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Testing a solution to a system of equations

Sal checks whether (-1,7) is a solution of the system: x+2y=13 and 3x-y=-11. Created by Sal Khan and Monterey Institute for Technology and Education.

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  • winston baby style avatar for user yanicknorman
    What is system?
    (2 votes)
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  • blobby green style avatar for user jeremy jones
    im stupid i dont get this
    (8 votes)
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    • male robot hal style avatar for user HRSH.
      Bro chill. A solution of an equation is when both sides (i.e., LHS and RHS) become equal. What do you need to do to make both sides equal? Well, you need to find some values for X and Y so that they become equal when you plug X values wherever X and Y are.

      Here, some of the solutions are given, but we need to check after plugging them in it makes both sides of the equation equal. (like 1 = 1 , 2 = 2, BUT if you get 1 = 2, or 3 = 4 it is clear that it is false and hence the values of X or Y or both are wrong and hence, not the solution[s] )
      (7 votes)
  • leaf green style avatar for user ashantiharrisalgebra
    would this work for a quadratic equation?
    (1 vote)
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  • marcimus pink style avatar for user joseline.ramirez
    can u make an example more easier
    (4 votes)
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    • stelly blue style avatar for user Kim Seidel
      The example in the video is about as simple as it gets. Neither equation has fractions or decimals.

      The video is show you how to determine if an ordered pair (a point) is a solution to a system of equation. Sal has one point that he is testing to see if it is a solution to the system. In order for this to be true, the point must work in both equations (i.e., the 2 sides of each equation come out equal). He does the test by substituting the values from the ordered pair into each equation and simplifying.
      -- The point works in the 1st equation.
      -- The point did not work in the 2nd equation.
      This tells us the point in on the line created by the first equation, but it is not a point on the line created by the 2nd equation. Remember, to be solution to the system, the point must work for both equations. Since it didn't, the point is not a solution to the system.

      Hope this helps.
      (3 votes)
  • duskpin seedling style avatar for user mila stevenson
    ok, so how do you solve an equation that only has y = blah blah blah, and also y = blah blah blah, that is the only thing i struggle with, and its been making me fail some homework.
    (2 votes)
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  • leafers tree style avatar for user Emii14
    What would be the right coordinates to satisfy both equations??
    (3 votes)
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  • duskpin seedling style avatar for user Sara Velasco
    I have perfectly parallel lines, so is there a solution? I can't figure out this problem.
    (2 votes)
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  • blobby green style avatar for user Leslie Ganzwa
    Does a single linear equation with two or more unknowns always have infinitely many solutions
    (2 votes)
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    • leafers ultimate style avatar for user Alex
      Yes, that's right. You could choose whatever values you like for all but one of the variables, and then final variable can always be made to fit.
      For example, if you had the equation ax + by + cz = k, then whatever you pick for y and z, you can solve for x to get x = (k - by - cz)/a, and the equation will be satisfied.
      (2 votes)
  • mr pants teal style avatar for user sophia gonzalez
    i dont understand math im confused
    (2 votes)
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  • male robot hal style avatar for user mwp07
    is it just me or am i just really dumb? nothing makes sense
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      By now you should be familiar with the concept of testing solutions to equations by using substitution. If you are asked if a point is a solution to an equation, we replace the variables with the given values and see if the 2 sides of the equation are equal (so is a solution), or not equal (so not a solution).

      A solution to a system of equations means the point must work in both equations in the system. So, we test the point in both equations. It must be a solution for both to be a solution to the system.

      Hope this helps.
      (2 votes)

Video transcript

Is negative 1 comma 7 a solution for the system of linear equations below? And they give us the first equation is x plus 2y is equal to 13. Second equation is 3x minus y is equal to negative 11. In order for negative 1 comma 7 to be a solution for the system, it needs to satisfy both equations. Or another way of thinking about it, x equals 7, and y-- sorry, x is equal to negative 1. This is the x coordinate. X equals negative 1, and y is equal to 7, need to satisfy both of these equations in order for it to be a Solution. So let's try it out. Let's try it out with the first equation. So we have x plus 2y is equal to 13. So if we're thinking about that, we're testing to see if when x is equal to negative 1, and y is equal to 7, will x plus 2y equals 13? So we have negative 1 plus 2 times 7-- y should be 7-- this needs to be equal to 13. And I'll put a question mark there because we don't know whether it does. So this is the same thing as negative 1 plus 2 times 7 plus 14. That does, indeed, equal 13. Negative 1 plus 14, this is 13. So 13 does definitely equal 13. So this point it does, at least, satisfy this first equation. This point does sit on the graph of this first equation, or on the line of this first equation. Now let's look at the second equation. I'll do that one in blue. We have 3 times negative 1 minus y, so minus 7, needs to be equal to negative 11. I'll put a question mark here because we don't know whether it's true or not. So let's see, we have 3 times negative 1 is negative 3. And then we have minus 7 needs to be equal to negative 11-- I put the question mark there. Negative 3 minus 7, that's negative 10. So we get negative 10 equaling negative 11. No, negative 10 does not equal a negative 11. So x equaling negative 1, and y equaling 7 does not satisfy the second equation. So it does not sit on its graph. So this over here is not a solution for the system. So the answer is no. It satisfies the first equation, but it doesn't satisfy the second. In order to be a solution for the system, it has to satisfy both equations.