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Non-Linear Systems

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Hi. My name is Rebecca Muller. During this session, we're going to look at non-linear systems. Now, you should already be familiar with solving linear systems. If not, you probably want to review that before moving on in this session. You may also recall when we solve linear systems that we had three methods we were able to use-- graphing, substitution, and elimination. We're going to revisit those methods when we're looking at non-linear systems in this session. Specifically, let's look at our subtopics. We're going to look at graphing nonlinear systems. We're then going to use substitution to solve nonlinear systems; and finally, look at using elimination to solve nonlinear systems. So let's begin with the idea of graphing. And the system I want to look at has the first equation, which is x squared plus y squared equals 4. The second equation will be x minus 2y equals 2. Now, if we're going to try to do graphing, we want to be able to recognize what the equations are going to give us graphically. So for the first equation, we have the typical format for an equation-- for a circle, that is-- that is centered at the origin. And the radius is going to equal the square root of the numerical value. That's going to be the square root of 4, which is 2. So we end up seeing that this first equation is going to give us a circle. For the second equation, we have the format-- a number times x plus a number times y equals a third number. This is the typical format, ax plus by equals c, which is going to be the equation of a line. So our second equation will be linear. Now, if we think about this graphically, let's just do a quick sketch as to what we could possibly end up with for solutions to the system, which would be values that are going to solve both equations simultaneously. So once more, the circle is going to be centered at the origin. And we have a radius of 2. So I can go from the origin out 2 units to the right, and then two units up, and then 2 units to the left, and then 2 units down. Now. I've got to draw in a circle here. And forgive me if it's not a perfect circle. You get the idea, OK? So here's that part. And then here we have this part. So, yeah, it kind of looks like a circle, right? Then we have a straight line. And to graph that straight line, what you may decide do is the intercept method. So I'm going to quickly just do a little T chart here. If I let x equal 0 in this equation, then I'm looking at negative 2y equals 2. And that's going to give us a value of y equals negative 1. So again, 0, negative 1 would be one of the points on this line. If I let my y value equal 0, then I'm going to end up substituting and getting x minus 0 equals 2. So x is going to equal 2. Right. Let's plot those two points. So I have the point 0, negative 1. Let's just do that. And then I have the point 2, 0, which is right here. And I know that this is linear-- so drawing in the straight line. We end up with our picture, where we have a circle that is being intersected by a straight line. But notice that we end up with two intersection points. Here's one. Here's the other one. Now, because of the fact that these two graphs, when I was graphing them, both, I noticed, went through the point 2, 0. We can identify one of the solutions to the system. That is, if x is equal to 2, and y is equal to 0, that's going to solve both equations. But notice there's a second point out here, which is not going to be so easily defined-- or determined, I should say. So we're going to need another method in order to solve these systems of nonlinear equations. Because one, we can have more than one solution, and two, can we graph it all the time and figure out what those solutions are? And obviously, even from this first example, the answer is no. I can't determine what the second solution would be to the system. So we'll have to look at other methods now. It's time for a quick quiz. Why might graphing not be the best method to use, to solve a non-linear system of equations? A. Graphs may not be as accurate as needed. B. Points of intersection may not be obvious, especially if not integer coefficients. C. Both A and B. Choose A, B or C. You're correct, both A and B. Sorry, both A and B describe reasons why this may not be the best method to use. With non-linear systems, it may be hard to get a completely accurate graph and that leads to points of intersection not being obvious, when using grid paper unless your point is exactly on a cross hair, it may be hard to tell what that point of intersection would be. Hence, typically graphing is not going to be the preferred method. So let's return to the system that we saw earlier. We had x squared plus y squared equals 4. And x minus 2y equals 2. Now for that system, we could use, remember, other