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### Course: 8th grade>Unit 4

Lesson 2: Systems of equations with graphing

# Systems of equations with graphing

When solving systems of linear equations, one method is to graph both equations on the same coordinate plane. The intersection of the two lines represents a solution that satisfies both equations. Other, more mathematical, methods may also be used. Created by Sal Khan.

## Want to join the conversation?

• how do you have a graph without lines
(15 votes)
• if the number before x is positive than the line looks like this /. If the number is negative, then the line looks like this\
(26 votes)
• Am I the only one studying in the summer
(19 votes)
• no i am too. im scared im gonna lose all my knowledge like in previous years.
(14 votes)
• most voted comment?
(17 votes)
• This: if the Variable (x) is by itself, there is always a 1 in front of it. They have 115 votes
(4 votes)
• How do you graph an equation when all it gives you is y=7
(10 votes)
• When y equals one value
y = 7
It is a ↔️ Horizontal line, it has a Slope of Zero , it includes all x values in its solution set, but only one y…
y = 7 the seven in this case

When x equals one value
x = 2
It's a ↕️ Vertical Line, it has an Undefined Slope, it includes all y values, but only one x…
x = 2 the two in this case.

(≧▽≦) Hope this helps!
(12 votes)
• I'm sooooo confused, I started this section after completing the last section of graphing and I 've never seen any of this before. Can some one tell me what section I need to do do be up to speed. This is the first I'm hearing of "slope intercept"......
(9 votes)
• If you have never heard of slope-intercept form, type "slope-intercept form" at the search bar at the top of the Khan Academy homepage. Slope-intercept form is easy though. The equation for slope-intercept form is:
``y=mx+b``

In this equation, 'm' is the slope and 'b' is the y-intercept. To graph a line from a slope-intercept equation, take the value of the slope and put it over 1. For example, if the slope was 5, the slope would be 5/1. Next graph the y-intercept, take the number that is the y-intercept, and graph that number on the graph. For example, if the y-intercept was 2 graph the number 2 on the y axis of the graph. Next, take the slope, in this case 5/1, and graph it on the coordinate plane. The slope equals:
``y/x``

This is also rise divided by run. Move five places up (the rise), and one place to the left (the run). After the two points have been plotted, use a ruler to connect the two points. Now you have the line! I hope this helped!

Have a Happy New Year!
(14 votes)
• Is anyone else still in school? (Besides me)
(8 votes)
• Nah my school ended a month ago
(8 votes)
• How do you know when you have to graph the line left or right?
(5 votes)
• The video shows graphing positive and negative sloped lines, and finding their intersection…

★Both Positive and Negative lines run from Left to Right.
(all linear equations do, except Vertical)

↗️ Positive Sloped lines move upward as they move Right, increasing x-inputs = increasing y-outputs.

↘️ Negative Sloped equations move downward as the move Right, increasing x-inputs = decreasing y-outputs.

★Slope Intercept Form
y = mx + b←positive slope
y = -mx + b←negative slope

If there is a negative sign infront of the coefficient for x, (the 'm'), then the ↘️ Slope is Negative, and the line will graph from left to right, downward.

★Any two linear equations with different slope values will intersect, if on the same plane, even if they are both positive, or both negative.

(≧▽≦) I hope this helps!
(12 votes)
• how would you graph linear systems with fractions?
(6 votes)
• slope is very easy!
(6 votes)
• When there is no coefficient attached to the variable (x or y), do you just assume it's equal to 1 for the rise/run?
(4 votes)
• Whoops never mind I just saw a comment answering that question, and for those who're wondering, the answer is yes.
(2 votes)

