also how do I do (x^-1 + y^-1)^-1 a negative times a negative is a positive right? Well the asnwer the book has is xy/x+y
x^-b is the same as 1/(x^b) (note the positive b), so that is equivalent to 1/(1/x + 1/y). First let's look at 1/x + 1/y. When adding or subtracting fractions they have to be in the same base; you'll usually want to use the least common denominator for that base, which in this case is xy. Multiply 1/x by y/y (which equals 1, so you're not changing the value) to get y/xy, and likewise multiply 1/y by x/x to get x/xy. You can now do y/xy + x/xy = (x+y)/xy. Now that we have that part done we take the inverse of it, which is the same as flipping the numerator and denominator and gives us xy/(x+y).
I am so fucking stressed out right now I can't even think.
If I have 20x * 2x is that 40x or 40x^3? I can't even think right now. And then 2x * 2x would be 4x but not 4x^2?
When you multiply or divide you do so for all parts. So 20x*2x = 20 * x * 2 * x = 40 * x^2 = 40x^2.
How does it work when I divide? 10x^2 / x^2
Is that 10?
When I was learning these things I found it helpful sometimes to just plug in a couple numbers to see what's what. For example:
20x *2x = ?
20(3) * 2 (3) = 60*6 = 360
40(3)^2 = 40 (9) = 360
Edit: Well, smasher explained it better. I think it might help for you to write down some relevant general cases on a single piece of a paper. Kind of like a cheat sheet.
For example you could put some these recurring kinds of problems you have on there like:
This problem seems pretty simple to me, all I have to do is add the first two fractions then work it from there. pq would be the denominator which would leave me with pq/p^2q + pq/pq^2 which would give me 2pq/pq^2 + p^2q from there I can multiply the other side by the denominator and I am left with something crazy and ridiculous and not even close to the answer.
Where did I go wrong?
Again this is such a simple problem my book assumed I would figure it out on my own and gave no examples.
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SmasherStarting to get dizzyRegistered Userregular
edited April 2009
You don't want to multiply all the fractions by the common denominator. Instead you want to multiply each one by whatever fraction it takes to get to the common denominator.
Fuck, yes I am. Wait isn't that what I did? Doesn't that still leave me with 1/pq?
Also this makes no sense to me either, if I am multiplying 7/x by x(x+10) I can cancel that first x out right? The book does it I think but it makes no sense because I should have 7x(7x+70)/x which would be 6x(7x+60) correct?
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SmasherStarting to get dizzyRegistered Userregular
Fuck, yes I am. Wait isn't that what I did? Doesn't that still leave me with 1/pq?
You multiplied both 1/p and 1/q by (pq)/(pq). That's how you ended up with denominators of p^2*q and p*q^2 respectively.
What's the common denominator of 1/p and 1/q? What do you have to multiply them by (it'll be different for each one) to get that common denominator?
Also this makes no sense to me either, if I am multiplying 7/x by x(x+10) I can cancel that first x out right? The book does it I think but it makes no sense because I should have 7x(7x+70)/x which would be 6x(7x+60) correct?
You can cancel out the x, yes. What you're doing doesn't work because you're both distributing the 7 into the parenthesis and leaving it outside them. Either of those is valid, but not both at the same time. If you do both you effectively multiply the whole thing by an extra 7 which changes the value of the equation. In other words you can have either 7x(x+10)/x or x(7x+70)/x among other possibilities, but not 7x(7x+70)/x.
I am having trouble figuring out how to multiply things like that and when they cancel out and when they do not.
Obviousl 7-x/x I can not cancel out the x but in 7x/x I can. But what about 7(x+10)/x+10? Does that become 7 or 7x+70/x+10 which becomes 7+7 which is 14. Did I screw that up?
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SmasherStarting to get dizzyRegistered Userregular
edited April 2009
Always remember the order of operations. PEMDAS or Please Excuse My Dear Aunt Sally are two common mnemonics to help remember them:
Those are listed from greatest to lowest precedence, with multiplication and division sharing a level (if there's a multiplication and division in the same expression, just go left to right) and likewise for addition and subtraction. Why are they in that order? Ultimately it's just convention, but it's convention that absolutely everybody uses so you have to understand it.
What does all that really mean in practical terms? Let's look at the basic operations. Take the expression x = 5*3+4*2. If you read it from left to right the value would be 38, but you have to remember the order of operations. That expression is equivalent to x = (5*3) + (4*2) = 23 because multiplication has a higher precedence than addition. Within a given precedence level the order goes from left to right; this doesn't matter for addition and multiplication because they're commutative (a*b*c = c*a*b, for example), but for division and subtraction it does. So 8/4/2 = (8/4)/2 = 1 rather than 8/(4/2) which would equal 4.
