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Number of solutions to a system of equations algebraically

Sal solves several examples where he reasons about the number of solutions of systems of equations using algebraic reasoning.

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  • blobby green style avatar for user aobi0885
    If both sides equal zero is that also a zero solution equation?
    (10 votes)
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    • stelly blue style avatar for user Kim Seidel
      If you are solving a system of equations and get to the point where you have: 0 = 0; this is an Identity (a situation that is always true). In a system of equations it means that the 2 equations are actually the same line (visualize one line sitting right on top of the other). So, the system has a solution set of all the possible points on that line.

      Contrast that with a solution like: 0 = 2. This is a contradiction (it is always false). This indicates that the 2 equations have no points in common. So, they would be parallel lines.
      (21 votes)
  • hopper cool style avatar for user connerking2
    y=−2x−4
    y=3x+3
    ​How many solutions does this system have? Pls help!
    (4 votes)
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    • aqualine ultimate style avatar for user 管墨涵
      This system of linear equations have only one solution.
      That is because this system of equations is written in slope-intercept form:
      y=mx+b,
      In which m is the slope and b is the y-intercept.
      So in the first equation, -2 is the slope.
      And in the second equation, 3 is the slope.
      And it becomes very obvious -- two lines with a DIFFERENT slope will always intersect at some point!
      So this system has only one solution.
      P.S. sorry for answering 9 months later!
      (8 votes)
  • male robot hal style avatar for user Pᴀʀᴀsᴇʟᴇɴᴇ Tᴀᴏ
    I'm purposefully trying to make a difficult question.
    What is the solution to this system of linear equations?
    y=mx-1
    y=(m-1)x-2
    Where m = Graham's Number
    I figure it's something VERY close to (0,-2) because the two lines will almost seem to be on top of one another in an almost vertical line.
    To be a little more accurate, I know the solution will have an x value slightly more negative than 0 (-0.000...1) and a y value slightly more negative than -2 (-2.000...1).
    How do I show my solution algebraically?
    Another question: Picking an arbitrary location on the y-axis, how would I show the distance from those two lines?
    (3 votes)
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    • blobby green style avatar for user Mariette
      I got a different answer to your first question.
      Through substitution, x = 3.
      So the solution to the system of equations y = mx - 1 and y = (m - 1)x - 2 is the ordered pair (3, y).
      To find y, we simplify again and see that:
      y = 3(Graham's Number) - 5
      So the lines will intersect at (3, y) where y is an extremely big number.
      (1 vote)
  • mr pink orange style avatar for user Mohammad Altamimi
    I have difficulty understanding this kind of equations:
    ​y=4x−7
    ​y=x−3
    even when I get a hint, the hint is talking about another equation!
    (5 votes)
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    • aqualine ultimate style avatar for user sugarkin246
      You have probably already figured this out by now, but you can use the second equation to solve for the first one :D Since the second equation says that y=x-3, you can substitute "x-3" into the first equation so that there's only one variable (x). Then you can solve the rest of the equation, I believe.
      (1 vote)
  • aqualine sapling style avatar for user Ybar.jasl25
    How did you get -36 on the first problem
    (3 votes)
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  • blobby green style avatar for user Arbaaz Ibrahim
    On the last question, Sal wrote the co-ordinates 4/5, -1 but, how does he know that it' exactly one solution?
    (3 votes)
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    • starky ultimate style avatar for user Gabriel Barberini
      If you mean the one before the last exercise, try to think this way:

      There were 2 equations:

      5x-2y=6
      5x+3y=1

      Those 2 equations forms a system of equations right?

      To validate if the system has indeed only one solution, all of the lines within the system must have a different y-intercept.

      Since there is 2 equations in the system, we can say that there are 2 lines as well.

      To check their y-intercept you can assume x is zero for all of them.

      5(0)-2y=6 -> -2y=6 -> y=-3
      5(0)+3y=1 -> 3y=1 -> y=1/3

      Since their y-intercept is different, there is only 1 solution to the system.

      But, there are many ways to solve something like this, what Saul did was following the exercise clue: "Albus takes several correct steps that lead to the equation 5y=-5", which means that Albus joined both lines "behaviors" (equations) and ended up with a single equation (behavior).

      And if you observe carefully, you'll notice that what he did was a subtraction.

      5x-2y=6
      -
      5x+3y=1
      =
      -5y=5
      -5y(-1)=5(-1)
      5y=-5

      And that leads us to a different way of knowing if a system equation has one or more solutions, by solving them instead of analyzing its behaviors.

      If you join the behaviors (by adding, subtracting or isolating a variable and then merging the equations) and get a result different from 0=0 it means that the equation has only one possible solution (which will be either an 'y' or a 'x', that means where the lines encounter each other, in the case of the exercise Albus gave you the 'y').
      (3 votes)
  • male robot hal style avatar for user josh
    At , even if the 0 had an x, like 0x = -20, it still wouldn't work, right? Because everything times 0 = 0, not -20.
    (3 votes)
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  • blobby green style avatar for user Amer Naseem
    5x - 9y = 16
    5x - 9y = 36
    Both 16 and 36 are equal to 5x - 9y. This means 16 = 36, but they aren't equal. I don't understand.
    (2 votes)
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  • leafers tree style avatar for user arnavnegi14
    Sooo, this is my logic:
    after solving for y, I get an equation, say K.x +C(where K is coefficeint of x and C is a constant). Then, the equations have-
    1. No solutions if K is same but C is different in both equation.
    2. infinite solutions if K and C are equal for both equations.
    3. and, one solution if K is different.
    Is this right?
    (2 votes)
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    • leaf grey style avatar for user Ben
      Yes, that is correct.
      K in this instance is called the 'slope' of a line. It's the number of steps that 'y' will grow at every step of 'x'. C is the y-intercept.
      When both equations are equal, you'll get infinite intersections, since the two lines overlap.
      When both equations have the same slope, but not the same y-intercept, they'll be parallel to each other and no intersections means no solutions.
      When both equations have different slopes than regardless of the y-intercept they'll intersect for certain, therefore it has exactly one solution.
      (4 votes)
  • starky ultimate style avatar for user Hammad Sheikh
    How would you know if a system has exactly one solution, and i didn't get how in (4/5, 1) represents exactly one solution
    (3 votes)
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Video transcript

