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Systems of equations with graphing: y=7/5x-5 & y=3/5x-1

Sal graphs the following system of equations and solves it by looking for the intersection point: y=7/5x-5 and y=3/5x-1. Created by Sal Khan.

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  • blobby green style avatar for user chloe cavallaro
    What do you mean when you say "when x is equal zero, y is equal to negative 5," That did not make sense to me. i could be because of the way I explain things when i grph but it did not make sense to me.
    (33 votes)
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  • duskpin sapling style avatar for user Arpitha Bose
    I'm sooo confused.
    How exactly do you plot it?
    (21 votes)
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    • blobby green style avatar for user NatalieB
      you take the y intercept and plot it on the y-axis then use the slope (example:4x or 2/3x) if its 4 then go four up and if its -4 go four down if its a fraction like 3/5 then use rise over run and go up 3 and over 5
      (2 votes)
  • male robot hal style avatar for user korean_default
    what if it is just a whole number, not a fraction
    (7 votes)
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  • ohnoes default style avatar for user manasa20006
    i thought you move 3 to the right and 5 up for the equation's rise over run
    (9 votes)
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  • duskpin ultimate style avatar for user Amberlynn
    So, how would you plot something like
    7x−y=7
    x+2y=6
    ​ I cannot figure out how to plot it.
    (4 votes)
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  • blobby green style avatar for user lilyannadavidson26
    this graphing stuff it soo.........🤫🤫🤫
    (6 votes)
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  • aqualine sapling style avatar for user Peter Prado
    What is the troll he's talking about in ?
    (4 votes)
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  • starky sapling style avatar for user Wilson, Isabel
    I'm really confused how he did it because all i saw him do was move around the dots. Can someone please help me?
    (2 votes)
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    • aqualine ultimate style avatar for user Hannah Alisse
      Yeah, that's pretty much it for this exercise: moving around dots, but the "math" part is the line that's connected to the dots. You're basically "drawing" a line by moving the dots. I'll see if I can explain it a different way to make it easier to understand.

      y = x
      This one's pretty simple to graph. It's just a diagonal line where at any point it's the same distance from the x axis as it is from the y axis. (I really wish I could show you graphs in these answers but I don't think I can. Sadness) All the other equations we'll talk about will look like this one, but changed a little bit.

      Let's try another.
      y = x + 1
      This is just like the other one, but it's shifted up one unit. If x = 1, then y = 1 + 1 or 2. If x = 7 then y = 7 + 1 which is 8.

      y = x + 4
      This one's the same as the other one, but shifted up 4 units.

      Now let's change the slope of the line.
      y = 2x
      Now the line is steeper. If x = 1 then y = 1*2 or 2 If x = 5 then y = 5*2 or 10

      y = ½x
      This one's like the other one, except instead of the line being steeper, it's less steep. Instead of multiplying by 2, we divide by 2. If x = 1 then y = 1/2. If x = 8 then y = 8/2 which equals 4

      Okay, let's do one last one.
      y = 3x + 2
      In this one, we have to shift it up, and make it steeper. If x = 1 then y = 1*3 + 2 which equals 5. If x = 7 then y = 7*3 +2 which is (seven times three is 21, plus 2...) 23

      I don't know if this was helpful at all but this concept is really hard to explain without being able to draw. Please let me know if this helped, or if you have any further questions. :)
      (8 votes)
  • male robot donald style avatar for user junney123
    I sometimes still have a hard time finding the slope. On :36 it says the Y intercept is equal to 5. Does that mean that the -5 is the Y intercept. If so than the number without the variable is always the Y intercept?
    (4 votes)
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  • primosaur ultimate style avatar for user Steve
    Can someone take a look at the Practice problems for this area? There's something not quite right in the explanations in this topic. Here's the problem, they ask to find the intercepting points which makes both equations true. One of the problems for example is :
    5x-10y=20 and 2x+4y=-16

    So all it says is solve for both equations. Well when I do it the way to get them both into y= mx+b form, I ultimately end up with :

    y=5/10x -2 and y=2/-4x -4

    And we end up with the fractions 5/10 so starting at 0, -2 we rise 5 and run 10...yet on the graph we can't plot anything to 10 cause it only goes up to 9. Frustrated and stuck, I go to get the answer, and the solution is using a specific subsititution method? How do you know WHEN to use this subsitition method, or elimination? or just when to put it in y=mx+b and graph it? Shouldn't all 3 methods get you the same answer though??
    (2 votes)
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    • stelly blue style avatar for user Kim Seidel
      You should always reduce your fractions.
      5/10 = 1/2.
      2/(-4) = -1/2

      Your system can be solved by graphing (though not always accurate, it depends on your graph), substitution and elimination. Substitution and elimination will always give you the same result. And, if you used graphing, it should also give you the same result. But, the accuracy of the graphing method is dependent upon how precise your graph is.

      Hope this helps.
      (5 votes)

Video transcript

Just in case we encounter any more trolls who want us to figure out what types of money they have in their pockets, we have devised an exercise for you to practice with. And this is to solve systems of equations visually. So they say right over here, graph this system of equations and solve. And they give us two equations. This first one in blue, y is equal to 7/5x minus 5, and then this one in green, y is equal to 3/5x minus 1. So let's graph each of these, and we'll do it in the corresponding color. So first let's graph this first equation. So the first thing I see is its y-intercept is negative 5. Or another way to think about it, when x is equal to 0, y is going to be negative 5. So let's try this out. So when x is equal to 0, y is going to be equal to negative 5. So that makes sense. And then we see its slope is 7/5. This was conveniently placed in slope-intercept form for us. So it's rise over run. So for every time it moves 5 to the right it's, going to move seven up. So if it moves 1, 2, 3, 4, 5 to the right, it's going to move 7 up. 1, 2, 3, 4, 5, 6, 7. So it'll get right over there. Another way you could have done it is you could have just tested out some values. You could have said, oh, when x is equal to 0, y is equal to negative 5. When x is equal to 5, 7/5 times 5 is 7 minus 5 is 2. So I think we've properly graphed this top one. Let's try this bottom one right over here. So we have when x is equal to 0, y is equal to negative 1. So when x is equal to 0, y is equal to negative 1. And the slope is 3/5. So if we move over 5 to the right, we will move up 3. So we will go right over there, and it looks like they intersect right at that point, right at the point x is equal to 5, y is equal to 2. So we'll type in x is equal to 5, y is equal to 2. And you could even verify by substituting those values into both equations, to show that it does satisfy both constraints. So let's check our answer. And it worked.