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donf · Jan 7, 2010, 08:00 AM

Can some one please demonstrate the process needed to resolve this equation.

3x+2y = 16

Thanks

J_9 · Jan 7, 2010, 08:02 AM

Remember, I just came off a 12 hour shift...

Are you solving for x or y?

Curlyben · Jan 7, 2010, 08:04 AM

Seems like you only have half of the problem here.
You can't really solve it at all.

Yes you can change the equation to make x or y the subject, but that's about it..

J_9 · Jan 7, 2010, 08:06 AM

Actually, it is solvable as it is... but I'm not thinking straight... can't do this one in my head and don't have a pencil and paper in the bed with me.

Curlyben · Jan 7, 2010, 08:10 AM

Oooops.
I was thinking simultaneous equations here.

Curlyben · Jan 7, 2010, 08:16 AM

I'll have a go then.

$3x+2y = 16$
(-3x from both sides)
$3x+2y-3x=16-3x$
so $2y = 16 - 3x$
(divide both sides by 2 to make y the subject)

$2y/2 = (16 - 3x)/2$
so
$y = 8 - 3/2x$

Plug this value back into the original and you get

$3x + 2(8-3/2x) = 16$
$3x + 16 - 3x =16$

Oh look at that ;)

J_9 · Jan 7, 2010, 08:22 AM

Okay, this is solving by substitution...

If I remember correctly, this can be called slopes of lines as well. There are several ways to solve this, but someone else will have to help right now.

Here are some good sites...

Slope and Systems of Equations

Solve Y 3x 19 Y 2x 1 By Substitution | TutorVista.com | web search

Unknown008 · Jan 7, 2010, 10:36 AM

Originally Posted by Curlyben

I'll have a go then.

$3x+2y = 16$
(-3x from both sides)
$3x+2y-3x=16-3x$
so $2y = 16 - 3x$
(divide both sides by 2 to make y the subject)

$2y/2 = (16 - 3x)/2$
so
$y = 8 - 3/2x$

Plug this value back into the original and you get

$3x + 2(8-3/2x) = 16$
$3x + 16 - 3x =16$

Oh look at that ;)

And what is that exactly! Lol.

donf, actually, no one can solve this equation the way it is because there is only one equation with two unknowns. However, you can make x or y the subject of the formula.

And if you want to make y the subject of formula, Ben gave you the answer, that is:

$y = 8 - \frac32 x$

galactus · Jan 7, 2010, 12:42 PM

I must agree with Unknown. There is only one equation and two variables. As is, it has infinite solutions and is just the standard form of a line equation. Infinite points lie on the line.
One can not solve an equation for a variable and then plug that back into the same equation. One just ends up going in circles. Take a look at your last line, CB. It is 16=16.

Yeah it can be solved for y or x in terms of the other:

$y=\frac{-3x}{2}+8$

$x=\frac{-2y}{3}+\frac{16}{3}$

But now anything value can be plugged in.

Unknown008 · Jan 7, 2010, 12:45 PM

Originally Posted by galactus

I must agree with Unknown. There is only one equation and two variables. As is, it has infinite solutions and is just the standard form of a line equation. Infinite points lie on the line.
One can not solve an equation for a variable and then plug that back into the same equation. One just ends up going in circles. Take a look at your last line, CB. It is 16=16.

I think Ben knew it... the wink he did... oh that wink, what does it mean? :p

galactus · Jan 7, 2010, 12:47 PM

Oh, I see. Ben was being facetious. I should have known, unknown :)

Unknown008 · Jan 7, 2010, 12:50 PM

Maybe... only he can tell :)

galactus · Jan 7, 2010, 12:56 PM

Well, I was not thinking on those lines at first, but Ben is onto something. Let's treat this as a Diophantine Equation.

Find the GCD(3,2) and we have:

$2=0\cdot 3 +2$

$3=1\cdot 2+1$

$2=2\cdot 1+0$

Now, apply the Euclidean algorithm:

$1=(3\cdot 1)+(-1\cdot 2)$

$x=16, \;\ y=-16$

The general solution is:

$x=16+2p, \;\ y=-16-3p$

This means the values of x and y will have this general form.

i.e. 16+2(1)=18, -16-3(1)=-19

3(18)+2(-19)=16

and on and on for any real values of p.

Rather histrionic, but fun.

morgaine300 · Jan 8, 2010, 03:32 AM

J-9, but what exactly is being "solved"? Plotting a point in like Ben did is only re-writing it another way, not solving anything.

$3x+16-3x = 16$

So, subtract 16 from both sides and you've got:
$3x - 3x = 16 - 16$

Which is just 0 = 0.

That solves nothing.

Substitution would be taking the 8-3/2x and substituting that y into a different equation. Plugging it back into itself gets you nowhere. It's a way to solve for the "crossover point" of two different equations. Perhaps that's what Don is after, but he didn't say so.

Don, since I don't think you're doing homework, what exactly is it you're trying to do?