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Current time:0:00Total duration:6:07

Video transcript

welcome to level two linear equations let's do a problem so 2x plus 3 is equal to minus 15 so the minus in there to make it a little bit tougher so the first thing we want to do whenever we do any linear equation is 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 tend to get my variables on the left-hand side of the equation well the only thing that my variables are already already on the left-hand side of the equation but I have this plus 3 that I somehow want to move to the right-hand side of the equation and the way I can you can put it in quotes move the 3 is I could subtract 3 from both sides of this equation and look at that carefully as to why you think that works because if I subtract 3 from the left-hand side clearly this negative 3 that I'm subtracting the original 3 will cancel out and become 0 and as long as I do whatever I do on the left hand side as long as I do it on the right hand side as well because whatever you do on on one hand side of the equal side you have to do on the other then I'm making a valid operation so this will simplify to 2 X because the 3s cancel out they become just 0 equals minus 15 minus 3 well that's minus 18 and now we're just at a level two level one problem and you can just multiply both sides of this equation times the reciprocal on the coefficient of 2 X I mean you could some people just say that we're dividing by two which is essentially what we're doing I like to always go with the reciprocal because if this 2 was a fraction it's easier to think about it that way but either way you either multiply by the reciprocal or divide by the number it's the same thing so 1/2 times 2x well that's just 1x so you get x equals and then minus 18 over 2 and minus 18 over 2 well that just equals minus 9 let's do another problem and actually well if we wanted to check it we could say well the original problem was 2x plus 3 equals minus 15 so we could say 2 x minus 9 plus 3 2 x minus 9 is minus 18 plus 3 well that equal to minus 15 which is equal to what the original equation said so we know it's right that's that's the neat thing about algebra you can always check your work let's do another problem I'm going to put some fractions in this time just to show you that it's it can get a little bit a little bit hairy so let's say I had minus 1/2 X plus 3/4 is equal to 5/6 so we'll do the same thing first we just want to get this 3/4 out of the left hand side of the equation and and and actually if you want to try working this out yourself you might want to pause the video and then play it once you're ready to see how I do it but anyway let me move forward assuming you haven't paused it if we want to get rid of this 3/4 all we do is we subtract 3/4 from both sides of this equation minus 3/4 well the left hand side the 2 3/4 will just cancel and we get minus 1/2 x equals and then on the right hand side we just have to do this we have to do this fraction addition or a fraction subtraction so the least common multiple of 6 & 4 is 12 so that's becomes 5 over 6 is 10 over 12 minus 3 over 4 is 9 over 12 so we get minus 1/2 X is equal to is equal to 1/12 hopefully I didn't make a mistake over here and if that step confuse you I went a little fast you might just want to review the adding and subtraction of fractions so going back to where we were so now all we have to do is well as the coefficient on the X term is minus 1/2 and this is now a level-one problem so to solve for X we just multiply both sides by the reciprocal of this minus 1/2 X and that's minus 2 over 1 times minus 1/2 X on that side and then that's x minus 2 minus 2 over 1 the left-hand side and you're used to this by now simplifies to X the right-hand side becomes minus 2 over 12 and we could simplify that further to minus 1 over 6 well let's check that just to make sure we did the right we got it right okay so let's try to remember that minus 1 over 6 and so the original problem was minus 1/2 X so here we can substitute the minus 1 over 6 plus 3/4 I just wrote only the left-hand side of the original problem so minus 1/2 times minus 1/6 well that's positive 1 over 12 plus 3/4 well that's the same thing as 12 the 1 stays the same plus 9 1 plus 9 is 10 over 12 and that is equal to 5 over 6 which is what our original problem was our original problem was this this stuff I wrote later so this 5/6 so the problem checks out so hopefully you're now ready to try some level 2 problems on your own I might add some other example problems but the only extra step here relative to level 1 problems is you'll have this constant term that you need to add or subtract from both sides of this equation and essentially turn it into a level 1 problem have fun