so let's get rid of them. First multiply both sides of the equation by Sw; since we're multiplying both sides of the equation by the same thing the equation will remain true. That leaves us with 7*Sw/(Sw+10) = 2. Now we multiply both sides by (Sw+10) to get rid of that denominator; now we have 7*Sw = 2*(Sw + 10). Multiply the right side out to get 7*Sw = (2*Sw) + 20. We're trying to solve for Sw, so subtract both sides by (2*Sw) to get rid of it on the right side of the equation and we have 5*Sw = 20. Finally we divide both sides by 5 to get Sw = 4.
That's what the problem is asking for, but let's keep going to check our work. Sb = Sw + 10, so Sb = 4 + 10 = 14.
Tp = Dp/Sw = 2/4 = .5 hours
Ts = Ds/Sb = 7/14 = .5 hours
The problem says it takes her the same time to bike to school as it does walk to the post office, which checks out with our figures, so we're done.
Ts=Tp
7/Sb = 2/(Sb - 10) // substitution
7*(Sb - 10)/Sb = 2 // Multiply both sides by (Sb - 10)
7*(Sb - 10) = 2*Sb // multiply both sides by Sb
7*Sb - 70 = 2*Sb // expand left side
5*Sb - 70 = 0 // subtract 2*Sb from both sides
5*Sb = 70 // add 70 to both sides
Sb = 14 // divide both sides by 5
Sw = Sb - 10 = 14 - 10 = 4
Should this shit be so confusing? I can't even comprehend how to think about these things I do not think I should be taking any math classes. People who are good at math actually piss me off, and that just makes me angry in class because so many things come so easily for them since they are good at math.
Could you clarify what you're referring to here?
When solving problems the goal is usually to get it in the form x = (something) where x is the variable you're solving for. That's the real purpose behind all the equation manipulation you see, and you should keep that in mind whenever you're doing a problem. Each step you take should be a step towards isolating x, like unwrapping a present someone's put in too many boxes.
Please give me a detailed explanation of how you did that multiplication and simplification. Don't combine any steps and explain everything, no matter how trivial it seems.
Cancel out an x on the top and bottom so I am now left with (x+10)(7)
7x+70
Onto more confusing stuff now (funny how math keeps getting worse and worse and I was never able to even do the first stuff) I have to do this shit:
(2a^1/2 / 6^1/3)^6
To start off I raise everything by the 6th power correct? So I get 64^1/2 / 216^1/3 which is now 8/14.5, I fucking hate math.
I must be incredibly horrible at math if the book doesn't even bother to show examples of things I find so confusing and complex that it takes me hours to figure out on my own, I am an idiot.
I guess I am just having problems with the power a power rules which my book refuses to explain and google only answers with confusing d x and n symbols that I do not understand. I need this explained to me like I am a 9 year old.
Fixed.
edit: As for powers, they get a little crazy when you're dealing with polynomials inside parenthesis. (1 + 2)^2 would be (1 + 2)(1 + 2). That also applies to fractions. (1/2 + n/5)^2 = (1/2 + n/5)(1/2 + n/5).
? Shouldn't the end result have an "a" in it somewhere?
Alright, first, I'm going to change the numbers to make a new problem:
((5a^(1/3)) / (8^(1/4)))^12
First step: the ^12 on the whole fraction can be divided into the numerator and denominator:
(5a^(1/3))^12 / ((8^(1/4)))^12
2nd step: move the ^12 on to each term individually:
(5^12 * (a^(1/3))^12) / ((8^(1/4))^12)
3rd step: combine the exponents:
(5^12 * a^(12/3)) / (8^(12/4))
4th step: simplify the exponents:
(5^12 * a^4) / (8^3)
Finally, calculate the exponents, and cancel out numbers as needed. Doesn't work with the numbers I picked, though.
"the square root of 2 times the cubed root of three" is the exact same as
"the square root of 2 times the cubed root of 3". I think you mistyped something.
Anyways one thing I did not understand and that I have asked here before I think is why can't I cancel out on x/x when it is written like x-x^-2/1-x^-2 I wrote the answer as x since x/1 = x but I the teacher saw it and yelled at me saying that is a mistake he told us not to make from day one. Well I always do that on problems and it is usually right. If I have 4(x-2)/x(x-2) the x-2 cancels out even though they are both apart of a term you can cancel it out. Why can't I do that when it is addition or subtraction?
