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Topic: Help me solve THIS!!! (Read 2176 times) |
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MrUdon
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Hello pplz that are going to read this. I am in Year 10 in some Australian school and in like high level math class. I just had an assignment handed to me with this question. It would make me feel real happy if you could lend me your brains and solve this question.
If x+y=a and 1/x+1/y=b, where x and y are both positive, show that (x-y)squared = ba(squared)-4a/b
P.S also plz teach me how to put power symbols please, and here is something i draw to make the question easier to read since i dont have a scaner
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Neelesh
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Re: Help me solve THIS!!!
« Reply #1 on: Oct 15th, 2005, 5:52am » |
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Since x+y = a and 1/x+1/y = b you can get the product "xy" in terms of a and b.
Further, note that
(x-y)2
= x2 + y2 - 2*x*y
= (x + y )2 -4*x*y
Since you know (x+y) and x*y, you can solve this.
As far as the symbol-representation is concerned, for simple power etc you can use tag "sup" . But there is a very good formula generator by towr (look at the bottom of any of his posts, you will find a link there)
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MrUdon
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Re: Help me solve THIS!!!
« Reply #2 on: Oct 15th, 2005, 5:57am » |
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Thanks but doesnt (x+y)2 = x2+2xy+y2 not x2+y2??
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Neelesh
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Re: Help me solve THIS!!!
« Reply #3 on: Oct 15th, 2005, 6:00am » |
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on Oct 15th, 2005, 5:57am, MrUdon wrote:| Thanks but doesnt (x+y)2 = x2+2xy+y2 not x2+y2?? |
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You probably missed that there was a "-2" in the second step which became "-4" in the third step - simply because
x2 + y2
= (x+y)2 - 2*x*y
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MrUdon
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Re: Help me solve THIS!!!
« Reply #4 on: Oct 15th, 2005, 6:05am » |
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Hmm ic wat u did......dam this stuff is too much for my level of understanding algebra><
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MrUdon
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Re: Help me solve THIS!!!
« Reply #5 on: Oct 16th, 2005, 2:39am » |
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Man sorry to keep askin but i still cant understnad this  If you could be kind, plz post and answer here.
Thank You
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towr
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Re: Help me solve THIS!!!
« Reply #6 on: Oct 16th, 2005, 7:36am » |
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1/x+1/y = b
multiply both sides by xy
y+x = xy b
divide both sides by b (and swap sides)
xy = (x+y)/b
since a=x+y
xy = a/b
As shown by Neelesh
(x - y)2 = (x + y)2 - 4 xy
use the equalities we found earlier
(x + y)2 - 4 xy = a2 - 4 a/b
bring factors together
a2 - 4 a/b = [b a2 - 4 a]/b
So, by transitivy of equality (=)
(x - y)2 = [b a2 - 4 a]/b
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MrUdon
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Re: Help me solve THIS!!!
« Reply #7 on: Oct 17th, 2005, 1:04am » |
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Man thanks towr  Helped me heaps
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towr
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Re: Help me solve THIS!!!
« Reply #8 on: Oct 17th, 2005, 2:24am » |
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I hope it helped you to understand it, rather than just helped you hand in your assignment in time. 
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MrUdon
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Re: Help me solve THIS!!!
« Reply #9 on: Oct 17th, 2005, 3:55am » |
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Yer I eventually worked it out, so yea I did understnad. Thanks again towr 
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MrUdon
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Re: Help me solve THIS!!!
« Reply #10 on: Oct 17th, 2005, 4:45am » |
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Ok sorry guys but i had came across a another yet "hard" problem. x+y+z=9 2x-y+z=9 x-y+2z=9 Solve x, y, z
I could have done this type of question with 2 equation but there is three and even more annoying, there is three vaiables.
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towr
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Re: Help me solve THIS!!!
« Reply #11 on: Oct 17th, 2005, 4:52am » |
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All you need to do is add and subtract
for instance
2x-y+z=9
x+y+z=9
------------ -
x - 2 y = 0
Already one variable less. You can replace either (but not both) of the the equations above the line by the one under it.
And you know x in terms of y, so you can replace it in the other formulas to eliminate x, they'll all only have two variables.
