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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 5 y 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 this line up here is going to look something like this its y-intercept is at 2/5 it's a 2/5 and it's going to have a negative 4/5 slope so it's gonna look something like this it's gonna look something something like that 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 can I manipulate this so it is essentially or 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 to direct algebraic manipulation clearly on the right hand side you have x 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 8 X 5 y times 2 is 10 y 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 of a will result in a system of linear equations with no Solutions so you're going to have no solutions is 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 had the exact same slope but it had a different y-intercept so let's do it let's think about it both ways so first I'll think about it let's try to 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 let's think about that a little bit so it looks pretty close if we multiply the left-hand side by negative one it's gonna 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 one 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 6x is -7 Y's we have 6 X's minus 7 wise if on the right hand side 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're dealing with the same same line slash equation slash constraint and like the previous scenario we would have an infinite number of solutions infinite infinite solutions on the other hand if negative a is not equal is not equal to 4 then there's no way that there's anything that will satisfy both of them here you're saying it whatever your X's and whatever your wise take six of the X's subtract 7 otherwise you get 4 if here you take 6 of the XS and subtract 7 otherwise 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 gonna have no no solutions or another way of saying this is if a does not equal negative for obviously if a is equal to negative for the negative a is going to be equal to four so as long as eight is not equal to negative four you're gonna have no no solutions so a can be any number except for except for negative four now another way to think about this is to visually put them both in slope-intercept form and and you'll see that they'll have the same slope and then you 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 this top one right over here if we subtract 6x from both sides you get negative seven y is equal to negative six x plus four divide both sides by negative seven you get Y is equal to 6/7 X minus four sevens 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 over 7x plus a over 7 so notice 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 they definitely already have the same slope they have the exact same slope if a is equal to if a is equal to negative 4 then these are going to be the exact same equations are gonna have the exact same y-intercept you're gonna 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