methods, perhaps, in order to solve it-- for instance, maybe substitution, maybe elimination. Well, when we think about the elimination, we would have to add together and get one of the variables to be eliminated. But notice that we do not have like terms. So we're definitely going to use substitution in order to solve this system and not elimination. To solve for x in the second equation is going to be fairly easy, and it's going to allow us to then substitute into the first equation with what we find. So let's do that now. Second equation, solving for x, we'll end up with x equals 2y plus 2. Now, we're going to take that and substitute into the first equation. So we're, instead of x squared, we'll have 2y plus 2 quantity squared, plus y squared, equals the value 4. Notice that we now have a single equation and a single variable. And that's the point of what we're doing here-- is to come up with-- from two equations and two unknowns into a single equation and a single unknown. We can now simplify and solve for y. We have to square the binomial, 2y plus 2. So we square the first term. That's going to be 4y squared; plus twice that product of the two terms. So that's going to be 2 times 2y, times 2, which is going to end up giving us 8y; plus the last term squared, which is 4; then plus y squared equals 4. Let's combine like terms. We have 4y squared plus y squared. That's going to give us 5y squared. We have an 8y, which doesn't have another term that's common to it. So that's just going to give us 8y. And I'll subtract 4 from both sides of the equation, which is going to give us equals 0. Now, this is a quadratic set equal to 0. We notice we can factor out the value of y. That's going to give us y times 5y plus 8 equals 0. And now setting each factor equal to 0, using the 0 product property, we have y equals 0, and then we have 5y plus 8 equals 0, which would give us-- subtracting 8 and then dividing by 5-- we have y equals negative 8/5. Now, remember, this is not an ordered pair. This is just the y-coordinate that we would end up having to have from an ordered pair. So to solve for x, let's start off with if y is equal to 0. We can go back to where we solve for x. We have that x is equal to 2 times the y value of 0, in this case, plus 2. And so 2 times 0 is 0, plus 2 is going to be 2. This is going to give us the ordered pair whose coordinates will be the x value of 2 and a y value of 0. If we have a y value equal to negative 8/5, then we can say that our x value is going to have to equal 2 times the negative 8/5, plus the 2. Let's simplify. We end up getting a negative 16/5 plus-- now, if I'm going to add 2 to negative 16/5, I want a common denominator-- a 5. So what is 2 over 1 if I think of it as over 5? Well, 2 is going to be the same thing as 10/5. Combining those two fractions, we end up with negative 6/5, which gives us an ordered pair where the x-coordinate is negative 6/5, and the y-coordinate is negative 8/5. So now I've come up with our two ordered pairs. And we basically should substitute those values into both equations to make sure that we end up with the correct solution. This is like any other linear system-- or in this case, a nonlinear system. We want to substitute into both equations and check that we end up getting a solution that's correct. I'll leave that correct-- that I'm checking to you, because I know that you can do it on your own. But I do want to go back to the graphical representation that we saw earlier. We had a circle, 4x squared plus y squared equals 2. We had a straight line for the graph of x minus 2y equals 2. When we looked at it graphically previously, we could see that the point 2, 0 was going to be a solution to the system. But we had that other point that we really couldn't figure out what it would be, exactly, even though we had at least a vague idea that it's something in the third quadrant. Well, now, because we've been able to solve this algebraically, we see that the point of intersection is actually going to have the coordinates negative 6/5, negative 8/5. So at least, even without substituting in, we can see that we are in the right ballpark, because we end up with the third quadrant solution here; and of course, the point that we had seen previously, which was on the x-axis. Typically, this section on nonlinear systems comes toward the end of a college algebra course. And it's pretty great, because it means you're able to review a lot of the different types of equations that you've already seen in previous sections of your text. This particular example I'm about to work with requires you to remember how to work with exponential equations. The first equation is y equals 2 to the x power plus 1. The second equation is y equals 4 times 2 to the x, minus 8. Now, remember that we have substitution and elimination as possible methods that we could use. We just saw an example