- 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
To solve a system of linear equations by graphing, you need to graph each equation separately. Find two points on each line and connect them. The point where the two lines intersect is the solution to the system of equations. Created by Sal Khan and Monterey Institute for Technology and Education.
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- 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?(31 votes)
- What if the two lines end up not intersecting? Are you still "solving the system using the graphing method"?(6 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.(3 votes)
- 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 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)
- I'm still confused...Why did it transform into a fraction?(3 votes)
- Sal is solving these by finding the intercepts, intercepts always have a 0 somewhere. x intercepts have y=0 (x,0) where x is a number, and y intercepts have x=0 (0,y). Doing this, he ends up with 3y=7. To isolate the y, you have to divide both sides by 3 which ends up as a fraction y=7/3. Same thing happens with x intercept.(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)
- 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?(3 votes)
- guys, donate to khan Academy so they can keep coming with super helpful VIDEOS :)
link to donate is below
- wow im failing thanks khan academy(0 votes)
- Talk to your teacher and find out why you are failing.
Do you know your basic math facts? I mean KNOW them instantly, everything from 0+0=0 all the way through 144÷12=12? Having instant recall of all the basic math facts is critical to being able to quickly and accurately do higher math.
Is there a foundation topic you didn't master? Perhaps you aren't comfortable with fractions or decimals? Going back and working over topics that you didn't "get" the first time around makes a huge difference. If there's a topic you really don't want to review - "I hated that the last time I did it! I don't want to do it again!" - that's a good clue that you need to review it.
Talk to your teacher. Work with a tutor. And be patient with yourself.(8 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.