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# Number of solutions to equations

CCSS Math: 8.EE.C.7a

## Video transcript

Determine the number of solutions for each of these equations, and they give us three equations right over here. And before I deal with these equations in particular, let's just remind ourselves about when we might have one or infinite or no solutions. You're going to have one solution if you can, by solving the equation, come up with something like x is equal to some number. Let's say x is equal to-- if I want to say the abstract-- x is equal to a. Or if we actually were to solve it, we'd get something like x equals 5 or 10 or negative pi-- whatever it might be. But if you could actually solve for a specific x, then you have one solution. So this is one solution, just like that. Now if you go and you try to manipulate these equations in completely legitimate ways, but you end up with something crazy like 3 equals 5, then you have no solutions. And if you just think about it reasonably, all of these equations are about finding an x that satisfies this. And if you were to just keep simplifying it, and you were to get something like 3 equals 5, and you were to ask yourself the question is there any x that can somehow magically make 3 equal 5, no. No x can magically make 3 equal 5, so there's no way that you could make this thing be actually true, no matter which x you pick. So if you get something very strange like this, this means there's no solution. On the other hand, if you get something like 5 equals 5-- and I'm just over using the number 5. It didn't have to be the number 5. It could be 7 or 10 or 113, whatever. And actually let me just not use 5, just to make sure that you don't think it's only for 5. If I just get something, that something is equal to itself, which is just going to be true no matter what x you pick, any x you pick, this would be true for. Well, then you have an infinite solutions. So with that as a little bit of a primer, let's try to tackle these three equations. So over here, let's see. Maybe we could subtract. If we want to get rid of this 2 here on the left hand side, we could subtract 2 from both sides. If we subtract 2 from both sides, we are going to be left with-- on the left hand side we're going to be left with negative 7x. And on the right hand side, you're going to be left with 2x. This is going to cancel minus 9x. 2x minus 9x, If we simplify that, that's negative 7x. You get negative 7x is equal to negative 7x. And you probably see where this is going. This is already true for any x that you pick. Negative 7 times that x is going to be equal to negative 7 times that x. So we already are going into this scenario. But you're like hey, so I don't see 13 equals 13. Well, what if you did something like you divide both sides by negative 7. At this point, what I'm doing is kind of unnecessary. You already understand that negative 7 times some number is always going to be negative 7 times that number. But if we were to do this, we would get x is equal to x, and then we could subtract x from both sides. And then you would get zero equals zero, which is true for any x that you pick. Zero is always going to be equal to zero. So any of these statements are going to be true for any x you pick. So for this equation right over here, we have an infinite number of solutions. Let's think about this one right over here in the middle. So once again, let's try it. I'll do it a little bit different. I'll add this 2x and this negative 9x right over there. So we will get negative 7x plus 3 is equal to negative 7x. So 2x plus 9x is negative 7x plus 2. Well, let's add-- why don't we do that in that green color. Let's do that in that green color. Plus 2, this is 2. Now let's add 7x to both sides. Well if you add 7x to the left hand side, you're just going to be left with a 3 there. And if you add 7x to the right hand side, this is going to go away and you're just going to be left with a 2 there. So all I did is I added 7x. I added 7x to both sides of that equation. And now we've got something nonsensical. I don't care what x you pick, how magical that x might be. There's no way that that x is going to make 3 equal to 2. So in this scenario right over here, we have no solutions. There's no x in the universe that can satisfy this equation. Now let's try this third scenario. So once again, maybe we'll subtract 3 from both sides, just to get rid of this constant term. So we're going to get negative 7x on the left hand side. On the right hand side, we're going to have 2x minus 1. And now we can subtract 2x from both sides. To subtract 2x from both sides, you're going to get-- so subtracting 2x, you're going to get negative 9x is equal to negative 1. Now you can divide both sides by negative 9. And you are left with x is equal to 1/9. So we're in this scenario right over here. We very explicitly were able to find an x, x equals 1/9, that satisfies this equation. So this right over here has exactly one solution.