where we use substitution. And certainly, in this case, I could use substitution, as I could easily substitute for y in the second equation. But just for a purpose of demonstration, I'm going to do this-- we're using the elimination method. And it's not going to be easy for me to eliminate the x value-- at least not off the bat. But it is going to be fairly easy for me to eliminate the y value by multiplying the first equation by negative 1. So I'm going to take the second equation and rewrite it-- y equals 4 times 2 to the x, minus 8. And then for the second equation, I'm going to now multiply by negative 1 to give me negative y equals negative 2 to the x, minus 1. When I combine those two equations using addition, y minus y is 0; 4 times 2 to the x minus 2 to the x is the same as 4 times 2 to the x, minus 1 times 2 to the x. And that's going to give us 3 times 2 to the x. And then negative 8 and then minus 1 is going to end up giving us minus 9. So now we have an exponential equation that has only a single variable in it. We're going to try to isolate the term that has the x in it. And I can do that in the first step by adding 9 to both sides of the equation. So 9 equals 3 times 2 to the x. Next, we can divide both sides of the equation by 3 and come up with 9 divided by 3 is 3 equals 2 to x power. In order to solve for x, we can take this exponential form of the equation and change it into its logarithmic form. That will give us an equation that solves for the exponent. So we'll end up having our log base 2 of 3 equals x. And now, if you wanted to get an approximate value, you could. But really, I'm interested in the exact values for this system. I'm going to take the value of x. And now I need to solve for y. So notice that my first equation, y equals 2 to the x plus 1, can be used to solve for y, in that y is going to have to equal 2 raised to the power where are we going to have log base 2 of 3, and then plus 1. Reviewing the properties of logarithms and exponents, we know that 2 raised to log base 2 of 3, we're going to have just simply the value 3 that pops out of that, as exponential and logarithmic functions are inverses of each other. And then we have our plus 1 to give us our y value of 4. So, at this point, what we have is the ordered pair, where the x-coordinate is log base 2 of 3, and the y-coordinate is 4. Now, we've substituted into the first equation to find this, so it's really easy to check by taking those coordinates and substituting into the second equation. In our second equation, we would have our y value of 4. And we're trying to see whether or not that's going to be equivalent to 4 times 2 raised to the log base 2 of 3, and then minus 8. Once more, we're noting that 2 raised to the power where we have log base 2 of 3 is simply going to be the value 3. So on the right hand side, we're going to have 4 times 3, and then minus 8. That's the same as 12 minus 8. And of course, that does equal the value of 4. So we are able to find the ordered pair, where the x coordinate is log base 2 of 3, and the y-coordinate is equal to 4. And that's a solution to this system of equations. Now let's look at a system that you can try to solve on your own. Here's the example. We have x squared plus 3xy equals 40. The second equation is 3x squared minus xy equals 40. Let's pause the tape. Give it a try. And then come back, and we'll work it together. OK, so let's see how you did. So when I look at the system to start with, remember we have options. We're definitely not going to try to graph this one. That would be difficult. We can, however, think about substitution maybe. But notice that it would be-- I could solve for y in the first equation maybe, and then substitute to the second one. But it's going to not be a very pretty way to do it. I'm instead going to note that I can use elimination. Now, be a little cautious here, because if you try to eliminate the x squared term, that's going to be great. You can do that by multiplying the first term-- first equation by negative 3, for instance. But notice that you would still be left with an x and a y in your equation. So you need to eliminate a variable when you're doing that. So, instead of trying to eliminate the x squared term, I'm going to look at eliminating the xy term. I can accomplish that if I multiply the second equation by positive 3. So just rewriting, we're going to have our first equation-- x squared plus 3xy equals 40. And I'm going to multiply the second equation by positive 3. That's going to give me the coefficients on the xy term to be opposite each other. Multiplying by 3 in the first term gives me 9x squared. Multiplying by 3 in the second term gives me a minus 3xy. And on the right hand side, 40 times 3 is 120. So let's add the two equations together now-- x squared plus 9x squared is going to give us 10x squared. That 3xy minus 3xy is going to equal 