Given a 2-variable equation and the x or y values of a solution, Sal finds the value of the other variable in the solution.
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- In the first question, when x = -5 and you have to find the value of Y, how come he decided to subtract 2y on either side? And why in the next one did he add 3x?(14 votes)
- You could have done it either way (subtract 2y or subtract 7y/ add 3x or subtract 5x) but by moving the smallest number (2y and -3x), you will keep the coefficient of the variable positive so that you do not have to divide by a negative number.(21 votes)
- So, if these are linear equations, what would be an example of a non-linear equation?
Would non-linear equations be calculus?(14 votes)
- No, the two functions that are not linear in Algebra I are Quadratic equations which form parabolas and Exponential equations which will approach a horizontal asymptote and increase quickly as you move either to the right (for a exponential growth and to the left for an exponential decay. You will learn other equations such as absolute value, cubic, square root, etc. before you reach calculus.(10 votes)
- this stuff does not make sents to me i only understand regual math(10 votes)
- At2:00. why is the subtraction of a negative still equal -40? Isn't it a negative minus a negative makes the negative positive for the second number or am I wrong?(4 votes)
- If the problem was:
- ( - 40), then you are correct. It becomes
But, the problem in the video is:
- 15 - 25.
Visualize this on a number line, or draw a number line.
You got to -15, then you need to take away 25. This moves you further left on the number line and you end up on -40.
Hope this helps.(5 votes)
- how do you solve a question like this 4x+-6+10(1 vote)
- Since this is an expression, the only thing you can do is simplify. If it were 4x + (-6) + 10 = 24, you would add like terms, which are -6 and 10. Your equation would become 4x + 4 = 24. Subtract 4 from both sides, becoming 4x = 20. Now, divide both sides by 4, and you get x = 5. Your answer is 5.
Hope this helps.(6 votes)
- when the equation was simplified to 5y=8x why did it become y= 8/5x?(2 votes)
- Can't you just cancel out -25 with +25 and then add +25 to +15 or do you have to subtract from left to right?(4 votes)
- Yes, you could add 25 to both sides instead of subtracting 15. If you add the 25, then you would subtract the 7y from both sides. If you do the math, you will be the same results for "y" as Sal got in the video.
The properties of equality are very flexible. They allow us to move items across the equals symbol by doing the same operation to both sides of the equation. As long as we do that, we have an equivalent equation.
Hope this helps.(2 votes)
- i got the question to find x (y was -2) in the equation y+2=−3(x−4), and one of the steps is to go from 0=-3(x-4) to 0=x-4. can anyone tell me where the -3 went?(4 votes)
- At2:03How did he do that i really don't get the asnwer or the plot of this problem? I'm just confused.(3 votes)
- The trick is to keep the y-values balanced on both sides of the equation seeing as there are two of those but khan academy got a whole other unit dedicated solely to this topic; Solving equations & inequalities under Algebra 1. You should understand this better if you got a strong foundation there.(3 votes)
- 2:36If y=-8 when x=-5 then shouldn't x automatically be 5 when y=8?(0 votes)
- No, you can't automatically assume something like that. The reason both (-5,-8) and (5,8) are solutions to this equation is because they show up on the equation's line. Sal later says the (simplified) equation for the question is y=8x/5. To make this answer more clear, another equation, such as y=x-3, when x=-5, y is equal to -8, but when x equals 5, y equals 2. Hope this helps(11 votes)
- [Voiceover] So this is an example from the Khan Academy exercise, graphing solutions to two variable linear equations. And they tell us to complete the table so each row represents a solution of the following equation. And they give us the equation, and then they want us to figure out, what does y equal when x is equal to negative five? And what does x equal when y is equal to eight. And to figure this out, I've actually copied and pasted this part of the problem onto my scratchpad, so let me get that out. And so this is the exact same problem, there's a couple of ways that we could try to tackle it. One way, is you could try to simplify this more, get all your xs on one side and all your ys on the other side. Or we could just literally substitute when x equals negative five, what must y equal? Actually, let me do it the second way, first. So if we take this equation, and we substitute x with negative five, what do we get? We get negative three times, well, we're gonna say x is negative five, times negative five, plus seven y is equal to five times, x is once again, it's gonna be negative five, x is negative five, five times negative five, plus two y. See, negative three times negative five is positive 15, plus seven y, is equal to negative 25 plus two y. And now, to solve for y, let's see, I could subtract two y from both sides, so that I get rid of the two y here on