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Current time:0:00Total duration:6:04

Topic D: Systems of linear equations and their solutions

Video transcript

we're asked to solve and graph the system of equations here and just as a bit of review solving a system of equations really is means figuring out the x and y value that will satisfy both of these equations and one way to do it is to use one of the equations to solve for either the X or the Y and then substitute for that value in the other ones that make sure that you're making use of both constraints so let's start with this bottom equation right here so we have Y minus X is equal to 5 it's pretty straightforward to solve for y here we just have to add X to both sides of this equation so add X and so the left-hand side these X's cancel out the negative x and the positive x and we're left with Y is equal to 5 plus X y is equal to 5 plucks plus X now the whole point of me doing that is now we can anytime we see a Y in the other equation we can we can replace it with a 5 plus X so the other equation was during this orange color 9x plus 3y is equal to 15 this second equation told us if we just rearranged it that Y is equal to 5 plus X so we can replace Y in the second equation with 5 Plex plus X that make sure we're making use of both constraints so let's do that we're going to replace Y with 5 plus X so this 9x plus 3y equals 15 becomes 9x plus 3 times y the second equation says y is 5 plus X so we're going to put 5 plus X there instead of a Y 3 times 5 plus X is equal to 15 is equal to 15 and now we can just solve for x we get 9x plus 3 times 5 is 15 plus 3 times X is 3x is equal to 15 so we can add the 9x and the 3x so we get 12 X plus 15 12 X plus 15 is equal to 15 is equal to 15 now we could subtract 15 from both sides subtract 15 from both sides just so you get only X terms on the left hand side these guys cancel each other out and you're left with 12 X is equal to 0 now you divide both sides by 12 and you get X is equal to 0 over 12 or X is equal to 0 so let me scroll down a little bit so X is equal to 0 X is equal to 0 now if x equals to 0 what is y well we could substitute into either one of these equations up here if we substitute x equals 0 in this first equation you get 9 times 0 plus 3y is equal to 15 or that's just 0 so you get 3y is equal to 15 divide both sides by 3 you get Y is equal to 15 over 3 or 5y is equal to 5 and we can test that that also satisfies this equation y 5 minus 0 is also equal to 5 so the value so the value X is equal to I'll do this in green X is equal to 0 Y is equal to 5 satisfies both of these equations so we've done the first part let's do the second part where we're asked to graph it and the second equation is pretty straightforward to graph we actually ended up putting it in MX plus B form right there and actually let me rewrite it let me just switch the X and the 5 so it really isn't that MX plus B form so Y is equal to X plus 5 so it's y-intercept is 5 1 2 3 4 5 and it's slope is 1 right there's a 1 implicitly being multiplied or the X is being multiplied by 1 so it looks like this the line let me see how well I can draw it the line will look like that has a slope of 1 you move back 1 you go down 1 you move forward 1 you go up 1 that's a pretty good job so that right here is this equation now let's graph that top equation and we just have to put it in MX plus B form or slope-intercept form and I'll do that in green so we have 9x plus 3y is equal to 15 one simplification we can do from the right off right from the get-go is every number here's the visible by three so let's just divide everything by three to make things simpler so we get 3x plus y is equal to five now we can subtract 3x from both sides we subtract 3x from both sides we are left with y is equal to negative 3x plus five so that's what this first equation gets turned into if you put it in slope-intercept form y is equal to negative 3x plus five so if you were to graph it the y-intercept is 5 0 5 and then it's slope is negative 3 so you move 1 in the x-direction you move down 3 in the y-direction move 2 you would move down 6 2 4 6 move two you go to 4 6 so this line is going to look something like this it's going to look something something like that right there and as you can see the solution to this system is the point of intersection of these two lines it's the combination of x and y that satisfy both of these remember this pink line or this red line is all of the X's and Y's that satisfy this equation y minus X is equal to 5 this green line is all of the X's and Y's or all the combinations of them that satisfy this first equation now the one x and y combination that satisfies both is their point of intersection and we figured it out algebraically using substitution that happens at X is equal to 0 Y is equal to 5 X is equal to 0 this is the x-axis Y is equal to 5 right there