0, which means that that term is going to be eliminated, and the y variable is going to be eliminated in our equation. On the right hand side, 40 plus 120 is 160. So now let's divide both sides of this equation by 10 to give us x squared equals 16. And then take the square root of both sides of the equation. And remember, that when you take the square root, you end up getting x equals plus or minus the square root of 16. That's going to give us a plus or minus 4. Now, if x is equal to the value of 4, then we can use that to solve for y. So we're going to end up taking that and substituting into one of our equations. Let's just use the first equation. So x squared is going to give us 16 plus 3 times x, which is going to be 3 times 4, which going to be 12; and then times y, equals a value of 40. Let's subtract 16 for both sides of the equation. So 12y is going to equal the value of 24. And our y value is going to equal 24 divided by 12, which is 2. So we come up with an ordered pair whose first coordinate is 4; second coordinate is 2. Now we have to consider what happens if our x value is equal to negative 4. I'll go back to the first equation. We take negative 4 and square it. Negative 4 squared is going to be positive 16. And then we're going to have plus 3 times negative 4. Now, that's going to be a negative 12; and then times y equals the value of 40. Subtracting 16 from both sides of the equation-- negative 12y equals 24. And dividing by negative 12, we come up with y equals negative 2. So that gives us another ordered pair, where the x-coordinate is going to be a negative 4, and the y-coordinate is going to be a negative 2. Now, we should check our work by substituting into the original equations. We have two ordered pairs. Let's see if both of them are indeed are going to be solutions to this system. So, if we check-- and let's start with the coordinates 4, 2. In our first equation, we're going to have our 16 when we square the 4. Plus we're going to have 3 times the value of 4, times the value of 2. Let's just go ahead and simplify here. We have 16 plus 12, times 2 is 24. And that equals 40. So the first equation checks. In our second equation, we'd have 3 times the value of 16, because we squared the 4; minus-- we're going to have the value of 4 multiplied times 2. And let's simplify. We're going to have 48 minus 8, which is equal to 40. And that checks. So we know for sure that the ordered pair 4, 2 is going to be a solution to the system. Let's now check when we have the coordinates negative 4, negative 2. And again, just going through the same procedures-- square the negative 4 in the first equation. That's going to give us 16. Then we're going to have plus 3 times our value of x, which is negative 4; times our value of y, which is negative 2. Well, notice that's going to give us 8 times 3, which is 24. So 16 plus 24 equals 40. And that checks. And in our second equation, we would have 3 times our negative 4 squared, which is going to be 16; minus-- we're going to have the product of negative 4 and negative 2. And simplifying, we get 48 minus negative 4, times negative 2 is 8. And that equals 40. And so now we see that our second point is also a solution. So we're able to solve for both of these ordered pairs. And those two points would be the places where, if we could draw the graphs of these equations, we would end up finding intersection points. So I hope you've done well on this example, and I hope that you understand the explanations that we've gone through here. So remember that when you're solving the nonlinear systems, you have ostensibly simply three methods. One is graphing. But notice the graphing is going to be very tough to do on a lot of these systems, and also it's probably less accurate than the other two methods, which are substitution and elimination. It's time now for you to try some problems on your own. Good luck with it. It's time for another quick quiz. Which method is best for solving the following system? Our first equation, y equals negative three fourths x. Our second equation, x squared plus y squared equals 25. Is the best method going to be A. Substitution B. Elimination C. Graphing. Choose A, B or C. You're right. Your best method is substitution. Sorry the best method is probably substitution. When we look at equation one, it's already solved for the variable y. That is going to be an easy substitution into the second equation. So that we could read x squared, plus negative three fourths x, quantity squared equals 25, giving us a single equation with a single variable in it. Therefore, we don't have to do any type of manipulation prior to getting into this format. If you try to use elimination, you can't add them together as given because the variables in the first equation are not squared. Graphing is problematic due to the fact that graphs can sometimes not be accurate and sometimes it's hard to tell where the points of intersection are going to be.
