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# Forming systems of equations with different numbers of solutions

## Video transcript

Fill in the blanks to form a system of linear equations in the variables x and y with infinitely many solutions. So you're going to have infinitely many solutions if essentially both of these equations are describing the same line. If they're both essentially the same equation, they are the same constraint. And we can graphically imagine that. Let's say that that's our y-axis, and this is our x-axis. And we could even try to graph this right over here. If we were to put this into slope-intercept form, you would have 5y is equal to 2 minus 4x. And then if we divide both sides by 5, you get y is equal to-- and I'll swap these-- negative 4/5 x plus 2/5. So this line up here is going to look something like this. It's y-intercept is at 2/5. And it's going to have a negative 4/5 slope. So it's going to look something like this. So that's what the line looks like. So we're going to get an infinite number of solutions for the system. If the second line, if when we graph it, is essentially the exact same line, it overlaps at every x and y that satisfy the first equation. Now, the easiest way to think about what these blanks should be is, well, how do I fill in the blanks here so it is really just a direct algebraic manipulation of this first equation? And we have a clue here. On the right-hand side of the first equation, we have a 2. On the right-hand side of the second equation, we have a 4. So if we want a direct algebraic manipulation, clearly on the right-hand side, you have multiplied by 2. And if you want this to be the same equation, you have to do that to both sides. So let's multiply both sides by 2. So 4x times 2 is 8x. 5y times 2 is 10y. So this equation and this equation are the exact same constraint. They represent the exact same line. They have an infinite number of solutions. Let's do another example. Which of the following choices of a will result in a system of linear equations with no solutions? So you're going to have no solutions as if you take the same combination of x and y's, but you get to a different number. Or another way of thinking about it is if you plotted this equation and it had the exact same slope, but it had a different y-intercept. So let's think about it both ways. So first, let's try to algebraically manipulate the second equation so that the left-hand side looks exactly the same as the left-hand side up here. And then we just have to make sure that the right-hand side is different. And then you will have no solutions. So let's think about that a little bit. So it looks pretty close. If we multiply the left-hand side by a negative 1, it's going to look just like the left-hand side over here. So let's do that. Let's multiply the left-hand and the right-hand side by negative 1. So it's essentially the same equation. So negative 6x times negative 1 is positive 6x. 7y times negative 1 is minus 7y. And a times negative 1 is negative a. So notice, now on the left-hand side we have the same combinations of x's and y's. We have 6 x's minus 7 y's. If on the right-hand side these two things were to be equal, then we'd have the scenario that we just saw. We would have the exact same line. If negative a was equal to 4, we would have the same line. So let me write that down. If negative a is equal to 4, then we are dealing with the same line/equation/constraint. And like the previous scenario, we would have an infinite number of solutions. On the other hand, if negative a is not equal to 4, then there's no way that there's anything that will satisfy both of them. Here you're saying that whatever your x is and whatever your y is, take 6 of the x's and subtract 7 of the y's, and you get 4. If here you take 6 of the x's and subtract 7 of the y's and you get a different number, then there's never going to be an x and a y that satisfy both of them. So if negative 8 is not equal to 4, then you're going to have no solutions. Or another way of saying this is if a does not equal negative 4, obviously, if a is equal to negative 4, then negative a is going to be equal to 4. So as long as a is not equal to negative 4, you're going to have no solutions. So a can be any number except for negative 4. Now, another way to think about this is to put them both in slope-intercept form. And you'll see that they'll have the same slope. And then you would just want to have them have different y-intercepts. And they're going to be parallel lines that don't overlap. So this top one right over here, if we subtract 6x from both sides, you get negative 7y is equal to negative 6x plus 4. Divide both sides by negative 7. You get y is equal to 6/7 x minus 4/7. So that's this first one right up here written in slope-intercept form. Now, the second one, I'll just start with exactly what they gave us. Let me add 6x to both sides. So this is the second one. So if I add 6x to both sides, I get 7y is equal to 6x plus a. And then if I divide both sides by 7, I get y is equal to 6/7 x plus a over 7. So notice in slope-intercept form, our first equation looks like that. And in terms of a, our second equation looks like that. So they definitely already have the same slope. They have the exact same slope. If a is equal to negative 4, then these are going to be the exact same equations. They're going to have the exact same y-intercept. You're going to have an infinite number of solutions. On the other hand, if a is anything other than negative 4, you're going to have a different y-intercept. And these two things are just going to be parallel.