- Systems of equations with graphing
- Systems of equations with graphing
- Systems of equations with graphing: y=7/5x-5 & y=3/5x-1
- Systems of equations with graphing: 5x+3y=7 & 3x-2y=8
- Systems of equations with graphing: chores
- Systems of equations with graphing: exact & approximate solutions
- Systems of equations with graphing
Sal solves the system of equations 5x + 3y = 7 and 3x - 2y = 8 by graphing. Created by Sal Khan and Monterey Institute for Technology and Education.
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- What if you get a fraction for the y-intercept when putting the equation(s) into slope-intercept form (y=mx+b)?(5 votes)
- There's no problem if you get a fraction for the y-intercept. For example, let's say I have a line whose slope is 2 and y-intercept is 3/5. The slope-intercept form of this line would be: y = 2x + 3/5(6 votes)
- What if the two lines end up not intersecting? Are you still "solving the system using the graphing method"?(5 votes)
- If the lines do not intersect at all, there is no solution. If they intercept only one, there is one solution. If there seems to only be one line, there are infinite (Neverending) solutions.(2 votes)
- what about problems that have something like y=3x?(4 votes)
- In such a situation, the y-intercept, and the x- intercept are both (0,0). You can tell this, because if you plug in 0 for either x or y, or get 0 for the other. EX:
y=3x → 0=3x → 0/3=x=0
In order to graph, you would draw a line extending from the origin, increasing at a rate of 3 units. Consider the video Algebra: graphing lines 1(4 votes)
- So you just graph the slope?(3 votes)
- Sal started with these two equations:
5x + 3y = 7
3x - 2y = 8
He didn't put them into point slope form, but played around with them and got some results. Is this called a certain form, and will this always work?(2 votes)
- How would you solve: y= x + 6
- You will need to get x by itself using distribution. You already have y, (y=x+6) So then you plug the equation for y in place of the y in the second equation.
Then you distribute:
Add like terms:
now replace the x (in 4=x+6) with x=0
check your answer,
Hope this was helpful!(2 votes)
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- at the beginning of the video sal mentioned that you could convert the equation into y=mx+b form, but how do you do this(2 votes)
- So the first equation is 5x+3y=7. To get this to y=mx=b form you have to get y by itself on one side. Which all you do is 5x-5x and you have to do that to the other side. You first move the 5x on the other side which would look something like this:3y=-5x+7. To get y by itself you divide 3y by 3. You then have to do the same to the other side which would look something like this:y=-3/5x+2 1/3.
For the second equation which is 3x-2y=8. You pretty much do the same thing on the other equation. You have to move the 3x to other side so you minus it by -3x which cancels it out. Then you have to do the to other side which would look something like this:-2y=-3x+8. Then you divide the -2y by -2 which would look something like this:y=1 1/2x-4.