Hi. My name is Rebecca Muller. During this session, we're going to look at non-linear systems. Now, you should already be familiar with solving linear systems. If not, you probably want to review that before moving on in this session. You may also recall when we solve linear systems that we had three methods we were able to use-- graphing, substitution, and elimination. We're going to revisit those methods when we're looking at non-linear systems in this session. Specifically, let's look at our subtopics. We're going to look at graphing nonlinear systems. We're then going to use substitution to solve nonlinear systems; and finally, look at using elimination to solve nonlinear systems. So let's begin with the idea of graphing. And the system I want to look at has the first equation, which is x squared plus y squared equals 4. The second equation will be x minus 2y equals 2. Now, if we're going to try to do graphing, we want to be able to recognize what the equations are going to give us graphically. So for the first equation, we have the typical format for an equation-- for a circle, that is-- that is centered at the origin. And the radius is going to equal the square root of the numerical value. That's going to be the square root of 4, which is 2. So we end up seeing that this first equation is going to give us a circle. For the second equation, we have the format-- a number times x plus a number times y equals a third number. This is the typical format, ax plus by equals c, which is going to be the equation of a line. So our second equation will be linear. Now, if we think about this graphically, let's just do a quick sketch as to what we could possibly end up with for solutions to the system, which would be values that are going to solve both equations simultaneously. So once more, the circle is going to be centered at the origin. And we have a radius of 2. So I can go from the origin out 2 units to the right, and then two units up, and then 2 units to the left, and then 2 units down. Now. I've got to draw in a circle here. And forgive me if it's not a perfect circle. You get the idea, OK? So here's that part. And then here we have this part. So, yeah, it kind of looks like a circle, right? Then we have a straight line. And to graph that straight line, what you may decide do is the intercept method. So I'm going to quickly just do a little T chart here. If I let x equal 0 in this equation, then I'm looking at negative 2y equals 2. And that's going to give us a value of y equals negative 1. So again, 0, negative 1 would be one of the points on this line. If I let my y value equal 0, then I'm going to end up substituting and getting x minus 0 equals 2. So x is going to equal 2. Right. Let's plot those two points. So I have the point 0, negative 1. Let's just do that. And then I have the point 2, 0, which is right here. And I know that this is linear-- so drawing in the straight line. We end up with our picture, where we have a circle that is being intersected by a straight line. But notice that we end up with two intersection points. Here's one. Here's the other one. Now, because of the fact that these two graphs, when I was graphing them, both, I noticed, went through the point 2, 0. We can identify one of the solutions to the system. That is, if x is equal to 2, and y is equal to 0, that's going to solve both equations. But notice there's a second point out here, which is not going to be so easily defined-- or determined, I should say. So we're going to need another method in order to solve these systems of nonlinear equations. Because one, we can have more than one solution, and two, can we graph it all the time and figure out what those solutions are? And obviously, even from this first example, the answer is no. I can't determine what the second solution would be to the system. So we'll have to look at other methods now. It's time for a quick quiz. Why might graphing not be the best method to use, to solve a non-linear system of equations? A. Graphs may not be as accurate as needed. B. Points of intersection may not be obvious, especially if not integer coefficients. C. Both A and B. Choose A, B or C. You're correct, both A and B. Sorry, both A and B describe reasons why this may not be the best method to use. With non-linear systems, it may be hard to get a completely accurate graph and that leads to points of intersection not being obvious, when using grid paper unless your point is exactly on a cross hair, it may be hard to tell what that point of intersection would be. Hence, typically graphing is not going to be the preferred method. So let's return to the system that we saw earlier. We had x squared plus y squared equals 4. And x minus 2y equals 2. Now for that system, we could use, remember, other methods, perhaps, in order to solve it-- for instance, maybe