I am not sure how much longer I am going to take this class, I am probably going to fail it and owe the government $400 which I worked way too hard for but there is nothing I can really do at this point I am just terrible at math.
Remember, you can multiply anything by 1 and not change it. You can add 0 to anything and not change it. You can do anything to one half of an equation, as long as you do the same thing to the other side. It sounds like you're trying to add 1 to something, and not change it, which doesn't work.
EDIT: Oh, and by the way, there are some horrible math textbooks and teachers out there. I've had good luck with Saxon.
One thing you should try to check the validity of your cancellations is substituting in test numbers. For example, if you think
x-x^2 / 1-x^2 should equal x,
try it with x=0:
0-0^2/1-0 = 0/1 = 0. It works for x=0!
Try it again with x=2, say:
2-2^2 / 1-2^2 = -2/-3. But that doesn't equal 2, so the cancellation isn't valid.
edit: Sorry, I didn't notice the -'s in your exponents. I stand by my advice, though.
Alright, I'm home now, so I'm going to take a crack at it. When you start with a fraction, you can indeed remove terms from the fraction, as long as you're consistent. However, this is a simplification. What you are *really* doing is separating the sections of the fraction. With
4(x-2) / x(x-2)
when you remove (x-2), what you are really doing is changing it to (4/x) * ((x-2)/(x-2)), saying ((x-2)/(x-2)) = 1, and substituting it to turn the equation into (4/x) * 1=(4/x).
These steps get annoying after a while, so they get compressed into crossing out terms.
Now, back to x-x^-2/1-x^-2. What you are trying to do is say
(x-x^-2) / (1-x^-2) = (x/1) - (x^-2/x^-2), which isn't right,
and then substituting 0 for (x^-2/x^-2), not 1.
As a bonus, to solve the problem, I would start with the most annoying term, and see if I can fix it. In this case, that's x^-2.
I'm going to start by rephrasing to:
(x-(1/x^2)) / (1-(1/x^2))
and multiplying by (x^2/x^2), in order to neutralize the x^-2. That's equal to 1, so we're allowed to multiply by it.
((x-(1/x^2))*x^2) / ((1-(1/x^2))*x^2)
That's equal to
(x^3-(x^2/x^2)) / (x^2-(x^2/x^2))
and crossing out (x^2/x^2),
(x^3-1) / (x^2-1)
In a general "learning how to do this stuff" manner, do not ignore the Internet. If you are having trouble with a concept, I'd suggest Googling the term, and checking out the first few sites. (If one confuses you, ignore it and move on.) Eventually you should find a site that will explain things well to you. Being able to pull from multiple sources will mean you won't have to worry so much about the book explaining a concept badly, and should allow you to find a method of learning that will fit you better.
Just accept the fact that you cannot cancel anything when you are not multiplying anything in the numerator and denominator.
for instance:
x-x^-2 / 1 - x^-2
the only way to cancel anything (this won't really SIMPLIFY the problem) is to take out a like term from both.. lets factor out x^-2
then we get
x^-2(x/x^-2 - 1) / x^-2(1/x^-2 - 1) (notice that if you multiply those x^-2's back in, you get exactly you first equation).
now we can cancel out the x^-2's and do some exponential debauchery.
x^3 - 1 / x^2 - 1
yay..
notice how we did absolutely nothing to solve or simplify this problem.. but at least we did SOMETHING with it.
lets maybe check our answers okay? lets plug in x = 2
2^3-1 / 2^2 - 1 = 7/3 <
now with our OG equation
x-x^-2 / 1 - x^-2
(2-1/4)/(1-1/4) = 2.3333333 = 7/3 <
yay they are equal. this means that i did my algebra right.
what you should take from this is that in algebra, no matter what you do to the equations. is that THEY SHOULD ALWAYS EQUAL EACH OTHER QUANTITATIVELY (IE PLUGGING A NUMBER IN FOR YOUR VARIABLE) ALWAYS ALWAYS ALWAYS ALWAYS.
ALWAYS
ALWAYS.
sorry for the caps but this is VERY important.
Cancellation is where you basically "undo" an operation somewhere else in the expression. (4(x-2))/(x(x-2)) = (4(x-2))/((x-2)x) due to the commutative property of multiplication. What is this expression saying? We start with 4, then we multiply by (x-2), then we divide by the product of (x-2) and x.
