Main content
Algebra 1
Course: Algebra 1 > Unit 6
Lesson 1: Introduction to systems of equations- Systems of equations: trolls, tolls (1 of 2)
- Systems of equations: trolls, tolls (2 of 2)
- Testing a solution to a system of equations
- Solutions of systems of equations
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: exact & approximate solutions
- Systems of equations with graphing
- Setting up a system of equations from context example (pet weights)
- Setting up a system of linear equations example (weight and price)
- Creating systems in context
- Interpreting points in context of graphs of systems
- Interpret points relative to a system
© 2023 Khan AcademyTerms of usePrivacy PolicyCookie Notice
Systems of equations with graphing: exact & approximate solutions
CCSS.Math: ,
Sal solves a system of two linear equations in standard form, and then approximates the solution of a system whose solution isn't clearly visible.
Want to join the conversation?
At2:27, Khan says that when y is zero, x is negative one. I don't get how he got that. Can anyone explain this in another way?(15 votes)
Sal is now plotting points that lie on the line defined by the second equation in the system of equations. the relation he is graphing is 6x - 6y = -6. One way to plot a line is to plot any two points that are on the line, and for an equation in standard from like this one, two easy points to find at the x and y intercepts--the values where x=0 and y=0. To find the x intercept, plug y=0 into 6x-6y=-6 and you get 6x-6(0)=-6 which simplifies to 6x=-6 or x=-1. Similarly, to find the y-intercept, let x=0. you get 6(0)-6y=6 which simplifies to -6y=-6 or y=1. So the points (-1,0) and (0,1) are on our line.
You can see a video explaining the process of finding intercepts
https://www.khanacademy.org/math/algebra/two-var-linear-equations/x-and-y-intercepts/v/x-and-y-intercepts(18 votes)
I didn't understand the concept(7 votes)
Pretty much, the concept is to replace the x and y values with simple values and graph the result. In the case of a fraction like 1/3X, the most simple value would probably be 3 as it would remove the fraction. The X and Y values are not set in stone for substitution.(6 votes)
- How much of a lesson should I get done each day on khan academy?(4 votes)
Normally, I try to complete one of the lesson boxes, as in the videos or exercises in one of the rectangles in the unit. It could just be a quiz that day or a unit test. Either that or I give myself an hour to do as much as possible.(4 votes)
So,like, I don't get this. For example, when Sal says that "When x is equal to zero, y would be equal to negative three" what does that mean?(0 votes)
As Sal states - He is picking different values of X and then calculating Y using one of the equations. In the one you referenced, Sal is using the first equation: -x-3y=9. If you use x=0, the equation becomes: -0-3y=9, then solve for Y.
-3y=9
y=-3
This creates one point for graphing the first line. The point is (0, -3). Sal repeats this process using other values of X to find 2 points for each line.
Hope this helps.(12 votes)
I am so confused. Next time, could you explain slower?(1 vote)- 1) You can slow down the video yourself. Click on the gear symbol in the lower right of the video window and adjust the video speed.
2) Use the pause button as soon as you start to get confused. Try reviewing the transcript to see if that helps eliminate your confusion.
3) You can watch the video as many times as you need to.(9 votes)
So how do you plot the dot when, for example, x=2/3?(4 votes)- Fractions sit in between the integers on a number line.
For x=2/3, it is located between 0 and 1. Split the space between 0 and 1 into 3 equal sections. 2/3 is 2 of those sections from 0.
Search for the lesson on "fractions on a number line" for more details.(2 votes)
So, how would you plot something like
7x−y=7
x+2y=6
I cannot figure out how to plot it.(3 votes)- You have a couple of options:
1) You can convert each equation to slope-intercept form, then graph using the y-intercept and the slope.
2) You can calculate 2 points for each line. Once you have 2 points for the line, you can draw the line. To find a point, pick a value for X or Y and put it into the equation. Then, calculate the other variable. For example: if y=0
7x-0=7
7x=7
x=1
You now have the point (1,0) that can be graphed.
Hope this helps.(4 votes)
I don't understand on how he found not the y-intercept, but the other part on graphing.(4 votes)
Okay...I'm thoroughly confused...instead of taking each equation, and making the X =0 ...then the Y=0... can't we just re-arrange the equation to make it in y=mx+b format? then graph them? What's the purpose or need to set x=0 and then y to 0?(2 votes)- There are multiple ways to graph an equation.
-- The video is using the intercepts method -- you find the X and Y intercepts and graph those 2 points, then draw the line.
-- You want to use the slope-intercept form of the equation to graph using the y-intercept and the slope.
-- You could also find 2 random points on the line by picking values for either X or Y and solving for the other variable.
