11 12 1

21 22 2

a x a y b

a x a y b

The solution will be one of three cases:

1. Exactly one solution, an ordered pair (x, y) 2. A dependent system with infinitely many solutions3. No solution

Two Equations Containing Two Variables

The first two cases are called consistent since there are solutions. The last case is called inconsistent.

With two equations and two variables, the graphs are lines and the solution (if there is one) is where the lines intersect. Let’s look at different possibilities.

11 12 1

21 22 2

a x a y b

a x a y b

Case 1: The lines intersect at a point

(x, y),the solution.

Case 2: The lines coincide and there are infinitely many solutions (all points

on the line).

Case 3: The lines are parallel so there

is no solution.

dependent

consistent consistent

independent

inconsistent

If we have two equations and variables and we want to solve them, graphing would not be very accurate so we will solve algebraically for exact solutions. We'll look at two methods. The first is solving by substitution.

The idea is to solve for one of the variables in one of the equations and substitute it in for that variable in the other equation.

Let's solve for y in the second equation. You can pick either variable and either equation, but go for the easiest one (getting the y alone in the second equation will not involve fractions).

Substitute this for y in the first equation.4 3y x

2 3 1

4 3

x y

x y

2 3 4 3 1x x Now we only have the x variable and we solve for it.

14 8x 4

7x Substitute this for x in one of the equations to find y.

Easiest here since we already solved for y.

4 54 3

7 7y

2 3 1

4 3

x y

x y

So our solution is 4 5

,7 7

This means that the two lines intersect at this point. Let's look at the graph.

x

y

4 5,

7 7

We can check this by subbing in these values for x and y. They should make each equation true.

4 52 3 1

7 7

4 54 3

7 7

8 151

7 716 5

37 7

Yes! Both equations are satisfied.

Now let's look at the second method, called the method of elimination.

The idea is to multiply one or both equations by a constant (or constants) so that when you add the two equations together, one of the variables is eliminated.

Let's multiply the bottom equation by 3. This way we can eliminate y's. (we could instead have multiplied the top equation by -2 and eliminated the x's)

Add first equation to this.12 3 9x y The y's are eliminated.

Substitute this for x in one of the equations to find y.

3 3

2 3 1x y

14 8x 4

4 37

y

2 3 1

4 3

x y

x y

4

7x

5

7y

So we arrived at the same answer with either method, but which method should you use?

It depends on the problem. If substitution would involve messy fractions, it is generally easier to use the elimination method. However, if one variable is already or easily solved for, substitution is generally quicker.

With either method, we may end up with a surprise. Let's see what this means.

3 6 15

2 5

x y

x y

Let's multiply the second equation by 3 and add to the first equation to eliminate the x's.3 3

3 6 15

3 6 15

x y

x y

0 0

Everything ended up eliminated. This tells us the equations are dependent. This is Case 2 where the lines coincide and all points on the line are solutions.

Now to get a solution, you chose any real number for y and x depends on that choice.

2 5x y

If y is 0, x is -5 so the point (-5, 0) is a solution to both equations.

3 6 15

2 5

x y

x y

If y is 2, x is -1 so the point (-1, 2) is a solution to both equations.

So we list the solution as:

2 5 where is any real numberx y x

Let's solve the second equation for x. (Solving for x in either equation will give the same result)

x

y

Any point on this line is a solution

Let's try another one: 2 5 1

4 10 5

x y

x y

Let's multiply the first equation by 2 and add to the second to eliminate the x's.

2 2

4 10 2x y

0 7

This time the y's were eliminated too but the constants were not. We get a false statement. This tells us the system of equations is inconsistent and there is not an x and y that make both equations true. This is Case 3, no solution.

x

y

There are no points of intersection

3333231

2232221

1131211

bzayaxa

bzayaxa

bzayaxa

The solution will be one of three cases:

1. Exactly one solution, an ordered triple (x, y, z) 2. A dependent system with infinitely many solutions3. No solution

Three Equations Containing Three Variables

As before, the first two cases are called consistent since there are solutions. The last case is called inconsistent.

Planes intersect at a point: consistent with one solution

With two equations and two variables, the graphs were lines and the solution (if there was one) was where the lines intersected. Graphs of three variable equations are planes. Let’s look at different possibilities. Remember the solution would be where all three planes all intersect.

