Solving Systems of Linear Equations by Graphing │Algebra

in this lesson we will discuss how to solve system of linear equations by graphing we will work through different examples starting with equations given in standard form and then in slope intercept form finally we will see two special cases one where the system of equations has no solution and another where it has infinitely many solutions the first step in solving system of equations graphically is to plot the graph of each equation in this example both equations are given in standard form and the most straightforward method for graphing such equations is The Intercept method so we will start by finding their intercept and intercept let's begin with the first equation the intercept occurs when is0 so replace with 0 and solve for 2 * 0 = 0 so we are left with 3 = 12 to isolate divide both sides of the equation by three cancel out the 3es 12 / 3 is 4 therefore the intercept of the first equation is 4 comma 0 the intercept occurs when is 0 so set to 0 and solve for 3 * 0 = 0 and we have 2 = 12 to isolate divide each side of the equation by two cancel out the twos 12 / 2 is 6 so the intercept of the first equation is 0a 6 now let's find the intercepts for the second equation replace with 0 to find the intercept 2 * 0 = 0 and we get = -4 so the intercept of the second equation is -4 comma 0 set to 0 to find the intercept divide both sides by -2 to isolate -4 / -2 is 2 giving us the intercept 0a 2 now that we have our intercepts for both equations the next step is to plot them on graph for the first equation plot the intercept which is at 4A 0 and the intercept which is at 0a 6 then connect the two points with straight line this line represents the first equation for the second equation plot -4 comma 0 which is the intercept and 0a 2 which is the intercept then connect them with straight line This is the line of the second equation the intersection point of these two lines which is 2 comma 3 is the solution to the system of equations we can verify this Solution by substituting it into both equations substitute two for and 3 for in the first equation 3 * 2 is 6 and 2 * 3 is also 6 which adds up to 12 in the second equation 2 * 3 = 6 and 2 - 6 = -4 as you can see both equations are satisfied confirming that the order pair 2A 3 is indeed the solution now let's solve the second example where the equations are given in slope intercept form to graph an equation in slope intercept form the most straightforward method is to use the intercept and the slope so the first step is to identify these values from the equations for the first equation the intercept is at 0a -2 and the slope is 1/2 for the second equation the intercept is at 0a 1 and the slope is -1 now let's begin by graphing the first equation first plot the intercept which is located at 0a -2 remember that slope is the ratio of rise over ran since the slope is 1 over two the rise is one unit up and the ran is 2 units to the right now starting from the intercept go up one unit then move to the right two units plot this as the second Point Let's plot one more Point starting from this point again go up one unit and to the right two units then plot this as the third point the advantage of the slope intercept form is that we can easily plot more points and make our graph more accurate next connect the points with straight line This represents the line of the first equation now let's plot the graph of the second equation as we did before we start by plotting the intercept which is located at 0a 1 its slope is -1 which is the same as -1 over 1 right so the rise is one unit down and the run is one unit to the right starting from the intercept go down one unit then move to the right one unit plot one more Point by moving from this point one unit down and one unit right now connect these points with straight line This is the line of the second equation the intersection point of the two lines is 2 comma -1 which is the solution to the system of equations if you have any questions please feel free to leave them in the comments below now let's consider the the two special cases what method do you suggest to graph these equations the first equation is given in slope intercept form so you can use the intercept and the slope while the second equation is given in standard form so you can use the intercept method the intercept of the first equation is the point 0a 5 and the slope is -4 over3 as we did in the previous example we will start by plotting the intercept which is is located at 0a 5 slope is rise over ran and in this case it is -4 over 3 you can place the negative sign in the numerator it will not change the value of the slope since they are equivalent fractions the key concept to remember is that when the rise is positive we move up and when it is negative we move down in this case since the rise is -4 we move four units down similarly when the run is positive we go to the right and when it is negative we go to the left in this case the ran is positive three so we go three units to the right now starting from the intercept move down four units then move to the right three units and plot the second Point let's add one more Point starting from this point move down four units and move to the right three units and plot the third points then connect these points with straight line now let's move on to graphing the second equation using the intercept method first let's find the intercept set to 0 and solve for 3 * 0 = 0 so we are left with 4X = -12 dividing both sides of the equation by four we get = -3 so the intercept is -3 comma 0 next let's find the intercept by setting to 0 and solving for 4 * 0 = 0 and we get 3 = -12 to isolate divide both sides by 3 giving us = -4 therefore the intercept is 0a -4 now plot the intercept which is located at -3 comma 0 and the intercept which is at 0 comma -4 then connect these two points with straight line notice that the two lines do not intersect they are parallel lines if the lines of the two equations are parallel then the system has no solution finally let's consider the case where the system of equations has infinitely many solutions please feel free to pause the video and give it try the first equation is given in standard form so first we will find its and intercepts to find the intercept set to 0 5 * 0 = 0 and we get 2 = 10 divide both sides by two to isolate 10 / 2 is 5 so the intercept is 5A 0 to find the intercept set to 0 2 * 0 = 0 and we have -5 = 10 to isolate divide both sides by 5 giving us = -2 therefore the intercept is 0a -2 next plot the points on the graph plot 5 comma 0 which is the intercept plot 0a -2 which is the intercept and then connect them with straight line moving on to the second equation which is in slope intercept form the intercept is at 0a -2 and the slope is 2 over 5 start by plotting the intercept slope is rise over ran and in this case it is 2 over 5 which means the rise is 2 units up and the ran is five units to the right starting from the intercept move up two units then go to the right five units and plot the second point you have likely notice that both points lie on the line of the first equation when you connect these points with Str straight line it perfectly aligns with the first line so the line of the first equation and the line of the second equation are the same they coincide perfectly if the lines of the two equations are coincident then the system has infinitely many solutions this means that any point on the lines satisfies both equations in conclusion when solving system of equations graphically first graph each equation on the same coordinate plane if the lines intersect the point of intersection is solution to the system of equations if they are parallel the system has no solution if they are coincident the system has infinitely many solutions thanks for watching if you found this video helpful please give it thumbs up and subscribe