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Icarus
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Re: Help me solve THIS!!!
« Reply #12 on: Oct 17th, 2005, 5:17pm » |
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As a general rule:
If you have only 1 variable, one equation is enough to have a unique (or near unique) solution.
If you have 2 variables, it requires 2 equations to come up with a unique (or near unique) solution. With only one equation, the two values are related, but there are usually infinitely many pairs of values that satisfy the equation.
If you have 3 variables, it requires 3 equations to come up with a unique (or near unique) solution. With only one or two equations, the 3 values are related, but there are usually infinitely many triples of values that satisfy the equations.
If you have 4 variables, then you need 4 equations, etc.
(When I say "near unique", mean that there may be more than one solution, but there are only a finite number. For example, the equation x2 = 1 has two solutions: x = 1 and x = -1.)
So the fact that you had 3 equations to deal with was not an added complication: it was necessary, since you had three variables.
A general process for solving sets of equations like this is: choose one of the equations, and solve it for one variable in terms of the others. Then substitute the result into the remaining equations. You now have one fewer equation, and one less variable. Repeat the process until you have only one equation left. Solve it, then substitute backwards until you find the values of the other variables. For example, your problem was:
x + y + z = 9
2x - y + z = 9
x - y + 2z = 9
Solve the first equation for x: x = 9 - y - z, and substitute into the remaining equations:
2(9 - y - z) - y + z = 9
(9 - y - z) - y + 2z = 9
and simplify to get
3y + z = 9
-2y + z = 0
Solve the second equation for z: z = 2y, and substitute into the first:
3y + (2y) = 9
So:
y = 9/5
z = 2y = 2(9/5) = 18/5
x = 9 - y - z = 9 - 9/5 - 18/5 = 18/5.
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This process is seldom the most efficient way to go about solving equations. But it is the only one that will work all the time (provided you can actually solve the intermediate equations). For linear equations (where you have no variables raised to powers or other complications), the adding trick that towr showed can be faster, particularly once you get good at it. For instance, an experience solver would quickly see that your 2nd and third equations together imply that x and z must be equal. (The left sides consist of the expression x - y + z with either x or z added to it, but in both cases the result is the same: 9. This can only be if x and z are the same).
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MrUdon
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Re: Help me solve THIS!!!
« Reply #13 on: Oct 18th, 2005, 1:30am » |
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Wow, now that u posted the answer, it looks so easy....i am feelin real dum rite now...LOL ahh anyway, thanks alot guys, I probably get A for assignment 
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Neelesh
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Re: Help me solve THIS!!!
« Reply #14 on: Oct 18th, 2005, 1:41am » |
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on Oct 18th, 2005, 1:30am, MrUdon wrote:| anyway, thanks alot guys, I probably get A for assignment |
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The point is not whether you get an A or A+ for your assignment, the (more important) point is that you understand and learn the problem solving techniques through these discussions - that will in a way give you enough ability to get even more A's..and that too without posting the problems on the forum 
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MrUdon
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Re: Help me solve THIS!!!
« Reply #15 on: Oct 18th, 2005, 1:44am » |
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......Yer ur rite. Dam, that was something real good u said..Well at least this was the first step of a long journy one might say...
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rmsgrey
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Re: Help me solve THIS!!!
« Reply #16 on: Oct 18th, 2005, 3:02pm » |
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on Oct 18th, 2005, 1:41am, Neelesh wrote:
The point is not whether you get an A or A+ for your assignment, the (more important) point is that you understand and learn the problem solving techniques through these discussions - that will in a way give you enough ability to get even more A's..and that too without posting the problems on the forum
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Give a man an answer and he'll get an A in a test. Teach him how to find his own answers and he'll pass tests for the rest of his life
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MrUdon
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Re: Help me solve THIS!!!
« Reply #17 on: Oct 19th, 2005, 12:15am » |
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Okay, I gotta a hard problem for you guys
31992-21991=
I have to find out what the last digit of the answer to this problem is, without actually calculating it
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BNC
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Re: Help me solve THIS!!!
