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Studying for a test? Prepare with these 14 lessons on Linear equations, functions, & graphs.
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Is 3 comma negative 4 a solution to the equation 5x plus 2y is equal to 7? 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-- and then we'll do that way first-- and the other way is if you had a graph of this equation, you could see 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 for x. So 5 times 3 plus 2 times y-- so y is negative 4-- plus 2 times negative 4 needs to be equal to 7. I'll put a question mark here, because we're not sure yet if it does. So 5 times 3 is 15, and then 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 give ourselves the graph of this equation, and 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. 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 going to do this to give ourselves a graph. So 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 going to have 0 plus 2y is equal to 7. y is going to be equal to 3.5. When x is equal to 1, you have 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-- well that's actually enough for us to graph. 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. I have 0, 3.5. When x is 0, y is 1, 2, 3.5. 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 connect the dots. I can do a better job than that though. So it will look something like that. And so if someone gave you this line, you'd say oh, well it's 3, negative 4 on this line, and let's assume that we drew it really nicely and this was all to scale. Let me try one last attempt at it. So 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 actually don't substitute it, because you don't know. Maybe you're a little bit off. But if you look at it 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 4. So this is a point 3 comma negative 4. Obviously, in general, you don't want to just rely off of inspecting graphs. Maybe this was 3, negative 3.9999, and you just couldn't tell looking at the graph. That's why you always want to just the 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.