Can someone explain Elimination/Addition?

Question

I thought I understood this until I actually tried to do it. Okay, so let me get this straight. Say, to get y alone you have to multiply each equation by the coefficient of the opposite equation? In my book, it shows the bottom number being negative. Why?

Here, perhaps if I give an actual problem it'll be easier to explain my confusion.

Here's an example:
(1) 5x+3y=4
(2) 7x-4y=22

In my textbook, they multiply (1) with 4 and equation 2 with 3.
This is where I get confused. In some of these example problems they use a negative for the bottom number and I cannot figure out why. Can someone explain to me what's going on at this step because my book doesn't do a good job in explaining it.

In another example they do
1) -12x+9y=5
2) -8x+6y=-11

and they multiply the first by -2 and the second equation with 3. WHY? I don't understand what's going on here at all!

This is another example.

Say we have:

x-3y=1
-x+y=3

Do I mutliply the top equation by 3 or -3. Does the second one get multiplied by -1? This is where I'm getting confused.

Also I forgot to mention, I don't want people to give me the answers to the example problems. I just want an explanation of the thought process involved in these problems.

Answer 1

I'm an algebra teacher. I can help youl

First aim to eliminate one of the variable terms by creating opposite terms. Understand that there are infinitely many ways to do this with any problem.

Let's look at the example. Here are your choices.
*****multiply eqtn1 by 2 and eqtn2 by -3 to eliminate the x term.
*****multiply eqtn1 by -2 and eqtn2 by 3 to eliminate the x term.
*****multiply eqtn1 by 4 and eqtn2 by -6 to eliminate the x term.
*****multiply eqtn 1 by 4 and eqtn2 by 6 to eliminate the y term.

Get it? Do you see it that you must have opposites before adding the two equations together to eliminate the variable term?

Let's look at the original example you provided.
*****multiply eqtn1 by 5 and multiply eqtn2 by -7 to eliminate the x term.
*****multiply eqtn1 by 8 and multiply eqtn2 by 6 to eliminate the y term.

There are infinitely many choices. So pick one and go with it.

BTW, the example you provided has no solution. After multiplying and adding, you'll get a false state, 0 = something other than zero. Therefore, the lines are parallel.

Hope this helps! :-)

PS. I just noticed that you had aditional details...

You say <

x-3y=1
-x+y=3

Do I mutliply the top equation by 3 or -3. Does the second one get multiplied by -1? This is where I'm getting confused.>

I say take a good look at this problem. This problem doesn't require multiplication by any number other than one because the x-terms are already ready for elimination. Therefore when you combine the two equations, the result will be -2y = 4.

Answer 2

You are trying to match up the coefficients so that they cancel out. You want the coefficients of x (or y) in the two equations to be additive inverses of each other (e.g., if one is 15, the other should be -15).

(1) 5x+3y=4
(2) 7x-4y=22

Multiplying 3y by 4 (12y), and then -4y by 3 (-12y) causes them to cancel when the equations are added together.

You could also have multiplied the first equation by 7 (35x), and the second one by -5 (-35x) to get x to cancel out.

=====================
In the second example, you say:

1) -12x+9y=5
2) -8x+6y=-11

"and they multiply the first by -2 and the second equation with 3. WHY? I don't understand what's going on here at all!"

-12x * -2 = 24x, and -8x * 3 = -24x. 24x and -24x will cancel out when added. Note that this also causes the Y-terms (9y * -2 = -18y, and 6y * 3 = 18y) to cancel out.

======================
In your third example, you say:

x-3y=1
-x+y=3

You don't have to multiply either equation by anything. If you add it up as-is, the x and -x will cancel out.

As an alternative, you could try to get the y values to cancel out. This would involve multiplying the second equation by 3, so that its 3y would cancel the -3y in the first equation.

You COULD multiply the first one by -1 (-3y * -1 = 3y) and the second one by -3 (y * -3 = -3y). You COULD also multiply the first one by 100 (-3y * 100 = -300y) and the second one by 300 (y * 300 = 300y). It doesn't really matter how you get to one coefficient being the negative of the other. The exact same answer will result in any case.

All you are trying to do is get the coefficients of y to be additive inverses, and when they start out as (-3 and 1), it's quickest to multiply the second one by 3 to get (-3 and +3).

Answer 3

On the third example the logic would be
the
x-3y=1
-x +y=3
this is 1(x)-3(y)=1
-1(x)+1(y)=3

1x+(-1x)=0
-3y+y=-2y
1+(3)=4
One variable is gone and you are left with -2y=4
get rid of the - on both sides and your done. No multiplying required in this case the x's took care of themselves.

As as long as you do EXACTLY the same thing to each term on BOTH sides of the equation you will end up with the answer.
Each problem will have a different set of multipliers as pointed out above. Just remember if it allows you to get rid of a variable it is correct.
hope it gives you the general idea and
cheers