« Reply #18 on: Oct 19th, 2005, 12:28am » |
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on Oct 19th, 2005, 12:15am, MrUdon wrote:Okay, I gotta a hard problem for you guys
31992-21991=
I have to find out what the last digit of the answer to this problem is, without actually calculating it |
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I'll suggest a first step, and hopefully you can deduct the rest yourself.
Lets look and the last digit of the first few powers of 2:
2, 4, 8, (1)6, (3)2, (6)4, (12)8, (25)6, (51)2 ...
See a pattern emarging?
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Barukh
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Re: Help me solve THIS!!!
« Reply #19 on: Oct 19th, 2005, 12:28am » |
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This thread gets really exciting...
MrUdon, could you please show us how would you try to attack this problem?
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Grimbal
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Re: Help me solve THIS!!!
« Reply #20 on: Oct 19th, 2005, 2:43am » |
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Once you have figured out a result using BNC's hint, and you want to check it, and only then, see the hidden text below.
As william would say, the important is not the answer which in this case is completely useless, the important is how you get to the result.
Oh, yes, another hint, the last digit of a sum or product only depends on the last digit of the numbers added or multiplied. At least for integers.
123*456 ends with 8 because 3*6 = 18.
I used the good ol Windows calculator. I typed:
3 x^y 1992 - 2 x^y 1991 = Mod 1000 =
I get 593.
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MrUdon
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Re: Help me solve THIS!!!
« Reply #21 on: Oct 19th, 2005, 5:51am » |
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Ok, heres wat i came with. I got the first 12 powers of both numbers. I came up with a pattern with both numbers. The pattern for power's of three was , 9 7 1 3. The pattern for powers of two was 4 8 6 2.
Then i came across an interesting pattern. There was a pattern hidden with the answers. Since the question is 31992-21991, we always have to make powers of 3 one digit higher. The pattern went 73377337.....
I got up to here then got stuck......
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towr
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Re: Help me solve THIS!!!
« Reply #22 on: Oct 19th, 2005, 6:44am » |
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Well, you noticed it repeats, right?
So 34-23 = 81-8 = 3 (mod 10)
(the (mod 10) means we only look at the last digit)
But since every answer repeats when you increase the exponent by 4 (or a multiple of 4), 38-27 must have the same answer.
38-27 = 34-23 (mod 10)
and we already know 34-23 = 3 (mod 10) from before. So 38-27 = 3 (mod 10)
But we can make bigger step as well, 100 is a multiple of 4, so
3108-2107 must have the same last digit as 38-27 .
etc
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MrUdon
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Re: Help me solve THIS!!!
« Reply #23 on: Oct 19th, 2005, 6:54am » |
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Ahh~ ic, man thats great, so anyway wats this "mod10 " thing?? ive never heard of it and could be usfull in the future so could u tell me how to use it plz?
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towr
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Re: Help me solve THIS!!!
« Reply #24 on: Oct 19th, 2005, 7:12am » |
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on Oct 19th, 2005, 6:54am, MrUdon wrote:| Ahh~ ic, man thats great, so anyway wats this "mod10 " thing?? ive never heard of it and could be usfull in the future so could u tell me how to use it plz? |
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A = B (mod N) means that the remainder of A divided by N and B divided by N is the same
So for (mod 10), this means that A and B have the same last digit, since the last digit of a (decimal) number is the remainder when you divide by 10.
123 divided by 10 is 12 with remainder 3 (because 12*10+3=123 and 0 <= 3 < 10 )
so 123 = 3 (mod 10)
123 divided by 7 is 17 with remainder 4 (because 17*7+4=123 and 0 <= 4 < 7 )
so 123 = 4 (mod 7)
53 divided by 7 is 7 with remainder 4 (because 7*7+4=53 and 0 <= 4 < 7 )
so 123 = 53 (mod 7), because both 123 and 53 have the same remainder (in this case 4).
It's also usefull when you look at times and dates. All our times are modulo 24 or modulo 12 (if you use AM/PM)
so it's 4:15 PM here, in 24 hours it's again 4:15, because 4+24 = 4 (mod 24).
The date is 19 october, in 721 days, it will be 10 october, because 19+721 = 10 (mod 365) and we don't have a leap year the next two years.
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| « Last Edit: Oct 19th, 2005, 7:22am by towr » |
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