how many solutions does the following  system of linear equations have and I   have my system right over here there's  a couple of ways to think about it one   way is to think about them graphically  and think about well are they the same   line in which case they would have  an infinite number of solutions are   they parallel in which case they never  intersect you'd have no solutions or do   they intersect exactly one place in case  you would have one solution but instead   we're going to do this algebraically  so let's try to actually just solve   the system and see what we get so the  first equation I'm going to leave that   unchanged 5x minus 9y is equal to  is equal to is 16 now this second   equation right over here let's say I  want to cancel out the X terms so let me   multiply the second equation by negative  one so I have a negative 5x that I can   cancel out with the 5x so if I multiply  the second equation by negative one I'm   going to have negative 5x plus 9y plus  9 Y is equal to negative 36 negative 36   now what I'm going to do is I'm going  to add the left side of the equations   and the right sides of the equations to  get a new equation so 5x minus 5x well   that's going to be 0 negative 9y plus 9y  well that's going to be 0 again I don't   even have to write it it's just going  to be 0 on the left-hand side and on   the right-hand side I'm going to have  16 minus 36 so negative 20 so now I'm   left with the somewhat bizarre-looking  equation that says that 0 is equal to   negative 20 now one way you might say  well how does this make any sense and   the way to think about it is well are  there any X Y values 4 which is 0 is   going to be equal to negative 20 well no  0 is never going to be equal to negative   20 and so it doesn't matter what X Y  values you can never find an X well X or   XY pair that's going to make 0 equal to  negative 20 in fact the X's and Y's have   disappeared from this equation there's  no way that this is going to be true so   we have no we have no solutions now if  you were to plot these if you were to   plot each of these lines you would see  that they are parallel lines and that's   why they have the same slope different  y-intercepts and that's why we have no   solutions they don't intersect let's do  another one of these this is this is fun   all right how many solutions does the  following system of linear equations   have so let's do the same thing I'm  going to keep the first equation the   same negative 6x plus 4y is equal to 2  and the second equation let me just see   if I can cancel out the X term so if I  have a negative 6x if I multiply this by   2 I'm going to have a positive 6x so I  can let's see I'm going to multiply this   whole equation both sides of it by 2 so  I'm going to have 6x 3 x times 2 is 6x   negative 2y times 2 is negative 4y and  that is going to be equal to negative   2 now let's do the same thing let's  add the left sides and let's add the   right sides so negative 6x plus 6x well  that's going to be 0 4y minus 4y that's   0 we just have a 0 on the left-hand  side now on the right-hand side we   have 2 plus negative 2 well that's 0 so  this is a little bit different it still   looks a little bit bizarre 0 equals  0 last time we had 0 is equal to we   add what 0 is equal to negative 20 now  we have 0 equals 0 so one way to think   about it is even though the X's and  Y's are no longer in this equation   okay well what XY pairs is it going to  be true for is that it's going to make   it true that 0 is equal to 0 well this  is going to be true no matter what x and   y are in fact x and y are not involved  in this equation anymore 0 is always   going to be equal to 0 so this is going  to have an infinitely this is going to   have infinitely many solutions here and  that's because these are the same lines   they just look a little bit different  algebraically but if you scale one of   them in the right way in fact if you  just multiply both sides of this one   the second one by negative 2 you're  going to get the top one and so they   actually represent the exact same  lines you have an infinitely many   solutions all right when trying to  find the solution to the following   system of linear equations Yvonne takes  several correct steps that lead to the   equation negative 5 is equal to 20 how  many solutions is the system of linear   equations have I don't even have to  look at the system right over here   the fact that she got the statement  that can never be true negative 5 is   never going to be equal to 20 tells  us that she has no those solutions   and once again if you were to plot  these graphically you would see that   these are parallel lines that's why they  have no solutions they never intersect   there's no XY pair that satisfies both  of these constraints let's do a let's   do a couple more of these when trying  to find the solution to the following   system of linear equations Alba's take  several correct steps that lead to the   equation 5y is equal to negative 5 and  say how many solutions does this system   of linear equations have well 5y equals  negative 5 we could divide both sides   by 5 and we get Y is equal to negative  1 and then if you substitute back in Y   is equal to negative 1 if you did it  in this first equation if Y is equal   to negative 1 all of this becomes  positive 2 you can subtract 2 from   both sides and you get 5x is equal to  4 or you'd get what X is equal to 4/5   or if you put negative 1 over here you  would get 5x minus 3 is equal to 1 you   could add 3 to both sides and you get  5x equals 4 again X is equal to 4/5 so   you have exactly one solution you  would have X is equal to 4/5 y is   equal to negative 1 let's do one more  when trying to find the solution of the   following system of linear equations  Levon take several correct steps that   lead to the equation 0 equals 0 so once  again I don't even need to look at this   over here zero equals zero is always  going to be true so this is going to   have an infinitely this is going  to be an infinitely many solutions