The same idea goes for exponents. Exponents have a higher precedence than the basic operations, so 2x^4y = 2(x^4)y rather than (2x)^(4y) or 2(x^4y) or whatever other combination it could be. Now to take this back to your examples.
7-x/x is equivalent to 7-(x/x) = 7-1 = 6; in that sense the x's do cancel and you're left with 1. I suspect what you meant by that was (7-x)/x; in this case the x's don't cancel. However, you can divide x into 7-x by dividing each part by x, so (7-x)/x = 7/x - x/x = 7/x - 1.
7(x+10)/x+10 = (7(x+10)/x)+10 = ((7x+70)/x)+10 = (7+70/x)+10 = 17+70/x. If instead you have 7(x+10)/(x+10) then the two (x+10) cancel out and you're left with 7.
I am having trouble figuring out how to multiply things like that and when they cancel out and when they do not.
Obviousl 7-x/x I can not cancel out the x but in 7x/x I can. But what about 7(x+10)/x+10? Does that become 7 or 7x+70/x+10 which becomes 7+7 which is 14. Did I screw that up?
Just remember that (7-x)/x can also be written as (1/x)*(7-x).
Now it just looks like regular multiplication using the distributive property.
If you multiply that out you should get (7/x) - 1.
Now for 7(x+10)/(x+10) yes, you can cancel out the (x+10)'s.
This problem seems pretty simple to me, all I have to do is add the first two fractions then work it from there. pq would be the denominator which would leave me with pq/p^2q + pq/pq^2 which would give me 2pq/pq^2 + p^2q from there I can multiply the other side by the denominator and I am left with something crazy and ridiculous and not even close to the answer.
Where did I go wrong?
Again this is such a simple problem my book assumed I would figure it out on my own and gave no examples.
Okay, so lets figure out what we are trying to do here: solving for f right?
We have 1/f = 1/p + 1/q ... okay lets try and get that f so that it is just "f" and not "1/f"
(1/f)^-1 = (1/p + 1/q)^-1
f = (1/p + 1/q)^-1 (if that negative 1 confuses you, try this:)
1/f = 1/p + 1/q
1/1/f = 1/(1/p + 1/q)
f = 1/(1/p + 1/q)
So now, lets take a look at the denominator on the RHS of the equation:
1/p + 1/q
lets get these both in terms of the same denominator so we can add them:
(q/q)*(1/p) + (p/p)*(1/q)
q/qp + p/qp
(q+p)/qp
now the equation looks like:
f = ((q+p)/qp)^-1 or f = 1/(q+p)/qp
which simplifies to
f = (qp)/(q+p)
and there you go (make sure you are writing this down on paper)
Posts
Fractions don't work that way.
I just multiplied it by 1 and got the result on the far right. Not much you can do to reduce things.
also how do I do (x^-1 + y^-1)^-1 a negative times a negative is a positive right? Well the asnwer the book has is xy/x+y
I am so fucking stressed out right now I can't even think.
If I have 20x * 2x is that 40x or 40x^3? I can't even think right now. And then 2x * 2x would be 4x but not 4x^2?
How does it work when I divide? 10x^2 / x^2
Is that 10?
When I am getting the LCD of an equation with rational expressions why does it have to be completely factored out?
If I have 2(x-3) * 8-3x/ x-3 does the x-3 cancel out?
x^-b is the same as 1/(x^b) (note the positive b), so that is equivalent to 1/(1/x + 1/y). First let's look at 1/x + 1/y. When adding or subtracting fractions they have to be in the same base; you'll usually want to use the least common denominator for that base, which in this case is xy. Multiply 1/x by y/y (which equals 1, so you're not changing the value) to get y/xy, and likewise multiply 1/y by x/x to get x/xy. You can now do y/xy + x/xy = (x+y)/xy. Now that we have that part done we take the inverse of it, which is the same as flipping the numerator and denominator and gives us xy/(x+y).
When you multiply or divide you do so for all parts. So 20x*2x = 20 * x * 2 * x = 40 * x^2 = 40x^2.
Yes.
20x * 2x = 40x^2, 20x + 2x = 22x.
2x * 2x = 4x^2
10x^2/x^2 = 10(x^2/x^2) = 10
When I was learning these things I found it helpful sometimes to just plug in a couple numbers to see what's what. For example:
20x *2x = ?
20(3) * 2 (3) = 60*6 = 360
40(3)^2 = 40 (9) = 360
Edit: Well, smasher explained it better. I think it might help for you to write down some relevant general cases on a single piece of a paper. Kind of like a cheat sheet.
For example you could put some these recurring kinds of problems you have on there like:
Ax + Bx = (A+B)x
Ax * By = (A*B)x*y
etc.
2x-4, x-2, 4
So 4(2x-4)(x-2).