All these methods are acceptable. So, if you prefer to graph using the slope intercept, do it. The only time this wouldn't be acceptable is if your teacher or a particular problem told you to use a different method.(3 votes)
- so if i had
-20x+12y= -24
5x-3y=6
what do i do, because i can't make some of them 0 and get an answer(2 votes)
This is an interesting case of a system of linear equations, because it doesn't result in one unique solution. Using the method from the video above, we can attempt to solve this system graphically. In the first equation:
when x=0:
12y=-24
y=-2 --> (0,-2) is the y-intercept
when y=0:
-20x = -24
x = 1.2 --> (1.2, 0) is the x-intercept
Connecting these dots and extending them would create a line that has the solutions for this equation. Now we just need to find where the line of solutions for the second equation intersects with the first equation. For the second equation:
when x=0:
-3y=6
y=-2 --> (0, -2) is the y-intercept
when y=0:
5x=6
x=6/5=1.2 --> (1.2, 0) is the x-intercept
Notice something odd? When you connect the dots for both of these equations, you get two equal lines laying on top of each other. This means that every point along that line is a solution to this problem. Therefore, there will be an infinite number of solutions. To answer the question we could write the equation of this solution line in slope-intercept form to represent all the values of x and y that would make this system true.
Taking a look at the line created before, b= -2 (y-intercept). To find slope, (use the points found from before) we can use m = (y2-y1)/(x2-x1) = (0- -2)/(1.2-0) = 5/3. Therefore, in slope-intercept form the line that contains all of the solutions to this problem is:
y=mx+b
y= (5/3)x - 2(3 votes)
2:27
Video transcript
- The following two equations
form a linear system. This is one equation; it has X and Y so it's gonna define a line. And then I have another
equation that involves X and Y, so it's gonna define another line. It says: "Graph the system of equations "and find its solution." So we're gonna try to find it visually. So let's graph this first one. To graph this line, I have
the little graphing tool here. Notice if I can figure out two points, I can move those points around and it's going to define our line for us. I'm gonna pick two X values and figure out the corresponding Y values
and then graph the line. So let's see how I could do this. So let's see; an easy one is what happens when X is equal to zero? Well if X is equal to zero,
everything I just shaded goes away and we're left
with -3y is equal to nine. So -3y equals nine. Y
would be negative three. So when X is equal to zero,
Y would be negative three. So let me graph that. When X is equal to zero, X is zero, Y is negative three. Now another easy point
actually instead of trying another X value, let's think
about when Y is equal to zero 'cause these equations
are in a standard form so it's easy to just test. Well what are the X and Y intercepts? So when Y is equal to zero, this term goes away,
and you have negative X is equal to nine, or X would
be equal to negative nine. So when Y is zero, X is negative nine. So when Y is zero, X is negative nine, or when X is negative nine, Y is zero. So I've just plotted this first equation. So now let's do the second
one. We'll do the same thing. What happens when X is equal to zero? When X is equal to zero, so this is going to be
our Y intercept now. When X is equal to zero, -6y
is equal to negative six. Well Y would have to be equal to one. So when X is zero, Y is equal to one. So when X is zero, Y is equal to one. Get one more point here. When Y is zero, when this term is zero, Y being zero would make
this entire term zero, then 6x is equal to negative six or X is equal to negative one. So when Y is zero, X is negative one or when X is negative one, Y is zero. When X is negative one, Y is zero. And so just like that,
I've plotted the two lines. And the solution to the
system are the X and Y values that satisfy both equations;
and if they satisfy both equations, that means
they sit on both lines. And so in order to be on both lines, they're going to be at
the point of intersection. And I see this point of
intersection right over here, it looks pretty clear that
this is the point X is equal to negative three and Y
is equal to negative two. So it's the point negative
three comma negative two. So let me write that down. Negative three comma negative two. And then I could check
my answer; got it right. Let's do another. Let's
do another one of these. Maybe of a different type. So over here it says: "A
system of two linear equations "is graphed below. "Approximate the solution of the system." Alright so here I just have
to just look at this carefully and think about where this point is. So let's think about first its X value. So its X value, it's about right there
in terms of its X value. It looks like, so this is negative one. This is negative two, so negative 1.5 is gonna be right over here. It's a little bit to the
left of negative 1.5, so it's even more negative,
I would say negative 1.6. And I'm approximating it, negative 1.6. Hopefully it has a little leeway in how it checks the answer. What about the Y value? So if I look at the Y value here, it looks like it's a little
less than one and a half. One and a half would be
halfway between one and two. It looks like it's a little
less than halfway between one and two, so I'd give it 1.4, positive 1.4. And let's check the answer,
see how we're doing. Yep, we got it right. Let's actually just do
one more for good measure. So this is another system. They've just written the equations in more of our slope intercept form. So let's see, Y is equal to
negative seven, X plus three. When X is equal to zero,
we have our Y intercept. Y is equal to three. So when X is equal to
zero, Y is equal to three. And then we see that our
slope is negative seven. When you increase X by one,
you decrease Y by seven. So when you increase X
by one, you decrease Y by one, two, three, four, five, six, and seven. When X goes from zero to one, Y went from three to negative four, it went down by seven,
so that's that first one. Now the second one: our Y intercept. When X is equal to zero,
Y is negative three, so let me graph that. When X is zero, Y is
equal to negative three. And then its slope is negative one. When X increases by
one, Y decreases by one. So the slope here is negative one. So when X increases by
one, Y decreases by one. And there you have it. You have
your point of intersection. You have the X-Y pair that
satisfies both equations. That is the point of intersection. It's gonna sit on both lines which is why it's the
point of intersection. And that's the point X equals one, Y is equal to negative four. So you have X equals one and Y is equal to negative four. And I can check my answer
and we got it right.