substitution, maybe elimination. Well, when we think about the elimination, we would have to add together and get one of the variables to be eliminated. But notice that we do not have like terms. So we're definitely going to use substitution in order to solve this system and not elimination. To solve for x in the second equation is going to be fairly easy, and it's going to allow us to then substitute into the first equation with what we find. So let's do that now. Second equation, solving for x, we'll end up with x equals 2y plus 2. Now, we're going to take that and substitute into the first equation. So we're, instead of x squared, we'll have 2y plus 2 quantity squared, plus y squared, equals the value 4. Notice that we now have a single equation and a single variable. And that's the point of what we're doing here-- is to come up with-- from two equations and two unknowns into a single equation and a single unknown. We can now simplify and solve for y. We have to square the binomial, 2y plus 2. So we square the first term. That's going to be 4y squared; plus twice that product of the two terms. So that's going to be 2 times 2y, times 2, which is going to end up giving us 8y; plus the last term squared, which is 4; then plus y squared equals 4. Let's combine like terms. We have 4y squared plus y squared. That's going to give us 5y squared. We have an 8y, which doesn't have another term that's common to it. So that's just going to give us 8y. And I'll subtract 4 from both sides of the equation, which is going to give us equals 0. Now, this is a quadratic set equal to 0. We notice we can factor out the value of y. That's going to give us y times 5y plus 8 equals 0. And now setting each factor equal to 0, using the 0 product property, we have y equals 0, and then we have 5y plus 8 equals 0, which would give us-- subtracting 8 and then dividing by 5-- we have y equals negative 8/5. Now, remember, this is not an ordered pair. This is just the y-coordinate that we would end up having to have from an ordered pair. So to solve for x, let's start off with if y is equal to 0. We can go back to where we solve for x. We have that x is equal to 2 times the y value of 0, in this case, plus 2. And so 2 times 0 is 0, plus 2 is going to be 2. This is going to give us the ordered pair whose coordinates will be the x value of 2 and a y value of 0. If we have a y value equal to negative 8/5, then we can say that our x value is going to have to equal 2 times the negative 8/5, plus the 2. Let's simplify. We end up getting a negative 16/5 plus-- now, if I'm going to add 2 to negative 16/5, I want a common denominator-- a 5. So what is 2 over 1 if I think of it as over 5? Well, 2 is going to be the same thing as 10/5. Combining those two fractions, we end up with negative 6/5, which gives us an ordered pair where the x-coordinate is negative 6/5, and the y-coordinate is negative 8/5. So now I've come up with our two ordered pairs. And we basically should substitute those values into both equations to make sure that we end up with the correct solution. This is like any other linear system-- or in this case, a nonlinear system. We want to substitute into both equations and check that we end up getting a solution that's correct. I'll leave that correct-- that I'm checking to you, because I know that you can do it on your own. But I do want to go back to the graphical representation that we saw earlier. We had a circle, 4x squared plus y squared equals 2. We had a straight line for the graph of x minus 2y equals 2. When we looked at it graphically previously, we could see that the point 2, 0 was going to be a solution to the system. But we had that other point that we really couldn't figure out what it would be, exactly, even though we had at least a vague idea that it's something in the third quadrant. Well, now, because we've been able to solve this algebraically, we see that the point of intersection is actually going to have the coordinates negative 6/5, negative 8/5. So at least, even without substituting in, we can see that we are in the right ballpark, because we end up with the third quadrant solution here; and of course, the point that we had seen previously, which was on the x-axis. Typically, this section on nonlinear systems comes toward the end of a college algebra course. And it's pretty great, because it means you're able to review a lot of the different types of equations that you've already seen in previous sections of your text. This particular example I'm about to work with requires you to remember how to work with exponential equations. The first equation is y equals 2 to the x power plus 1. The second equation is y equals 4 times 2 to the x, minus 8. Now, remember that we have substitution and elimination as possible methods that we could use. We just saw an example where we use substitution. And certainly, in this case, I could