The two equations in y=mx+b form would now look something like this y=-3/5x+2 1/3 and y=1 1/2x-4. If I'm wrong then please tell me and I hope this was useful.😁(2 votes)
What about if you had 3 or more variables? How would you figure those out by graphing?(2 votes)
- If you go forward a couple more videos in the Developmental Math 2 section you will start to see videos that deal with 3 variables. It's titled, "Systems of Three Variables".(2 votes)
- At the end of the 2000 WNBA regular season the Houston comets had 22 more victories than loss is the number of victories they had were three less than six times the number of losses how many regular season games did the Houston comets play during the 2000 WNBA season?(2 votes)
Solve the system of linear equations by graphing, and they give us two equations here. 5x plus 3y is equal to 7, and 3x minus 2y is equal to 8. When they say, "Solve the system of linear equations," they're really just saying find an x and a y that satisfies both of these equations. And when they say to do it by graphing, we're essentially going to graph this first equation. Remember, the graph is really just depicting all of the x's and y's that satisfy this first equation, and then we graph the second equation that's depicting all of the x's and y's that satisfy that one. So if we were looking for an x and a y that satisfies both, that point needs to be on both equations or it has to be on both graphs. So it'll be the intersection of the two graphs. So let's try to see if we can do that. So let's focus on this first equation, and I want to graph it. So I have 5x plus 3y is equal to 7. There's a couple of ways we could graph this. We could put this in slope-intercept form, or we could just pick some points here. You just really need two points to graph a line. So let me just set some points over here. Let's say x and y. When x is equal to 0, what does y need to be equal to? So when x is equal to 0, we have-- let me do it over here-- we have 5 times 0, plus 3 times y, is equal to 7. That's just 0 over there. So you have 3y is equal to 7. Divide both sides by 3, you get y is equal to 7/3, which is the same thing as 2 and 1/3 if we want to write it as a mixed number. Now let's set y equal to 0. So if we set y as equal to 0, we get 5x, plus 3 times 0, is equal to 7. Or in this part right over here, it just becomes 0. So we have 5x is equal to 7. Divide both sides by 5, and we get x equals 7/5, which is the same thing as 1 and 2/5. So let's graph both of these points, and then we should be able to graph this line, or at least a pretty good approximation of that line. So we have the point, 0, 2 and 1/3. So that's that point right over there. So I'll call it 0, 7/3 right over there, and then we have the point, 7/5, 0, or 1 and 2/5, 0. So 1 and 2/5. 2/5 is a little less than a half. So 1 and 2/5, 0. So our line is going to look something like this. I just have to connect the dots. It's always hard to draw the straight line. I'll draw it as a dotted line. So it would look something like this. Normally, when you have to solve a system of equations by graphing, they normally give you a little bit cleaner numbers, but we'll try our best and see if we can see where these two lines intersect. So now, let's worry about the second one right over there. So we have 3x minus 2y is equal to 8. So I'll do the same thing. So 3x minus 2y is equal to 8. We'll just look at the x- and the y-intercepts. So first, the y-intercept. When x is equal to 0, this whole thing boils down to 3 times 0 minus 2y is equal to 8. That's just 0. So you have negative 2y is equal to 8. Divide both sides by negative 2, we get y is equal to negative 4. So the y-intercept is 0, negative 4. Right over here, and we mark it 0, negative 4. And then, let's set y equal to 0. So when y is equal to 0, this term right over here just becomes 0. So we get 3x minus 2 times 0, so that's just 0. So 3x is equal to 8. Divide both sides by 3, you get x is equal to 8/3. And 8/3 is the same thing as 2 and 2/3. So it puts it right about there. That's the point, 0, 8/3. Now let me try my best to graph it. Connect these two dots. So let me do my very best. I drew a dotted line there. It goes something like that. And just eyeballing it, it looks like these two lines intersect right over there. I'm hoping that this will give us a clean answer. And this is the point 2, negative 1. So that is the point, 2, negative 1. Right? The x value here is 2. y value is negative 1. Now, that's what we got just by eyeballing it, and clearly these are hand-drawn graphs. Not very precise. Let's verify or let's see if 2, negative 1 does satisfy both of these equations, if it's an x and y value that satisfies both and lies on both the graphs. So if you put 2, negative 1 in this first equation, you get 5 times 2, plus 3 times negative 1, and we're going to see if that is equal to 7. So this is 10 plus negative 3. So that's the same thing as 10 minus 3. Does that equal 7? And yeah, it does. 10 minus 3 is equal to 7. So 2, negative 1 is definitely on that graph or definitely satisfies that equation. And then, let's do it with the other one. So if we do 2, negative 1, you have 3 times 2, minus 2 times negative 1. And we're testing to see if that is equal to 8. So 3 times 2. 3 times 2 is 6. And then 2 times negative 1 is negative 2. But then, we're subtracting that, so it's 6 plus 2. 6 plus 2 is equal to 8. And it definitely does equal 8. So we have the coordinate or we have the point, 2, negative 1 satisfying both equations. So we've solved the system of linear equations by graphing.