I think. Better wait for someone to confirm since my brain is hurting right now, and might not be producing the right stuff.
That makes 4(x-2) your lcd.
This problem seems pretty simple to me, all I have to do is add the first two fractions then work it from there. pq would be the denominator which would leave me with pq/p^2q + pq/pq^2 which would give me 2pq/pq^2 + p^2q from there I can multiply the other side by the denominator and I am left with something crazy and ridiculous and not even close to the answer.
Where did I go wrong?
Again this is such a simple problem my book assumed I would figure it out on my own and gave no examples.
Are you trying to solve for f?
Also this makes no sense to me either, if I am multiplying 7/x by x(x+10) I can cancel that first x out right? The book does it I think but it makes no sense because I should have 7x(7x+70)/x which would be 6x(7x+60) correct?
You multiplied both 1/p and 1/q by (pq)/(pq). That's how you ended up with denominators of p^2*q and p*q^2 respectively.
What's the common denominator of 1/p and 1/q? What do you have to multiply them by (it'll be different for each one) to get that common denominator?
You can cancel out the x, yes. What you're doing doesn't work because you're both distributing the 7 into the parenthesis and leaving it outside them. Either of those is valid, but not both at the same time. If you do both you effectively multiply the whole thing by an extra 7 which changes the value of the equation. In other words you can have either 7x(x+10)/x or x(7x+70)/x among other possibilities, but not 7x(7x+70)/x.
Obviousl 7-x/x I can not cancel out the x but in 7x/x I can. But what about 7(x+10)/x+10? Does that become 7 or 7x+70/x+10 which becomes 7+7 which is 14. Did I screw that up?
Parenthesis
Exponents
Multiplication/Division
Addition/Subtraction
Those are listed from greatest to lowest precedence, with multiplication and division sharing a level (if there's a multiplication and division in the same expression, just go left to right) and likewise for addition and subtraction. Why are they in that order? Ultimately it's just convention, but it's convention that absolutely everybody uses so you have to understand it.
What does all that really mean in practical terms? Let's look at the basic operations. Take the expression x = 5*3+4*2. If you read it from left to right the value would be 38, but you have to remember the order of operations. That expression is equivalent to x = (5*3) + (4*2) = 23 because multiplication has a higher precedence than addition. Within a given precedence level the order goes from left to right; this doesn't matter for addition and multiplication because they're commutative (a*b*c = c*a*b, for example), but for division and subtraction it does. So 8/4/2 = (8/4)/2 = 1 rather than 8/(4/2) which would equal 4.
The same idea goes for exponents. Exponents have a higher precedence than the basic operations, so 2x^4y = 2(x^4)y rather than (2x)^(4y) or 2(x^4y) or whatever other combination it could be. Now to take this back to your examples.
7-x/x is equivalent to 7-(x/x) = 7-1 = 6; in that sense the x's do cancel and you're left with 1. I suspect what you meant by that was (7-x)/x; in this case the x's don't cancel. However, you can divide x into 7-x by dividing each part by x, so (7-x)/x = 7/x - x/x = 7/x - 1.
7(x+10)/x+10 = (7(x+10)/x)+10 = ((7x+70)/x)+10 = (7+70/x)+10 = 17+70/x. If instead you have 7(x+10)/(x+10) then the two (x+10) cancel out and you're left with 7.
Just remember that (7-x)/x can also be written as (1/x)*(7-x).
Now it just looks like regular multiplication using the distributive property.
If you multiply that out you should get (7/x) - 1.
Now for 7(x+10)/(x+10) yes, you can cancel out the (x+10)'s.
Lets make y = (x+10) then the equation becomes:
7y/y which obviously = 7.
It all comes down to practice Fizban.
Okay, so lets figure out what we are trying to do here: solving for f right?
We have 1/f = 1/p + 1/q ... okay lets try and get that f so that it is just "f" and not "1/f"
(1/f)^-1 = (1/p + 1/q)^-1
f = (1/p + 1/q)^-1 (if that negative 1 confuses you, try this:)
1/f = 1/p + 1/q
1/1/f = 1/(1/p + 1/q)
f = 1/(1/p + 1/q)
So now, lets take a look at the denominator on the RHS of the equation:
1/p + 1/q
lets get these both in terms of the same denominator so we can add them:
(q/q)*(1/p) + (p/p)*(1/q)
q/qp + p/qp
(q+p)/qp
now the equation looks like:
f = ((q+p)/qp)^-1 or f = 1/(q+p)/qp
which simplifies to
f = (qp)/(q+p)
and there you go (make sure you are writing this down on paper)
1/1/f = 1/(1/p + 1/q)
All I did there was put both sides of the equations over 1 (inversed) so:
the inverse of f is 1/f
the inverse of 1/f is f.
the inverse of 1/q + 1/p is
1/(1/q+1/p)