Math Problem, Zelaron Come Together

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[BlueCube]

Your method ends up 2x + y = 5. Which is what you started with.



Also, your first line finishes the equation:

x + y = 3 ; 2x + y = 5 | x = 2 ; y = 1

If we're given x = 2 and y = 1, then we're finished right there - problem solved.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[Medieval Bob]

Um... you did take Algebra in high school, right?

There's nothing mysterious or groundbreaking about substitution and manipulation.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[BlueCube]

Graphing works for that. But beyond that, you're starting out with

Ax + By = C
Dx + Ey = F




The proper way is to do this:

y = -Ax/B + C/B
y = -Dx/E + F/E

-Ax/B + C/B = -Dx/E + F/E
-(-Ax/B + C/B) = -(-Dx/E + F/E)
Ax/B - C/B = Dx/E - F/E
Ax/B - Dx/E = C/B - F/E

Then you'd solve it from there. Actually we can go farther:

BE(Ax/B - Dx/E) =BE(C/B - F/E)
EAx - BDx = EC - BF

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Then you could just throw x into either of the starter statements. Doing that with your question:

==================

1x + 1y = 3
2x + 1y = 5

1*1x - 1*2x = 1*3 - 1*5
1x-2x = 3 - 5
-x = -2
x = 2

And you can find y easily here:

2 + 1y = 3
y = 1

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There is no way to get either x or y without getting rid of one somehow. You can't just throw the number away, it must be set as something equal. That's subbing.

[Mantralord]

Mathematics is not communism in berry form, but alas, a seagull displacing us in God's will.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[Xenn]

Graphing, matrices, subtraction; we learned 5 ways other than subbing to solve systems like that.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[Medieval Bob]

*sigh*

x + y = 3 ; 2x + y = 5

Either solve for a variable, plug it in the other equation, and find the value of that variable, or add or subtract the equations from each other. There is no other way to work the problem (besides variations and maps... ex: graph)

Solve for Var:
x + y = 3 --> y = 3 - x
2x + y = 5 = 2x + (3 - x) = x + 3 = 5 --> x = 2
x + y = 3 = 2 + y --> y = 1

Subtract:
x + y - (2x + y) = 3 - 5 = -2 = -x --> x = 2
x + y = 3 = 2 + y --> y = 1

Simple as 3.14159

*sigh* I hate puns...

[wilma]

ELIMINATION
2x+y=5 2x (x+y=3) = 2x+2y=6
- x+y=3 2x+2y=6
x=2 - 2x+y=5
y=1

OR SUBSTITUTION

x=3-y 2x+y=5 -----> 2(3-y)+y=5 ---------> 6-2y+y=5 -------> -y=-1 or y=1


OR COMPARISON

y=5-2x and y=3-x ------> 5-2x=3-x x=2

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[Lenny]

A friend of mine has his own little way where you use ratios and shit.

He has got it to work with many a weird thing like your:

x+y=3
2x+y=5

We've been taught to rearrange it, the method is what you'd probably call subbing.

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I'll ask him how he does the ratio thing.

Or someone could have fun with a few ratios, and work backwards from the answer of x=2, y=1, which Shinto kindly gave at the start of the equation, meaning we didn't have to work it out.

[Titusfied]

Even if there is another way, solving simulateous equations is the best, and most importantly easiest way to do so. Why create more work for yourself when the solution is already at a trivial way to solve?

[Medieval Bob]

The general motivation is to find new ways to do a bulk of things... Perhaps to revolutionize something... If, however, the method's you're discovering have been well known and established for hundreds of years, you should look elsewhere.

[Shining Knights]

It matters not whether you win or lose; what matters is whether I win or lose.

[Xenn]

well, my friend has said, if cant be done any other way. And i would LOVE to prove him wrong. And who knows, maybe if by finding another way will solve something else.

It can be done by matrices, let me go dig up my algebra notebook because I seem to have forgotten how..

Ah nevermind I remembered:

So x+y=3 and 2x+y=5. Matrix is
[1 1 3]
[2 1 5] now mR1(-2)+R2 (multiply row 1 by -2 {the inverse of 2} and add the results to row 2) You end up with:

[1 1 3]
[0 -1 -1] now mR2(1) + R1, you get:

[1 0 2]
[0 -1 -1] now mR2(-1) to get:

[1 0 2]
[0 1 1]

Row 1: x=2
Row 2: y=1

Done.

This is just one of the matrix methods I learned. A couple are easier, one is a bitch load harder. Also, you can do this on your calculator in 3 seconds.

[Medieval Bob]

That doesn't have a damn thing to do with matricies.

You simply organized the numbers into columns and rows, and did the exact same thing I did before, which was add or subtract the equations.

[Xenn]

That doesn't have a damn thing to do with matricies.

You simply organized the numbers into columns and rows, and did the exact same thing I did before, which was add or subtract the equations.

Well to be fair I didn't read what you did :D

And it is a matrix because it's in standard matrix gaussian form (or whatever the fuck it's called). There are other methods that involve adding and subtracting matrices also that work, but they're more annoying to do.

[Medieval Bob]

Yes, but if you put y before x in the second equation, it doesn't work anymore... It's just manipulating the numbers with brackets around them.

[Xenn]

Yes, but if you put y before x in the second equation, it doesn't work anymore... It's just manipulating the numbers with brackets around them.

Part of the pre-requisites are that the equations be in standard form. The same method works with every system of equation, so it's a valid approach.