use substitution, as I could easily substitute for y in the second equation. But just for a purpose of demonstration, I'm going to do this-- we're using the elimination method. And it's not going to be easy for me to eliminate the x value-- at least not off the bat. But it is going to be fairly easy for me to eliminate the y value by multiplying the first equation by negative 1. So I'm going to take the second equation and rewrite it-- y equals 4 times 2 to the x, minus 8. And then for the second equation, I'm going to now multiply by negative 1 to give me negative y equals negative 2 to the x, minus 1. When I combine those two equations using addition, y minus y is 0; 4 times 2 to the x minus 2 to the x is the same as 4 times 2 to the x, minus 1 times 2 to the x. And that's going to give us 3 times 2 to the x. And then negative 8 and then minus 1 is going to end up giving us minus 9. So now we have an exponential equation that has only a single variable in it. We're going to try to isolate the term that has the x in it. And I can do that in the first step by adding 9 to both sides of the equation. So 9 equals 3 times 2 to the x. Next, we can divide both sides of the equation by 3 and come up with 9 divided by 3 is 3 equals 2 to x power. In order to solve for x, we can take this exponential form of the equation and change it into its logarithmic form. That will give us an equation that solves for the exponent. So we'll end up having our log base 2 of 3 equals x. And now, if you wanted to get an approximate value, you could. But really, I'm interested in the exact values for this system. I'm going to take the value of x. And now I need to solve for y. So notice that my first equation, y equals 2 to the x plus 1, can be used to solve for y, in that y is going to have to equal 2 raised to the power where are we going to have log base 2 of 3, and then plus 1. Reviewing the properties of logarithms and exponents, we know that 2 raised to log base 2 of 3, we're going to have just simply the value 3 that pops out of that, as exponential and logarithmic functions are inverses of each other. And then we have our plus 1 to give us our y value of 4. So, at this point, what we have is the ordered pair, where the x-coordinate is log base 2 of 3, and the y-coordinate is 4. Now, we've substituted into the first equation to find this, so it's really easy to check by taking those coordinates and substituting into the second equation. In our second equation, we would have our y value of 4. And we're trying to see whether or not that's going to be equivalent to 4 times 2 raised to the log base 2 of 3, and then minus 8. Once more, we're noting that 2 raised to the power where we have log base 2 of 3 is simply going to be the value 3. So on the right hand side, we're going to have 4 times 3, and then minus 8. That's the same as 12 minus 8. And of course, that does equal the value of 4. So we are able to find the ordered pair, where the x coordinate is log base 2 of 3, and the y-coordinate is equal to 4. And that's a solution to this system of equations. Now let's look at a system that you can try to solve on your own. Here's the example. We have x squared plus 3xy equals 40. The second equation is 3x squared minus xy equals 40. Let's pause the tape. Give it a try. And then come back, and we'll work it together. OK, so let's see how you did. So when I look at the system to start with, remember we have options. We're definitely not going to try to graph this one. That would be difficult. We can, however, think about substitution maybe. But notice that it would be-- I could solve for y in the first equation maybe, and then substitute to the second one. But it's going to not be a very pretty way to do it. I'm instead going to note that I can use elimination. Now, be a little cautious here, because if you try to eliminate the x squared term, that's going to be great. You can do that by multiplying the first term-- first equation by negative 3, for instance. But notice that you would still be left with an x and a y in your equation. So you need to eliminate a variable when you're doing that. So, instead of trying to eliminate the x squared term, I'm going to look at eliminating the xy term. I can accomplish that if I multiply the second equation by positive 3. So just rewriting, we're going to have our first equation-- x squared plus 3xy equals 40. And I'm going to multiply the second equation by positive 3. That's going to give me the coefficients on the xy term to be opposite each other. Multiplying by 3 in the first term gives me 9x squared. Multiplying by 3 in the second term gives me a minus 3xy. And on the right hand side, 40 times 3 is 120. So let's add the two equations together now-- x squared plus 9x squared is going to give us 10x squared. That 3xy minus 3xy is going to equal 0, which means that that term is going to be eliminated, and the y