Solving Systems of Equations with Three Variables

Uploaded by MATHRoberg on 27.09.2010

Hi, I'm Kendall Roberg, and today we're going to learn how to solve systems of equations involving three variables.
Now here, we have three separate equations; each of them has an x, y, and z variable in it.
The trick to solving systems with three variables is to break them down into systems with just two variables, and finally break them down into an equation with just one variable.
So, let's look at the first two equations.
I noticed that the coefficients of x are 4 and 2.
If I multiply the second equation by -2, let's see what happens.
So times -2, and I'll draw an arrow showing where the equation will be written...
-2 times 2 is -4x...
plus 6y...
minus 10z...
equals 28.
The top equation I'm simply going to rewrite again.
I'm not gooing to do anything to it.
4x + 2y + 3z = 1.
Now when I add these two equations, we notice that 4 and -4 are opposites.
Whenever we have opposite coefficients, we can eliminate a variable.
So let's do that to eliminate x.
4x + -4x is 0x...
Then we're left with 8y...
minus 7z...
equals 29.
Now this is great; I now have an equation with two variables in it.
But one equation with two unknowns is unsolvable.
So, we need to find another one.
Now, I used these two equations to produce this first equation.
I need to use two different equations to produce a separate equation that doesn't have x in it.
So let's use the bottom two.
This time, I'm going to multiply the middle equation by -3.
So I have -6x...
plus 9y...
minus 15z...
equals 42.
Now the reason I did that is because now -6x is an opposite coefficient to 6x.
So I'm going to rewrite the third equation; I didn't do anything to it.
When I add these two equations, I get 0x again... plus 8y...
And right here I get minus 11z...
And finally...41.
Now the good news is let's look at these two equations.
I now have two equations with only y and z in them.
That's great...we can solve that.
Let me rewrite the two equations I have right now.
Now, 8y and 8y are identical; they have the same coefficient.
If I subtract these two equations, I'm left with 0y, and then I have -7z minus -11z...
Well that's the same as -7z plus 11, so I'm left with 4z.
And down here, it looks like I'm left with -12.
Dividing both sides by 4...I find that z = -3.
Now I know what z is, and I could go back to any of these equations to determine what y is.
I can do that by substituting the value -3 for z.
So let's do that to...well, let's use this equation.
8y - 7, and instead of writing z I'm going to write -3...
equals 29.
So I have 8y + 21 = 29.
Subtracting 21 from both sides and dividing by 8, I can see that y = 1.
So at this point, I know that z = -3 and y = -1.
I just need to determine what x is.
So if we go back to our original three equations, we can substitute in our z and our y and figure out what x is.
Let's use the first equation.
4x + 2, and instead of writing y, I'm going to write 3z...but instead of writing z I'm going to write -3...and that all equals 1.
Now, let's do some algebra to figure out what x is.
4x - 7 = 1, add 7 to both sides, and we have 4x = 8...
Divide both sides by 4, and we finish with x = 2.
Now, the best way to write our answer for x, y, and z is in the form of an ordered triple.
That is, we write (x, y, z).
Similar to the ordered pair, where we'd write the (x,y) coordinates.
So in this case, (x, y, z) = (2, 1, -3).