“video link turned into a printable worksheet transcript ( Search in video 0:00 in this lesson we will discuss how to 0:02 solve a system of linear equations by 0:04 graphing we will work through different 0:06 examples starting with equations given 0:08 in standard form and then in slope 0:10 intercept form finally we will see two 0:13 special cases one where the system of 0:16 equations has no solution and another 0:19 where it has infinitely many solutions 0:21 the first step in solving a system of 0:23 equations graphically is to plot the 0:25 graph of each equation in this example 0:28 both equations are given in standard 0:30 form and the most straightforward method 0:32 for graphing such equations is The 0:34 Intercept method so we will start by 0:37 finding their x intercept and y 0:39 intercept let's begin with the first 0:42 equation the x intercept occurs when y 0:45 is0 so replace y with 0 and solve for x 0:49 2 * 0 = 0 so we are left with 3 x = 12 0:56 to isolate X divide both sides of the 0:58 equation by three cancel out the 3es 12 1:02 / 3 is 4 therefore the x intercept of 1:06 the first equation is 4 comma 0 the Y 1:10 intercept occurs when x is 0 so set X to 1:13 0 and solve for y 3 * 0 = 0 and we have 1:18 2 y = 12 to isolate y divide each side 1:22 of the equation by two cancel out the 1:25 twos 12 / 2 is 6 so the Y intercept of 1:30 the first equation is 0a 1:33 6 now let's find the intercepts for the 1:36 second equation replace y with 0 to find 1:40 the x intercept 2 * 0 = 0 and we get x = 1:45 -4 so the x intercept of the second 1:48 equation is -4 comma 1:51 0 set X to 0 to find the Y 1:55 intercept divide both sides by -2 to 1:58 isolate y -4 / -2 is 2 giving us the Y 2:04 intercept 0a 2 now that we have our 2:07 intercepts for both equations the next 2:10 step is to plot them on a graph for the 2:12 first equation plot the x intercept 2:15 which is at 4A 0 and the Y intercept 2:18 which is at 0a 6 then connect the two 2:21 points with a straight line this line 2:23 represents the first equation for the 2:26 second equation plot -4 comma 0 which is 2:30 the x intercept and 0a 2 which is the Y 2:34 intercept then connect them with a 2:36 straight line This is the line of the 2:38 second equation the intersection point 2:41 of these two lines which is 2 comma 3 is 2:44 the solution to the system of 2:46 equations we can verify this Solution by 2:49 substituting it into both equations 2:52 substitute two for x and 3 for y in the 2:56 first equation 3 * 2 is 6 and 2 * 3 is 3:00 also 6 which adds up to 3:03 12 in the second equation 2 * 3 = 6 and 3:08 2 - 6 = 3:10 -4 as you can see both equations are 3:13 satisfied confirming that the order pair 3:15 2A 3 is indeed the solution now let's 3:19 solve the second example where the 3:21 equations are given in slope intercept 3:23 form to graph an equation in slope 3:26 intercept form the most straightforward 3:28 method is to use the Y intercept and the 3:31 slope so the first step is to identify 3:34 these values from the equations for the 3:37 first equation the Y intercept is at 0a 3:40 -2 and the slope is 3:43 1/2 for the second equation the Y 3:45 intercept is at 0a 1 and the slope is -1 3:50 now let's begin by graphing the first 3:52 equation first plot the Y intercept 3:56 which is located at 0a -2 remember that 3:59 slope is the ratio of rise over ran 4:02 since the slope is 1 over two the rise 4:05 is one unit up and the ran is 2 units to 4:07 the right now starting from the Y 4:10 intercept go up one unit then move to 4:13 the right two units plot this as the 4:16 second Point Let's plot one more Point 4:19 starting from this point again go up one 4:22 unit and to the right two units then 4:25 plot this as the third point the 4:28 advantage of the slope intercept form is 4:30 that we can easily plot more points and 4:32 make our graph more accurate next 4:35 connect the points with a straight line 4:37 This represents the line of the first 4:39 equation now let's plot the graph of the 4:42 second equation as we did before we 4:45 start by plotting the Y intercept which 4:47 is located at 0a 1 its slope is -1 which 4:51 is the same as -1 over 1 right so the 4:55 rise is one unit down and the run is one 4:58 unit to the right 5:00 starting from the Y intercept go down 5:02 one unit then move to the right one unit 5:05 plot one more Point by moving from this 5:07 point one unit down and one unit right 5:11 now connect these points with a straight 5:13 line This is the line of the second 5:15 equation the intersection point of the 5:17 two lines is 2 comma -1 which is the 5:21 solution