variable is going to be eliminated in our equation. On the right hand side, 40 plus 120 is 160. So now let's divide both sides of this equation by 10 to give us x squared equals 16. And then take the square root of both sides of the equation. And remember, that when you take the square root, you end up getting x equals plus or minus the square root of 16. That's going to give us a plus or minus 4. Now, if x is equal to the value of 4, then we can use that to solve for y. So we're going to end up taking that and substituting into one of our equations. Let's just use the first equation. So x squared is going to give us 16 plus 3 times x, which is going to be 3 times 4, which going to be 12; and then times y, equals a value of 40. Let's subtract 16 for both sides of the equation. So 12y is going to equal the value of 24. And our y value is going to equal 24 divided by 12, which is 2. So we come up with an ordered pair whose first coordinate is 4; second coordinate is 2. Now we have to consider what happens if our x value is equal to negative 4. I'll go back to the first equation. We take negative 4 and square it. Negative 4 squared is going to be positive 16. And then we're going to have plus 3 times negative 4. Now, that's going to be a negative 12; and then times y equals the value of 40. Subtracting 16 from both sides of the equation-- negative 12y equals 24. And dividing by negative 12, we come up with y equals negative 2. So that gives us another ordered pair, where the x-coordinate is going to be a negative 4, and the y-coordinate is going to be a negative 2. Now, we should check our work by substituting into the original equations. We have two ordered pairs. Let's see if both of them are indeed are going to be solutions to this system. So, if we check-- and let's start with the coordinates 4, 2. In our first equation, we're going to have our 16 when we square the 4. Plus we're going to have 3 times the value of 4, times the value of 2. Let's just go ahead and simplify here. We have 16 plus 12, times 2 is 24. And that equals 40. So the first equation checks. In our second equation, we'd have 3 times the value of 16, because we squared the 4; minus-- we're going to have the value of 4 multiplied times 2. And let's simplify. We're going to have 48 minus 8, which is equal to 40. And that checks. So we know for sure that the ordered pair 4, 2 is going to be a solution to the system. Let's now check when we have the coordinates negative 4, negative 2. And again, just going through the same procedures-- square the negative 4 in the first equation. That's going to give us 16. Then we're going to have plus 3 times our value of x, which is negative 4; times our value of y, which is negative 2. Well, notice that's going to give us 8 times 3, which is 24. So 16 plus 24 equals 40. And that checks. And in our second equation, we would have 3 times our negative 4 squared, which is going to be 16; minus-- we're going to have the product of negative 4 and negative 2. And simplifying, we get 48 minus negative 4, times negative 2 is 8. And that equals 40. And so now we see that our second point is also a solution. So we're able to solve for both of these ordered pairs. And those two points would be the places where, if we could draw the graphs of these equations, we would end up finding intersection points. So I hope you've done well on this example, and I hope that you understand the explanations that we've gone through here. So remember that when you're solving the nonlinear systems, you have ostensibly simply three methods. One is graphing. But notice the graphing is going to be very tough to do on a lot of these systems, and also it's probably less accurate than the other two methods, which are substitution and elimination. It's time now for you to try some problems on your own. Good luck with it. It's time for another quick quiz. Which method is best for solving the following system? Our first equation, y equals negative three fourths x. Our second equation, x squared plus y squared equals 25. Is the best method going to be A. Substitution B. Elimination C. Graphing. Choose A, B or C. You're right. Your best method is substitution. Sorry the best method is probably substitution. When we look at equation one, it's already solved for the variable y. That is going to be an easy substitution into the second equation. So that we could read x squared, plus negative three fourths x, quantity squared equals 25, giving us a single equation with a single variable in it. Therefore, we don't have to do any type of manipulation prior to getting into this format. If you try to use elimination, you can't add them together as given because the variables in the first equation are not squared. Graphing is problematic due to the fact that graphs can sometimes not be accurate and sometimes it's hard to tell where the points of intersection are going to be.