Solving Systems of Equations By Graphing

in this lesson we're gonna solve system of equations by graphing so the first equation is is equal to two minus three and the second one is negative two over three plus five so what you want to do is you want to graph the two equations and you want to find out where the point of intersection is that's going to be the solution but let's put some marks on the graph first let's go up to 5 on the y-axis and let's do the same for the x-axis the advantage of drawing big graph is you can clearly see where the solution is going to be if you draw graph with the marks closely spaced to each other it might be difficult to see the answer so the bigger the graph the better it's going to be now each of these equations is in slope-intercept form so starting with the first equation we can see that the slope is 2 and the y-intercept is negative 3. so we need to go down at least three units so for the first equation the intercept is negative three that's gonna be the first point to find the next point use the slope the slope is two slope is rise over run so you have to go up two units and travel one unit to the right that will take us to the next point which is one comma negative one now let's do it again let's go up two over one so that's going to take us to the point two comma one and then if we go up two over one we're going to be at three comma three up two over one then four five and that's enough for this graph and my lines are not perfectly straight but let's do that again you get the picture now let's graph the second equation the intercept is five so we're going to start at value of five the slope is negative two over three so the rise is negative two the run is positive three so starting from the first point we need to travel three units to the right that's the run and then we need to go down two units that will take us to the point three comma three which we can see it's the point of intersection but let's continue graphing it now we need to go three to the right down two but we're out of space for that so let's just use these two points and so that's going to be the graph of the second line so the point of intersection is three comma three so therefore that's the solution to the equation now let's check it now let's solve this by substitution let's replace with two minus three so therefore two minus three is equal to negative two over three plus five and let's get rid of the fraction by multiplying everything by three so two times three that's equal to six negative three times three is negative nine now what is negative two over three times three well if we write it out three is the same as three over one you can cancel the three and you'll get negative two if you don't see it that way you can multiply across negative two times three is negative six and three times one is three negative six divided by 3 is negative 2. so you get the same answer whether if you cancel or if you multiply across so this is just going to be negative 2x 5 times 3 is 15. now let's add 9 to both sides and let's add 2x to both sides six plus two is eight fifteen plus nine is twenty four twenty four divided by eight is three now let's plug that into this equation two times three minus three two times three is six six minus three is three so equals three so therefore the answer is indeed three comma three which is the point of intersection between these two graphs