the right. So let me subtract two y, subtract two y from both sides. And then if I want all my constants on the right hand side, I can subtract 15 from both sides. So let me subtract 15 from both sides. And I'm going to be left with 15 minus 15, that's zero, that's the whole point of subtracting 15 from both sides, so I get rid of this 15 here. Seven y minus two y. Seven of something minus two of that same something is gonna be five of that something. It's gonna be equal to five y, is equal to negative 25 minus 15. Well, that's gonna be negative 40. And then two y minus two y, well, that's just gonna be zero. That was the whole point of subtracting two y from both sides. So you have five times y is equal to negative 40. Or, if we divide both sides by five, we divide both sides by five, we would get y is equal to negative eight. So when x is equal to negative five, y is equal to negative eight. Y is equal to negative eight. And actually we can fill that in. So this y is going to be equal to negative eight. And now we gotta figure this out. What does x equal when y is positive eight? Well, we can go back to our scratchpad here. And I'll take the same equation, but let's make y equal to positive eight. So you have negative three x plus seven, now y is going to be eight, y is eight, seven times eight is equal to five times x, plus two times, once again, y is eight, two times eight. So we get negative three x plus 56, that's 56, is equal to five x plus 16. Now, if we wanna get all of our constants on one side, and of all of our x terms on the other side, well, what could we do? Let's see, we could add three x to both sides. That would get rid of all the xs on this side, and put 'em all on this side. So we're gonna add three x to both sides. And, let's see, if we want to get all the constants on the left hand side, we'd wanna get rid of the 16, so we could subtract 16 from the right hand side, if we do it from the right, we're gonna have to do it from the left as well. And we're gonna be left with, these cancel out, 56 minus 16 is positive 40. And then, let's see, 16 minus 16 is zero. Five x plus three x is equal to eight x. We get eight x is equal to 40. We could divide both sides by eight, and we get, five is equal to x. So this right over here is going to be equal to five. So let's go back, let's go back, now. So when y is positive eight, x is positive five. Now they ask us, "Use your two solutions "to graph the equation." So let's see if we can do, oh, whoops, let me, let me use my mouse now. So to graph the equations. So when x is negative five, y is negative eight. So the point negative five comma negative eight. So that's right over there. So let me move my browser up so you can see that. Negative five, when x is negative five, y is negative eight. And when x is positive five, and we see that up here, when x is positive five, y is positive eight. When x is positive five, y is positive eight. And we're done. We can check our answer, if we like. We got it right. Now, I said there was two ways to tackle it, I kind of just did it, I guess you could say, the naive way. I just substituted negative five directly into this and solved for y. And then I substituted y equals positive eight directly into this, and then solved for x. Another way that I could have done it, that actually probably would have been, or, it would for sure, would have been the easier way to do it, is ahead of time to try to simplify this expression. So what I could have done, right from the get-go, is said, "Hey, let's put all my xs on one side, "and all my ys on the other side." So this is negative three x plus seven y is equal to five x plus two y. Now let's say I wanna get all my ys on the left and all my xs on the right. So I don't want this negative three x on the left, so I'd wanna add three x. Adding three x would cancel this out, but I can't just do it on the left hand side, I have to do it on the right hand side as well. And then, if I wanna get rid of this two y on the right, I could subtract two y from the right, but, of course, I'd also wanna do it from the left. And then what am I left with? So negative three x plus three x is zero, seven y minus two y is five y. And then I have five x plus three x is eight x. Two y minus two y is zero. And then if I wanted to, I could solve for y, I could divide both sides by five and I'd get y is equal to 8/5 x. So, this right over here represents the same exact equation as this over here, it's just written in a different way. All of the xy pairs that satisfy this, would satisfy this, and vice versa. And this is much easier. Because if x is now negative five, if x is negative five, y would be 8/5 times negative five, well, that's going to be negative eight. And when y is equal to eight, well, you actually could even do this up here, you could say five times eight is equal to eight x, and then you could see, well five times eight the same thing as eight times five, so x would be equal to five. So I think this would actually have been a simpler way to do it. You see it all, I was able to do the entire problem in this little white space here, instead of having to do all of this, slightly, slightly hairier, algebra.