Substituion + Elimination Method for math!? step by step I don't know how to do it !?

Question

I am really having difficulties with solving system of equtions in math right now and I'm in grade 11... I don't know the steps for solving system of equations using the substitution method... I can do matrices because I know the steps , I just can't do substituion and elmination... can someone please give me an example and go by steps using words as well as the math... I really need to know how to do it ?

it would be greatly appreciated... also I would like to know how to do somethign that is not in order... like this, how do i rearange it to solve ... ?

x+8z = 23y
z-2y=13x
y=8x+2z

Answer 1

You know matrices? WoW! I'm in grade 11 too... I don't know matrices, but I sure know substitution and elimination.

Lets take this example:

2x – 3y = –2 ---- (i)
4x + y = 24 ---- (ii)

First step is to make one of the coefficients i.e. of x or y, the same.

In this case, we'll make the y coefficients same (it's always better to add than subtract!)

Multiplying (ii) by 3, we get:
12x + 3y = 72

Now add (i) to this new equation:

12x + 3y = 72
2x – 3y = –2 (+)
---------------------
14x + 0 = 70

Solving this now,

x = 70/14 = 5

Now what you have to do is to substitute x = 5 in equation (i) or (ii)

I'm choosing (i) because there are no BIG numbers!

Substituting x = 5 in (i), we get:

2*5 - 3y = -2
10 - 3y = -2
3y = 12
y = 4

There you go! x = 5 and y = 4!! Ain't that simple?



Hope it helps!

Answer 2

The idea behind the the substitution/elimination method, is that you solve one equation at a time. Each time you solve one equation, you plug the results into the remaining equations.

Here's how this applies to your example. Start with:
x+8z = 23y
z-2y=13x
y=8x+2z

Now "solve" the first equation. To do that, you pick any of the variables (doesn't matter which), and try to get it so it appears on the left side by itself, and nowhere on the right side. Let's pick "x". We start with:

x+8z = 23y

We want to get rid of the "+8z" that's next to the "x". To do that, we use the most important principle in algebra: If you do an operation on the left side of the equation, you must do that same operation on the right side as well (and vice versa). In this case, the thing we'll do is: subtract "8z" from both sides:

x+8z – 8z = 23y – 8z

But since (8z–8z) is zero, what we have is:
x+0 = 23y – 8z

or:
x = 23y – 8z

So now we have "solved" the first equation.

The next step is substitution. To do this, we look at the remaining two equations, and wherever we see an "x", we replace it with the expression "23y – 8z". We're allowed to do that because we just proved that "x" and "23y – 8z" are the same thing. So:

Change this:
z-2y=13x
to this:
z–2y = 13(23y – 8z)

And change this:
y=8x+2z
to this:
y = 8(23y – 8z) + 2z

Now the nice thing about that is, those two equations used to have THREE variables each (x, y and z); but now they each have only TWO variables (y and z). We've "eliminated" the x.

Now we repeat the process for the second equation. Choose a variable that's in the second equation (there are now only two to choose from: y and z), and solve for one of those. Let's choose "z". We start with:

z–2y = 13(23y – 8z)

Add "2y" to both sides:
z–2y+2y = 13(23y – 8z) + 2y
z + 0 = 13(23y – 8z) + 2y
z = 13(23y – 8z) + 2y

Now we have "z" by itself on the left (that's good), but there's also a "z" on the right (that's bad). Here's an example of some steps to get rid of that z. I'm going to go through them kind of fast, and won't be able to explain everything in detail; you may need to follow up with your book or your teacher.

We have:
z = 13(23y – 8z) + 2y

Distributive law on right side:
z = 13•23y – 13•8z + 2y

Associative law:
z = (13•23)y – (13•8)z + 2y
z = 299y – 104z + 2y

Commutative law:
z = 299y + 2y – 104z

Distributive law:
z = (299 + 2)y – 104z
z = 301y – 104z

Add "104z" to both sides:
z + 104z = 301y – 104z + 104z
z + 104z = 301y – 0
z + 104z = 301y

Distributive law:
(1 + 104)z = 301y
105z = 301y

Divide both sides by 105:
105z / 105 = 301y / 105

On the left side, "105" cancels out in numerator and denominator, so:
z = 301y / 105
or:
z = (301/105)y

Whew. So now we have solved for "z". Now we do like we did before, and replace "z" with "(301/105)y" in the one remaning equation. (Note that each time you do a subsitution, you only substitute in the equations BELOW the one you just solved. We're done with Equation 1 for now.)

When we last saw Equation 3 it looked like:
y = 8(23y – 8z) + 2z

Now we replace every "z" with "(301/105)y":
y = 8(23y – 8(301/105)y) + 2(301/105)y

And notice that we have now eliminated "z" from equation 3. We previously eliminated "x" from that equation, so no the only variable left is "y". That's a good thing, because once you're down to a single variable, you can (usually) solve the equation and get a real number for the answer. Again, I'll go through this quickly:

y = 8(23y – 8(301/105)y) + 2(301/105)y

Simplify products:
y = 8(23y – (344/15)y) + (86/15)y
Combine like terms (distributive law):
y = 8(23 – 344/15)y + (86/15)y
y = 8(1/15)y + (86/15)y
y = (8/15)y + (86/15)y
Combine like terms:
y = (8/15 + 86/15)y
Simplify:
y = (94/15)y
Subract (94/15)y from both sides:
y – (94/15)y = (94/15)y – (94/15)y = 0
Combine like terms:
(1 – 94/15)y = 0
(-79/15)y = 0
Divide both sides by (-79/15):
(-79/15)y/(-79/15) = 0/(-79/15) = 0
y = 0

Whew. So now we know that y=0.
Now we work backwards to find “z” and “x”.

Go back up to Equation 2, which has just “y” and “z” in it:
z = (301/105)y

Since we now know that y=0, substitute that:
z = (301/105)0
z = 0

And now we know z=0. Finally, go back to Equation 1:
x = 23y – 8z

Substitute our known values for y and z:
x = 23(0) – 8(0)
x = 0 – 0
x = 0

So that’s our final solution:
x = 0
y = 0
z = 0

You can see that the substitution/elimination method can involve a lot of steps; that’s why it’s very often easier to use the matrix method. However, there are some systems of equations (called “non-linear” systems) for which the matrix method won’t work, and you must use substitution/elimination

Answer 3

If it's not in order you transpose them. If it crosses to the other side of the equal sign (=) the it will be negative.

I remember this lesson. Its a trial and error question so you have to constantly repeat and change the number you are multiplying to...

You can Do that!
I believe you can!