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Video transcript

is three comma negative for a solution to the equation 5x plus 2y is equal to seven so there's two ways to think about it one you could just substitute this x and y value into this equation to see if it satisfies it and we'll do that way first and the other way is if you had a graph of this equation you could see if this if this point sits on that graph which would also mean that it is a solution to this equation so let's do it the first way so we have 5x plus 2y is equal to 7 so let's substitute instead of X let us put in 3 let's put in 3 for X so 5 times 3 plus plus 2 times y so Y is negative 4 plus 2 times negative 4 needs to be equal to needs to be equal to 7 I'll put a question mark here because we don't we're not sure yet if it does so 5 times 3 5 times 3 is 15 and then 2 times negative 4 2 times negative 4 is negative 8 so the left-hand side it simplifies to 15 minus 8 and this needs to be equal to 7 and of course 15 minus 8 does equal 7 so this all works out this is a solution so we've answered the question but I also want to show you this way we just did it by substitution if we had the graph of this equation we could also do it graphically so let's let's give ourselves the graph of this equation I'll do that by setting up a table there's multiple ways to graph this you could put it in a slope intercept form and all of the rest but I'll just set up a table of x and y values X and y values and I'll graph it and then given the graph I want to see if this actually sits on it and obviously it will because we've already shown that this works in fact we could try the point 3 negative 4 and that actually is on the graph we could do it on our table but I won't do that just yet I'm just gonna do this to give ourselves a graph so let's say let's try it when X is equal to 0 we have 5 times 0 plus 2 times y is equal to 7 so when X is equal to 0 Y is going to be so you're gonna have 0 plus 2y is equal to 7 Y is going to be equal to 3 point 5 3.5 when X is equal to 1 when X is equal to 1 you have 5 plus 2y is equal to 7 you're going to 5 plus 2y is equal to 7 if you subtract 5 from both sides you get 2y is equal to 2 you get Y is equal to 1 so when X is 1 Y is 1 and when X is well let's try let's try it well that's actually enough for us to graph we could keep doing but we could keep doing more points we could even put the point 3 negative 4 there but let's just try to graph it in this very rough sense right here so let me draw my x-axis and then this right over here is my y-axis and let me draw some points here so let's say that this is y is 1 2 3 4 this is negative 1 negative 2 negative 3 negative 4 I could keep going down in that direction this is 1 let me do it a little bit 1 2 3 4 and I could just keep going on and on in the positive x-direction so let's plot these points at 0 3.5 when x is 0 Y is 1 2 3 point 5 when X is 1 Y is 1 when X is 1 Y is 1 and so if we were to draw this line I'll do it as a dotted line just so that I can make sure I connected the dots I could do a better job than that though so it will look it will look something something like that and so if you if someone gave you this line you said oh well it's 3 negative 4 on this line and let's assume that we drew it really let's assume that we drew it really nicely and this was all to scale let me try one last attempt at it let me do one last attempt at it so it's going to look something it's going to look something like that and if someone asked is 3 negative 4 on it you could visually do it but it's always hard when you're when you actually don't substitute it because you don't know maybe you're a little bit off but if you're here if you look at over here you say when X is equal to 3 what is y well you go down here and it looks like Y is equal to negative four so this is a point three comma 3 comma negative 4 obviously you won't you in general you don't want to just rely on off of inspecting graph maybe this was like three negative three point nine nine nine nine and you just couldn't tell looking at the graph that's why you always want to just substitute and make sure that it really does equal that this equality really does hold true at that point not just looking at the graph but it's important to realize that the graph really is another representation to all of the solutions of this equation