Question

Solve using two different methods. Explain which method you found more efficient.
a. 3x-9y=3 and 6x-3y=-24

Answer 1

This is not a chemistry question. But I will help you either way

Method 1: Substitution
(rearrange eq.1 or eq.2 to get it in the form y = ____ )
Thus, arranging the first equation, 3-3x and then dividing by 9, you get
\[y = \frac{ 3 }{ 9 }-\frac{ 3 }{ 9 }x\]
obviously, you have to do some reduction of both fractions. Now, notice that you have y = _____
Plug in that new equation into your second equation for every y.
It's like saying y=4
3x+y=0
then you have
3x+4=0
You plug it in and solve for x!
That's what you're doing here, getting everything in one vartiable to solve for another. Once you have your value of either "x" or "y", then just plug in that value for equation you used to sbstituted and get your second variable.


Method 2: Elimination
For this method, you want to do the same thing, except this method may require just a bit more thinking than just substituting. You have two equations, what your mission is is to eliminate either an "x" or a "y" by multiplying one equation by some number so that you can add (or subtract) and cancel out one. For your \[\left(\begin{matrix}-6x+18y=-6 \\ 6x-3y=-24\end{matrix}\right)\]
What you might notice is that the top equation is different, that's because I multiplied EVERYTHING by (-2). That way I can eliminate, x. Now I can eliminate x, since -6+6=0
To get:
18y+(-3y)=-30
Obviously, you can combine like terms, and you should get 15y=-30
divide by 15 and you get your y value. And now, all you do is plug in y to any of your equations and solve for "x".

As far as which one is more efficient, I think you should make the judge yourself based on which one you feel more comfortable working with.

Answer 2

Sorry I thought I was in Mathematics still