Planes intersect in a line: consistent system called dependent with an infinite number of solutions

Three parallel planes: no intersection so system called inconsistent with no solution

No common intersection of all three planes: inconsistent with no solution

We will be doing elimination to solve these systems. It’s like elimination that you learned with two equations and variables but it’s now the problem is bigger.

732

10223

42

zyx

zyx

zyx

Your first strategy would be to choose one equation to keep that has all 3 variables, but then use that equation to “eliminate” a variable out of the other two. I’m going to choose the last equation to “keep” because it has just x.

coefficient is a 1 here so it will be easy to work with

732

10223

42

zyx

zyx

zyx 732 zyxkeep over here for later use

Now use the third equation multiplied through by whatever it takes to eliminate the x term from the first equation and add these two equations together. In this case when added to eliminate the x’s you’d need a -2.

7322 zyx

1055 zy

1055 zy

put this equation up with the other one we kept

14642 zyx42 zyx

732

10223

42

zyx

zyx

zyx 732 zyx

7323 zyx 21963 zyx10223 zyx

1174 zy

1055 zy

We won’t “keep” this equation, but we’ll use it together with the one we “kept” with y and z in it to eliminate the y’s.

Now use the third equation multiplied through by whatever it takes to eliminate the x term from the second equation and add these two equations together. In this case when added to eliminate the x’s you’d need a 3.

So we’ll now eliminate y’s from the 2 equations in y and z that we’ve obtained by multiplying the first by 4 and the second by 5

732

10223

42

zyx

zyx

zyx 732 zyxkeep over here for later use

1174 zy

1055 zy

We can add this to the one’s we’ve kept up in the corner

1055 zy 402020 zy553520 zy

1515 z1z

1z

Now we are ready to take the equations in the corner and “back substitute” using the equation at the bottom and substituting it into the equation above to find y.

732

10223

42

zyx

zyx

zyx 732 zyxkeep over here for later use1055 zy

Now we know both y and z we can sub them in the first equation and find x

10155 y

1z

55 y 1y

71312 x

75 x 2x

These planes intersect at a point, namely the point (2, -1, 1).The equations then have this unique solution. This is the ONLY x, y and z that make all 3 equations true.

I’m going to “keep” this one since it will be easy to use to eliminate x’s from others.

Let’s do another one:

143

52

032

zyx

zyx

zyx52 zyx

032

10242

zyx

zyx

10 zy

10 zy

we’ll keep this one

143

15363

zyx

zyx

1622 zy

Now we’ll use the 2 equations we have with y and z to eliminate the y’s.

If we multiply the equation we kept by 2 and add it to the first equation we can eliminate x’s.

If we multiply the equation we kept by 3 and add it to the last equation we can eliminate x’s.

143

52

032

zyx

zyx

zyx 52 zyx

10 zy

10 zy

we’ll multiply the first equation by -2 and add these together

1622 zy

Oops---we eliminated the y’s alright but the z’s ended up being eliminated too and we got a false equation.

2022 zy1622 zy

40

This means the equations are inconsistent and have no solution. The planes don’t have a common intersection and there is not any (x, y, z) that make all 3 equations true.

Let’s do another one:

454

22

12

zyx

zyx

zyxlet's “keep” this one since it will be easy to use to eliminate x’s from others.

12 zyx

22

2422

zyx

zyx

033 zywe’ll keep this one

454

4844

zyx

zyx

If we multiply the equation we kept by -2 and add it to the second equation we can eliminate x’s.

033 zy

Now we’ll use the 2 equations we have with y and z to eliminate the y’s.

033 zy

If we multiply the equation we kept by -4 and add it to the last equation we can eliminate x’s.

454

22

12

zyx

zyx

zyx 12 zyx

033 zy

multiply the first equation by -1 and add the equations to eliminate y.

033 zy

033 zy

033 zy033 zy

00

Oops---we eliminated the y’s alright but the z’s ended up being eliminated too but this time we got a true equation.

This means the equations are consistent and have infinitely many solutions. The planes intersect in a line. To find points on the line we can solve the 2 equations we saved for x and y in terms of z.

454

22

12

zyx

zyx

zyx 12 zyx033 zyzz

First we just put z = z since it can be any real number. Now solve for y in terms of z.

033 zy

zy

Now sub it -z for y in first equation and solve for x in terms of z.

12 zzxzx 1

The solution is (1 - z, - z, z) where z is any real number.

For example: Let z be 1. Then (0, -1, 1) would be a solution.Notice is works in all 3 equations. But so would the point

you get when z = 2 or 3 or any other real number so there are infinitely many solutions.