to the system of equations if 5:24 you have any questions please feel free 5:26 to leave them in the comments below now 5:28 let's consider the the two special cases 5:31 what method do you suggest to graph 5:33 these 5:35 equations the first equation is given in 5:37 slope intercept form so you can use the 5:40 Y intercept and the slope while the 5:42 second equation is given in standard 5:44 form so you can use the intercept method 5:47 the Y intercept of the first equation is 5:49 the point 0a 5 and the slope is -4 5:53 over3 as we did in the previous example 5:56 we will start by plotting the Y 5:58 intercept which is is located at 0a 5 6:02 slope is rise over ran and in this case 6:05 it is -4 over 3 you can place the 6:08 negative sign in the numerator it will 6:10 not change the value of the slope since 6:12 they are equivalent 6:13 fractions the key concept to remember is 6:16 that when the rise is positive we move 6:18 up and when it is negative we move down 6:21 in this case since the rise is -4 we 6:24 move four units down similarly when the 6:28 run is positive we go to the right and 6:30 when it is negative we go to the left in 6:33 this case the ran is positive three so 6:36 we go three units to the right now 6:39 starting from the Y intercept move down 6:42 four units then move to the right three 6:44 units and plot the second Point let's 6:47 add one more Point starting from this 6:49 point move down four units and move to 6:52 the right three units and plot the third 6:55 points then connect these points with a 6:57 straight line now let's move on to 7:00 graphing the second equation using the 7:02 intercept method first let's find the x 7:06 intercept set y to 0 and solve for x 3 * 7:11 0 = 0 so we are left with 4X = -12 7:16 dividing both sides of the equation by 7:18 four we get x = -3 so the x intercept is 7:24 -3 comma 7:26 0 next let's find the y intercept by 7:29 setting X to 0 and solving for y 4 * 0 = 7:34 0 and we get 3 y = -12 to isolate y 7:39 divide both sides by 3 giving us y = -4 7:44 therefore the Y intercept is 0a 7:48 -4 now plot the x intercept which is 7:51 located at -3 comma 0 and the Y 7:54 intercept which is at 0 comma 7:57 -4 then connect these two points with a 8:00 straight line notice that the two lines 8:03 do not intersect they are parallel lines 8:06 if the lines of the two equations are 8:08 parallel then the system has no 8:10 solution finally let's consider the case 8:13 where the system of equations has 8:15 infinitely many solutions please feel 8:17 free to pause the video and give it a 8:20 try the first equation is given in 8:22 standard form so first we will find its 8:26 x and y 8:27 intercepts to find the X intercept set y 8:30 to 0 5 * 0 = 0 and we get 2 x = 10 8:37 divide both sides by two to isolate x 10 8:40 / 2 is 5 so the x intercept is 5A 0 to 8:47 find the Y intercept set X to 0 2 * 0 = 8:52 0 and we have -5 y = 10 to isolate y 8:57 divide both sides by 5 giving us y = -2 9:03 therefore the Y intercept is 0a 9:07 -2 next plot the points on the graph 9:11 plot 5 comma 0 which is the x intercept 9:14 plot 0a -2 which is the Y 9:18 intercept and then connect them with a 9:20 straight line moving on to the second 9:23 equation which is in slope intercept 9:25 form the Y intercept is at 0a -2 and the 9:30 slope is 2 over 5 start by plotting the 9:33 Y 9:34 intercept slope is rise over ran and in 9:37 this case it is 2 over 5 which means the 9:40 rise is 2 units up and the ran is five 9:43 units to the right starting from the Y 9:46 intercept move up two units then go to 9:49 the right five units and plot the second 9:51 point you have likely notice that both 9:54 points lie on the line of the first 9:56 equation when you connect these points 9:58 with a Str straight line it perfectly 10:00 aligns with the first line so the line 10:03 of the first equation and the line of 10:05 the second equation are the same they 10:08 coincide perfectly if the lines of the 10:10 two equations are coincident then the 10:13 system has infinitely many 10:15 solutions this means that any point on 10:17 the lines satisfies both equations in 10:21 conclusion when solving a system of 10:22 equations graphically first graph each 10:25 equation on the same coordinate plane if 10:27 the lines intersect the point of 10:29 intersection is a solution to the system 10:31 of equations if they are parallel the 10:34 system has no solution if they are 10:37 coincident the system has infinitely 10:39 many solutions thanks for watching if 10:42 you found this video helpful please give 10:44 it a thumbs up and 10:50 subscribe)”