Quick aside: take a/b. What happens if we multiply the denominator (bottom part) by c? The quotient becomes smaller by a factor of c. In other words, a/(bc) = (a/b)/c. To convince yourself this is true, imagine some particular a/b, and then imagine that you divide by twice as much as b instead (so that you have a/(2b)). The latter quotient will be half as much as the former; this generalizes to when the 2 is any other value (except 0, but that's a whole other can of worms I mention only for technical accuracy and you shouldn't worry about).
Back to our original example. Because of the above we can rewrite (4(x-2))/((x-2)x) as ((4(x-2))/(x-2))/x. We now have "multiply 4 by (x-2) then divide by (x-2) then divide by x". Since multiplication and division are inverse operations, multiplying and then dividing (or vice versa) some number by the same quantity results in the original number. That means we can remove them from the equation without affecting the value, leaving us with 4/x.
While it's not usually referred to as canceling, you can do a similar thing with addition and subtraction. If you have some money in your wallet, you loan a fiver to your friend, then he pays you back, the cash in your wallet is x - 5 + 5 = x. This is because addition and subtraction are also inverse operations; adding then subtracting (or vice versa) the same quantity leaves the original value.
Now let's look at your first expression, x-x^-2/1-x^-2. To simplify this explanation I'm going to use the substitution a = x^-2, giving us (x-a)/(1-a). Just substitute x^-2 wherever you see 'a' and the explanation will be identical. This expression reads "take x and subtract a, then divide that by the result of subtracting a from 1". What you were trying to do was "undo" the result of subtracting a by then dividing by a. Since subtraction and division are not inverse operations you can't do that. Addition and subtraction are inverses, and multiplication and division are inverses. Among those four operations those are the only pairs of inverses, and thus the only pairs of operations that can cancel.
Just about everyone says this, but the problem is that math textbooks are extremely information-dense, and everything is relentlessly cumulative. You have to re-read each section several times over before everything starts making sense, and then you have to figure out what you've missed so you can review it.
Let's just say I didn't expect to be factoring quadratics--much less understand them--when I started on this networking degree 2 years ago.
Maybe I can help with this part. You can't cancel stuff out that is added/subtracted. You can only cancel out multiply/divide stuff.
1 + 2 / 3 * 1 + 3 / 3
First, put in the parentheses. They help mark which units can and can't be canceled.
(1 + 2) / 3 * (1 + 3) / 3
Now that there are parentheses in place it is easy to see that you can't cancel just the 1 in either group of parentheses because it is part of a single unit.
3 / 3 * 4 / 3
Now that you've simplified you can cancel the numerator three on the left with the denominator 3 on the right
1/3 * 4/1
4/3
It works the same with your original equation:
x (x + 10) * 7 / x + 10
Put in parentheses (and signs if you want them)
x * (x + 10) * 7 / (x + 10)
x* 7 * (x + 10)
(x + 10)
Now you can cancel the x + 10 numerator with the x + 10 denominator.
x * 7
7x
If I have a simple equation like 9(3x+5)=0 can I divide 0 by 9 to get rid of the nine? It seems like I should be able to but it just feels like cheating, ecspecialy when I do it like this. If I want to simplify 1000+2000+3000=0 can I divide both sides by 1000? I know that isn't really an equation that looks good but it is just a rough example.
I just wanted to post again because I am feeling great about math, I haven't felt this good about anything for a while so I am pretty excited. I just dominated some quadratic equations with negative exponents and fractional exponents, fuck yeah!
Yes you can. The only time you can't divide in the proper manner (more on that later) is when you're trying to divide by 0 (as opposed to dividing 0 by something else), which is an invalid operation for several reasons I won't go into here.
I don't want to answer this question without a better example equation. That's not because of how it looks or that it's incorrect, but because there's a few things you could have been trying to demonstrate with it and one of them is wrong.
Here, simplify/solve this equation instead: 1000a + 2000a + 3000a = 0. I'm looking for your process rather than the answer, so don't skimp on that.
1000+2000+3000=6000
6000a=0
6000a/6000=a
0/6000=0
a=0
The problem I was trying to describe was pretty complex and you are right, the reason I had trouble describing it is because there is no universal answer to it, it all depends on the context of the number. I think I figured it out since I got the right answer.
See man, all it takes is just a little bit of familiarity.