Today we're going to talk about solving a system of equations by graphing. Remember, a solution to a system of equations is the set of values for the variables of your system that satisfy all of the equations in the system. And when you're looking at a graph, that solution is going to be the intersection point of all of the lines or the equations in the system.
So there must be a single intersection point of all of the lines for the equations in the system. And so one problem with the graphing method is that if your graph is not completely accurate, then you might find approximate answers for the solution or for the variables in your system, but it may not be the exact solution.
So we're going to see how to use the graphing method by doing some examples. So from my first example, I've got a system of two equations, and I'm going to graph each one to find the intersection point, which, again, is the solution to the system. So my first equation is in slope-intercept form, so I know the y-intercept is going to be 5 and the slope is going to be negative 4.
So I'm going to start by putting a point at 5 on my y-axis. And my slope is negative 4 or negative 4 over 1, which means that my rise is down 4 and my run is over to positive 1. Do that a couple of times. And now I can connect my points with a line.
My second equation is also in slope-intercept form. Here, my y-intercept is negative 4, and my slope is 1 over 2. So now I'm going to start by putting a point at negative 4 on my y-axis, then using my slope as rise of 1 and run of 2. I'll go up 1 and over 2. And again, I'll do that a couple of times. And then I can connect with a line.
So I see that my two lines intersect at the point 2, negative 3, which means that my solution has a value of 2 for x and a value of negative 3 for y. So I'm going to verify that by substituting these values into both equations to make sure they yield true statements for both equations.
So with my first equation, this will become negative 3 for y is equal to negative 4 times 2 for my value of x plus 5. Simplifying, negative 4 times 2 will give me a negative 8, and negative 8 plus 5 does indeed equal negative 3. So x equals 2 and y equals negative 3 does satisfy my first equation.
Let's check the second equation. So then I'll have a negative 3 for y again is equal to 1 over 2 times my value for x, which is 2 and minus 4. 1/2 times 2 is just going to give me 1, and 1 minus 4 does equal negative 3. So x equals 2 and y equals negative 3 satisfies my second equation, which means that the solution 2, negative 3 is the solution to my system.
So let's look more closely at what it means to be the solution to a system of equations by looking at its graph. So I have got two equations that are graphed here-- y equals x, and y equals negative 3 over 2x plus 5. And so these two equations make up a system and, again, their solution is going to be the intersection point of those two lines.
So the intersection point is at 2, 2, where the value of x is 2 and the value of y is 2. So x equals 2 and y equals 2 would be the solution to this system of equations. And we can verify that by substituting these values for x and y into both of our equations to make sure that they both yield true statements.
So in my first equation, that's pretty easy because if I substitute 2 in for y and 2 in for x, it's clear to see that this yields a true statement, 2 equals 2. For my second equation, I'm going to substitute 2 in for my y again, and to in for my x. And that is going to give me 2 is equal to negative 3 over 2 times 2 plus 5.
So now simplifying this equation, negative 3 over 2 times 2 is going to give me negative 3. And if I add 5 to that, negative 3 plus 5 is going to give me 2. So again, I see that in my second equation, substituting 2 for x and 2 for y yields another true statement. So since it satisfies both equations, this is my solution to this system of equations.
So here's my second example. I've got a system of three equations. And I'm going to graph each one to find the intersection point. So my first equation is in slope-intercept form. There's no constant value at the end, so my y-intercept is going to be 0, and my slope is going to be negative 1. So I'm going to start by plotting 0 on my y-axis. And then using my slope of negative 1, I'll go down 1 over 1 a couple of times. And I can connect with a line.
My second equation starts at a y-intercept of negative 3 and has a slope of 2, or 2 over 1. So I'll start at negative 3 on my y and go up 2 and over 1 a couple of times. And I can connect with a line. And my third equation has a y-intercept of 1 and a slope of 1 over 4. So I'll start at 1 on my y-axis, and from there I'll go up 1 and over 4, connecting with a line.
I've got my three lines graphed that represent my system of equations. However, even though I can see that some of the lines do intersect, there's not a single intersection point of all three lines. And a solution to a system of equations has to satisfy all of the equations in this system, which on a graph means that it must intersect-- that all three lines must intersect or all of the lines in the system must intersect at a single point. So the solution to this system is, in fact, no solution.
So here's my third example. I've got a system of equations and I, again, am going to graph each equation and find where the intersection point or points are. So my first equation is in slope-intercept form. So I see that the y-intercept is negative 5 and the slope is 2.
So to graph, I'm going to start by putting a value at negative 5 on my y-axis, a point at negative 5 on my y-axis. And then from there, I'll use my slope of 2, which is the same as 2 over 1, and so my rise is 2, I'll go up 2, and my run is 1, so I'll go over 1. I'll do that a couple more times. And then I can connect these points with a line.
My second equation I see is in point-slope form. So a point on the line is going to be at 1, negative 3. And again, my slope is going to be 2. So I'm going to start by putting a point at the point 1, negative 3, So 1, negative 3. And then I'll, again, use my slope of 2 to find other points on the line. And when I do that, I see that each point on my second line was also on my first line.
And so my second equation is actually identical to my first equation. And so these two lines are identical, which means they share an infinite number of intersection points. And when you have lines in a system that have an infinite number of intersection points, there are going to be, therefore, an infinite number of solutions to the system. So the solution to this system of equations is going to be infinite number of solutions.
Let's go over key points from today. Solving a system of equations on a graph involves determining the intersection point of all lines in the system. A system of equations has no solution if all the lines that represent the equations do not intersect at a common point. And a system of equations has infinite solutions if all the lines that represent the equations are identical. So I hope that these key points and examples helped you understand a little bit more about solving a system of equations by graphing. Keep using your notes and keep on practicing and soon you'll be a pro. Thanks for watching.
