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solving equations of the form AX+B=C. Created by Sal Khan.
Video transcript
Welcome to Level 2 Linear Equations! Let's do a problem. 2X plus 3 is equal to minus 15. I took a minus is there to make it a little bit tougher. So the first thing we wanna do whenever we do any linear equations , we want to get all of the variable terms on one hand side of the equation and all the constant terms on the other side It doesn't really matter, although, I intend to get my variables on the left hand side of the equation. Well, the only thing that my variables are already on the left hand side of the equation, but have this plus three, that I somehow wanna move to the right hand side of the equation. And the way I can "move" the 3 is, I could subtract 3 from the both sides of the equation, and look at that carefully as to why you think that works. Because if I subtract 3 from the left side, clearly, this negative 3 that I am subtracting, the original 3 will cancel out and becomes zero. And as long as I do whatever I do on the left hand side as long as I do on the right hand side as well, because, whatever you do on one hand side, equal side, you have to do to the other, then I'm making a valid operation. So this would simplify to 2 X, cause 3 is cancelled out, and comes zero, equals minus 15, minus 3 Well, that's minus 18. And now we're just on the level 2 to level 1 problem And you could just multiply both sides of this equation times their reciprocal on the coefficient of 2 X. Some people would just say that we're dividing by 2, which is essentially, what we're doing. I like to always go to the reciprocal, because if these 2 was a fraction, it's easier to think about that way. But either way, you either multiply it by the reciprocal or divide by the number. 1/2 times 2 x is just 1 x, so you get x equals minus 18 over 2 minus 18 over 2 is just equal to minus 9. Let's do another problem. And actually, if we wanna check it we could say that the original problem was 2x plus 3 equals minus 15. We could say 2 times minus 9 plus 3. 2 times minus 9 is minus 18 plus 3. Well, that is equal to minus 15. which is equal to what the original equation said. So we know it's right! That the neat thing about algebra, you can always check that your work. Let's do another problem. I'm gonna put some fractions in this time just to show you that it can get a little bit hairy. So, let's say I had minus 1/2 x plus 3/4 is equal to 5/6 So we will do the same thing. First we just wanna get the 3/4 out of the left hand side of the equation. Actually if you wanna try to work this out yourself, you could pause the video now. Let's go ahead. if we wanna get rid og the 3/4, all we do is subtarct 3/4 from both sides On the left hand side the 3/4 just cancel out, and we get minus 1/2 x equal. On the right hand side we have do the fraction subtraction. the least common multiple of 6 and 4 is 12. that becomes 10 minus 9. We get minus 1/2 x is equal to 1/12. I hope I did not make a mistake. If that step confuse you -- I went a little fast -- you might just want to review the addition and subtraction of fraction. Going back to where we were. now all we have to do is the coefficient on the x term is minus 1/2, and this is now a level 1 problem, so to solve for x we just multiply x by the reciprocal of minus 1/2 x. That is minus 2 over 1, on both sides. The left hand side simplifies to x. The right hand side becomes minus 2 over 12, and we can simplify that to minus 1/6. Let's just check that to make sure that we got it right. We got minus 1/6 and the original problem was minus 1/2 x, so we substitute minus 1/6, plus 3/4. I only wrote the left hand side of the original problem. So minus 1/2 times minus 1/6 that is positive 1/12 and then plus 3/4. That is the same thing as 1/12 plus 9/12. 1 plus 9 is 10 over 12. That is equal to 5/6 which is what our orignal problem was. This other stuff was what I wroter later So the problem checks out. Hopefully you are now ready to try some level 2 problem on your own. I might add some more example problems. The only extra step here relative to level 1 problems is that we have this constant term that you need to add or subtract from both sides of this equation, and you essentially turn it into a level 1 problem. Have fun!