Solving systems of equations in 3 variablesPresentation Transcript
Systems of Linear Equations 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 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. 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. Now we only have the x variable and we solve for it. Substitute this for x in one of the equations to find y. Easiest here since we already solved for y .
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.
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. The y 's are eliminated. Substitute this for x in one of the equations to find y. 3 3
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.
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. So we list the solution as: Let's solve the second equation for x . (Solving for x in either equation will give the same result) Any point on this line is a solution
Let's try another one: Let's multiply the first equation by 2 and add to the second to eliminate the x 's. 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
Systems of Linear Equations 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 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. 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
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
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 keep over here for later use We can add this to the one’s we’ve kept up in the corner
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 . keep over here for later use 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.
Let’s do another one: we’ll keep this one 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. 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.
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. 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: we’ll keep this one If we multiply the equation we kept by -2 and add it to the second equation we can eliminate x ’s. 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 -4 and add it to the last equation we can eliminate x ’s. let's “keep” this one since it will be easy to use to eliminate x ’s from others.
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. 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 .
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.