## Video transcript

Let's say I have the equation y is equal to x plus 3. And I want to graph all of the sets, all of the coordinates x comma y that satisfy this equation right there. And we've done this many times before. So we draw our axis, our axes. That's my y-axis. This is my x-axis. And this is already in mx plus b form, or slope-intercept form. The y-intercept here is y is equal to 3, and the slope here is 1. So this line is going to look like this. We intersect at 0 comma 3-- 1, 2, 3. At 0 comma 3. And we have a slope of 1, so every 1 we go to the right, we go up 1. So the line will look something like that. It's a good enough approximation. So the line will look like this. And remember, when I'm drawing a line, every point on this line is a solution to this equation. Or it represents a pair of x and y that satisfy this equation. So maybe when you take x is equal to 5, you go to the line, and you're going to see, gee, when x is equal to 5 on that line, y is equal to 8 is a solution. And it's going to sit on the line. So this represents the solution set to this equation, all of the coordinates that satisfy y is equal to x plus 3. Now let's say we have another equation. Let's say we have an equation y is equal to negative x plus 3. And we want to graph all of the x and y pairs that satisfy this equation. Well, we can do the same thing. This has a y-intercept also at 3, right there. But its slope is negative 1. So it's going to look something like this. Every time you move to the right 1, you're going to move down 1. Or if you move to the right a bunch, you're going to move down that same bunch. So that's what this equation will look like. Every point on this line represents a x and y pair that will satisfy this equation. Now, what if I were to ask you, is there an x and y pair that satisfies both of these equations? Is there a point or coordinate that satisfies both equations? Well, think about it. Everything that satisfies this first equation is on this green line right here, and everything that satisfies this purple equation is on the purple line right there. So what satisfies both? Well, if there's a point that's on both lines, or essentially, a point of intersection of the lines. So in this situation, this point is on both lines. And that's actually the y-intercept. So the point 0, 3 is on both of these lines. So that coordinate pair, or that x, y pair, must satisfy both equations. And you can try it out. When x is 0 here, 0 plus 3 is equal to 3. When x is 0 here, 0 plus 3 is equal to 3. It satisfies both of these equations. So what we just did, in a graphical way, is solve a system of equations. Let me write that down. And all that means is we have several equations. Each of them constrain our x's and y's. So in this case, the first one is y is equal to x plus 3, and then the second one is y is equal to negative x plus 3. This constrained it to a line in the xy plane, this constrained our solution set to another line in the xy plane. And if we want to know the x's and y's that satisfy both of these, it's going to be the intersection of those lines. So one way to solve these systems of equations is to graph both lines, both equations, and then look at their intersection. And that will be the solution to both of these equations. In the next few videos, we're going to see other ways to solve it, that are maybe more mathematical and less graphical. But I really want you to understand the graphical nature of solving systems of equations. Let's do another one. Let's say we have y is equal to 3x minus 6. That's one of our equations. And let's say the other equation is y is equal to negative x plus 6. And just like the last video, let's graph both of these. I'll try to do it as precisely as I can. There you go. Let me draw some. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. And then 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. I should have just copied and pasted some graph paper here, but I think this'll do the job. So let's graph this purple equation here. Y-intercept is negative 6, so we have-- let me do another [? slash-- ?] 1, 2, 3, 4, 5, 6. So that's y is equal to negative 6. And then the slope is 3. So every time you move 1, you go up 3. You moved to the right 1, your run is 1, your rise is 1, 2, 3. That's 3, right? 1, 2, 3. So the equation, the line will look like this. And it looks like I intersect at the point 2 comma 0, which is right. 3 times 2 is 6, minus 6 is 0. So our line will look something like that right there. That's that line there. What about this line? Our y-intercept is plus 6. 1, 2, 3, 4, 5, 6. And our slope is negative 1. So every time we go 1 to the right, we go down 1. And so this will intersect at-- well, when y is equal to 0, x is equal to 6. 1, 2, 3, 4, 5, 6. So right over there. So this line will look like that. The graph, I want to get it as exact as possible. And so we're going to ask ourselves the same question. What is an x, y pair that satisfies both of these equations? Well, you look at it here, it's going to be this point. This point lies on both lines. And let's see if we can figure out what that point is. Just eyeballing the graph here, it looks like we're at 1, 2, 3 comma 1, 2, 3. It looks like this is the same point right there, that this is the point 3 comma 3. I'm doing it just on inspecting my hand-drawn graphs, so maybe it's not the exact-- let's check this answer. Let's see if x is equal to 3, y equals 3 definitely satisfies both these equations. So if we check it into the first equation, you get 3 is equal to 3 times 3, minus 6. This is 9 minus 6, which is indeed 3. So 3 comma 3 satisfies the top equation. And let's see if it satisfies the bottom equation. You get 3 is equal to negative 3 plus 6, and negative 3 plus 6 is indeed 3. So even with our hand-drawn graph, we were able to inspect it and see that, yes, we were able to come up with the point 3 comma 3, and that does satisfy both of these equations. So we were able to solve this system of equations. When we say system of equations, we just mean many equations that have many unknowns. They don't have to be, but they tend to have more than one unknown. And you use each equation as a constraint on your variables, and you try to find the intersection of the equations to find a solution to all of them. In the next few videos, we'll see more algebraic ways of solving these than drawing their two graphs and trying to find their intersection points.