1. Simultaneous equation...

Hiya, really stumped on this one, have got to do this without a calculator. i understand how to do simultaneous equations, i think, but can't get the solution on this question, using these numbers. couldn't see how the numbers could be nicely divided/multiplied to get numbers that i can work with. please help
what's the best/easiest way to solve it?

'
23x + 47y + 105 = 0
7x - 5y + 3 = 0

2. Originally Posted by artem1s
Hiya, really stumped on this one, have got to do this without a calculator. i understand how to do simultaneous equations, i think, but can't get the solution on this question, using these numbers. couldn't see how the numbers could be nicely divided/multiplied to get numbers that i can work with. please help
what's the best/easiest way to solve it?

'
23x + 47y + 105 = 0
7x - 5y + 3 = 0
One way is to eliminate the variable y and solve for x (the usual way). I'm going to give a second. The second equation (a diophantine equation) can be solved involving a parameter. Guess one solution

$\displaystyle (x,y) = (1,2)$

Then the solution of the second is

$\displaystyle (1)\;\;\;x = 1 + 5t,\;\;\;y = 2 + 7t.$

Substitute into the first and simplfying gives

$\displaystyle 444t + 222 = 0$ or $\displaystyle t = - \frac{1}{2}$ . Subs into (1) gives

$\displaystyle x = - \frac{3}{2}, \;\;\;x = - \frac{3}{2}.$

3. Hiya, thanks for the reply. I've not done parameters or anything like that yet, so... i struggled for about half an hour trying to figure out what was going on there, kind-of understood it, but i think i'll leave that until when it appears in my text book

I actually put down the wrong question, sorry. that last one was easy enough using elimination.

This other one though, i just found the resulting numbers too large to be whittled down to their answers (-3 and 1/6, 3/5)

i can work through it, and get a large numbers that do, when simplified, result in the right number. (6282 / 10470 = 3/5 ) I just didn't think the book wanted me messing around with long division or something, perhaps i'd missed an important step in simplifying it earlier on. I think i should only be using elimination or substitution, or that other way where its something like

3x + 4 = 6x + 1

Not sure, just hoping someone will point out i'm missing something extremely simple ^^, heres the question. please help

60y = 150x + 511
12x + 65y -1 = 0

4. Originally Posted by artem1s
This other one though, i just found the resulting numbers too large to be whittled down to their answers (-3 and 1/6, 3/5)

i can work through it, and get a large numbers that do, when simplified, result in the right number. (6282 / 10470 = 3/5 ) I just didn't think the book wanted me messing around with long division or something, perhaps i'd missed an important step in simplifying it earlier on. I think i should only be using elimination or substitution, or that other way where its something like

3x + 4 = 6x + 1

Not sure, just hoping someone will point out i'm missing something extremely simple ^^, heres the question. please help

60y = 150x + 511
12x + 65y -1 = 0
Hello Artem1s,

I'm really not sure what your question is because you did come up with the correct solution for the system. Did you use substitution, elimination or some other technique to solve it?

[1] $\displaystyle 60y = 150x + 511$
[2] $\displaystyle 12x + 65y -1 = 0$

I solved [1] for y getting
[3] $\displaystyle y=\frac{5}{2}x+\frac{511}{60}$

Then, I substituted that into [2], getting

$\displaystyle 12x+65\left(\frac{5}{2}x+\frac{511}{60}\right)=1$

$\displaystyle 12x+\frac{325}{2}x+\frac{33215}{60}=1$

$\displaystyle \frac{349}{2}x=-\frac{6631}{12}$

$\displaystyle 4188x=-13262$

$\displaystyle x=-\frac{19}{6}$

Substitute this value back into [3] to find $\displaystyle y=\frac{3}{5}$

5. Hello, artem1s!

We can use Elimination (to avoid working extensively with fractions).
The size of the resulting numbers should not be a problem . . . should it?

$\displaystyle \begin{array}{cccc}150x - 60y &=& \text{-}511 & {\color{blue}[1]}\\ 12x + 65y &=& 1 & {\color{blue}[2]}\end{array}$

$\displaystyle \begin{array}{ccccccc}\text{Multiply {\color{blue}[1]} by 13:} & 1950x - 780y &=&\text{-}6643 \\ \text{Multiply {\color{blue}[2]} by 12:} & 144x + 780y &=& 12 \end{array}$

Add: .$\displaystyle 2094x \:=\:-6631 \quad\Rightarrow\quad\boxed{ x \:=\:-\frac{19}{6}}$

Substitute into [2]: .$\displaystyle 12(\text{-}\tfrac{19}{6}) + 65y \:=\:1 \quad\Rightarrow\quad\boxed{ y \:=\:\frac{3}{5}}$

6. thanks for the replies,

I can grasp the elimination method, thats easy enough, i just don't have a clue how you get from ;

.

-6631/2094

to

- 19/6

sorry, maybe you guys are just really good at it and are just seeing it as easy simplification, but i don't know how you get from a to b.

7. Originally Posted by artem1s
thanks for the replies,

I can grasp the elimination method, thats easy enough, i just don't have a clue how you get from ;

.

-6631/2094

to

- 19/6

sorry, maybe you guys are just bloody good at it and are just seeing it as easy simplification, but i don't know how you get from a to b.
Artem1s,

There's really no quick and dirty way to do this unless you have a calculator handy. You could try to factor the numerator and denominator to see if there are any common factors you can divide out, but there's a pretty big prime factor here.

$\displaystyle -\frac{6631}{2094}=-\frac{19 \cdot 349}{6 \cdot 349}=-\frac{19}{6}=-3 \frac{1}{6}$

Or, you could just use long division.

Code:
                 -3
___________________
- 2 0 9 4 | 6 6 3 1
6 2 8 4
-----------
3 4 9
The above yields a quotient of -3 with a remainder of 349. Thus $\displaystyle -3 +^-\frac{349}{2094}=-3 \frac{1}{6}=-\frac{19}{6}$

8. That stuff seems Nasty. i'd best brush up on my long division :/

thanks for the help guys!