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# Systems of equations with graphing: 5x+3y=7 & 3x-2y=8

CCSS.Math:

## Video transcript

solve the system of linear equations by graphing and they give us two equations here 5x plus 3y is equal to 7 and 3x minus 2y is equal to 8 when they say solve the system of linear equations they really just saying find an x and a y that satisfies both of these equations and when they say to do it by graphing we're essentially going to graph this first equation remember the graph is really just depicting all of the X's and Y's that satisfy this first equation and then we graph the second equation that's depicting all the X's and Y's that satisfy that one so if we were looking for an x and y that satisfies both though that that point needs to be on both equations and so or it has to be on both graphs so it'll be the intersection of the two graphs so let's try to see if we can do that so let's focus on let's focus on this first equation and I want to graph it so I have 5x plus 3y is equal to is equal to 7 there's a couple of ways we could graph this we could put this in slope intercept form or we could just pick some points here you just really need two points to graph a line so let's just set let me just set some points over here let's say x and y when X is equal to 0 what does y need to be equal to so when X is equal to 0 we have let me do it over here we have 5 times 0 plus 3 times y is equal to 7 that's just 0 over there so you have 3y is equal to 7 divide both sides by 3 you get Y is equal to 7/3 Y is equal to 7/3 which is the same thing as 2 & 1 thirds if we want to write it as a mixed number now let's set y equal to 0 so if we set Y is equal to 0 so when y is equal to 0 we get 5x plus 3 times 0 is equal to 7 or this part right over here just becomes 0 so you have 5x is equal to 7 divide both sides by 5 and we get x equals 7/5 X is equal to seven fifths which is the same thing as 1 & 2 fifths so let's graph these points then we should be able to graph this line or at least a pretty good approximation of that line so we have the point 0 2 and 1/3 so 0 and then we're going to go up 2 and then 1/3 so that's that point right over there so that's 0 I'll call that 0 7/3 right over there and then we have the point seven fifths 0 or 1 and 2/5 comma 0 so 1 and 2/5 2/5 is a little less than 1/2 so 1 in two fifths come comma zero so our line is going to look something like this I just have to connect the dots I'll try to draw it it's always hard to draw the straight line I'll draw it as a dotted line so it would look something like this normally when you have to solve a system of equation by graphing it so they normally give you a little bit cleaner numbers but we'll we'll try our best and see if we can see where this where these two points these two lines intersect so now let's worry about the second one right over there so we have 3x minus 2y is equal to 8 so I'll do the same thing so 3x minus 2y is equal to 8 we'll just look at the X and the y intercepts so first a y-intercept when X is equal to 0 when X is equal to 0 this whole thing boils down to 3 times 0 minus 2y is equal to 8 that's just 0 so you have negative 2y is equal to 8 divide both sides by negative 2 we get Y is equal to negative 4 so Y is equal to negative 4 so the y-intercept is 0 negative 4 right over here we mark it 0 negative 4 and then let's set y equal to 0 so when y is equal to 0 this term right over here just becomes 0 so we get 3x minus 2 times 0 so that's just 0 so 3x is equal to 8 divide both sides by 3 you get X is equal to 8/3 X is equal to 8/3 and 8/3 is the same thing as 2 and 2/3 so 2 and 2/3 so it puts it right about right about there that's the point 0 8/3 now let me try my best to graph it connect these two dots so let me do my very best do a dotted line there it goes something goes something like that and just eyeballing it it looks like these two it looks like these two lines intersect right over there I'm hoping that this will give us a clean answer and this is the point 2 comma negative 1 so that is the point 2 comma 2 comma negative 1 right the x value here is 2 y value is negative 1 now that's what we got just by kind of eyeballing it and clearly these are hand-drawn graphs not very precise let's verify or let's see if 2 comma negative 1 does satisfy both of these equations if it's an x and y value that satisfies both and lies on both graphs so if you put 2 comma negative 1 in this first equation you get 5 times 2 plus 3 times negative 1 and we're going to see if that is equal to 7 so this is 10 10 plus negative 3 so that's the same thing as 10 minus 3 it does that equal 7 and yeah it does 10 minus 3 is equal to 7 so 2 comma negative 1 is definitely on that on that graph or it definitely satisfies that equation and then let's do it with the other one so if we do 2 comma negative 1 you have 3 times 2 minus 2 times negative 1 and we're testing to see if that is equal to 8 so 3 times 2 3 times 2 is 6 and then 2 times negative 1 is negative 2 but then we're subtracting that so it's 6 plus 2 6 plus 2 is equal to 8 and it definitely does equal 8 so we have the coordinate or we have the X we have the the the point 2 comma negative 1 satisfying both equations so we've solved the system of linear equations by graphing