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Systems of Linear Equations

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Presentation on theme: "Systems of Linear Equations"— Presentation transcript:

1 Systems of Linear Equations
Two Equations Containing Two Variables The first two cases are called consistent since there are solutions. The last case is called inconsistent. The solution will be one of three cases: 1. Exactly one solution, an ordered pair (x, y) 2. A dependent system with infinitely many solutions 3. No solution

2 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. independent inconsistent dependent consistent consistent 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.

3 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. Easiest here since we already solved for y. Now we only have the x variable and we solve for it. Substitute this for x in one of the equations to find y.

4 So our solution is This means that the two lines intersect at this point. Let's look at the graph. We can check this by subbing in these values for x and y. They should make each equation true. Yes! Both equations are satisfied.

5 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) 3 3 Add first equation to this. The y's are eliminated. Substitute this for x in one of the equations to find y.

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

7 Any point on this line is a solution
Let's solve the second equation for x. (Solving for x in either equation will give the same result) Now to get a solution, you chose any real number for y and x depends on that choice. If y is 0, x is -5 so the point (-5, 0) is a solution to both equations. If y is 2, x is -1 so the point (-1, 2) is a solution to both equations. Any point on this line is a solution So we list the solution as:

8 Let's multiply the first equation by 2 and add to the second to eliminate the x's.
Let's try another one: 2 2 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. There are no points of intersection

9 Systems of Linear Equations
Three Equations Containing Three Variables As before, the first two cases are called consistent since there are solutions. The last case is called inconsistent. 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 solutions 3. No solution

10 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 at a point: consistent with one solution

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

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

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

14 We will be doing elimination to solve these systems
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. coefficient is a 1 here so it will be easy to work with 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.

15 keep 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. put this equation up with the other one we kept

16 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. 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.

17 keep over here for later use
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 We can add this to the one’s we’ve kept up in the corner

18 keep over here for later use
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. Now we know both y and z we can sub them in the first equation and find x 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.

19 Let’s do another one: I’m going to “keep” this one since it will be easy to use to eliminate x’s from others. If we multiply the equation we kept by 3 and add it to the last equation we can eliminate x’s. If we multiply the equation we kept by 2 and add it to the first equation we can eliminate x’s. we’ll keep this one Now we’ll use the 2 equations we have with y and z to eliminate the y’s.

20 This means the equations are inconsistent and have no solution
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. we’ll multiply the first equation by -2 and add these together Oops---we eliminated the y’s alright but the z’s ended up being eliminated too and we got a false equation.

21 Let’s do another one: let's “keep” this one since it will be easy to use to eliminate x’s from others. If we multiply the equation we kept by -2 and add it to the second equation we can eliminate x’s. If we multiply the equation we kept by -4 and add it to the last equation we can eliminate x’s. we’ll keep this one Now we’ll use the 2 equations we have with y and z to eliminate the y’s.

22 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. multiply the first equation by -1 and add the equations to eliminate y. Oops---we eliminated the y’s alright but the z’s ended up being eliminated too but this time we got a true equation.

23 First we just put z = z since it can be any real number
First we just put z = z since it can be any real number. Now solve for y in terms of z. Now sub it -z for y in first equation and solve for x in terms of z. 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.


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