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welcome to the presentation on level 4 linear equations so let's start doing some problems so let's say I had the situation let me give me a couple of problem if I said 3 over X is equal to let's just say 5 so what we want to do this this problem is a little unusual from everything we've ever seen because here instead of having X in the numerator we actually have X in the denominator so I personally don't like having X's in my denominator so we want to get outside of the denominator or into a numerator or at least not in the denominator as soon as possible so one way to get a number out of the denominator is if we were to multiply both sides of this equation by X you see that on the left hand side of the equation these 2 X's will cancel out and on the right side you'll just get 5 times X so this equals the 2 X's cancel out and you get 3 is equal to 5x now we could also write that is 5x is equal to 3 and then we can think about this two ways we either just multiply both sides by 1/5 or you could just do that as dividing by 5 if you multiply both sides by 1/5 the left hand side becomes X and the right hand side 3 times 1/5 is equal to 3/5 so what did we do here this is just like this actually turned into a level-two problem or actually a level-one problem very quickly all we had to do is multiply both sides of this equation by X and we've got the X's out of the denominator let's do another problem let's have let me say X plus 2 over X plus 1 is equal to say 7 so here instead of having just an accident no mater we have a whole X plus one in the denominator but we're going to do it the same way to get that X plus 1 out of the denominator we multiply both sides of this equation times X plus 1 over 1 times the side since we do it on the left hand side we also have to do it on the right hand side and this is just 7 over 1 times X plus 1 over 1 on the left hand side the X plus ones cancel out and you're just left with X plus 2 it's over 1 but we can just ignore the 1 and then that equals 7 times X plus 1 and that's the same thing as X plus 2 and remember it's 7 times the whole thing X plus 1 so we actually have to use a distributive property and that equals 7 X plus 7 so now it's turned into a I think this is a level 3 linear equation and now all we do is we say well let's get all the X's on one hand one side of the equation and let's get all the constant terms like the 2 and the 7 on the other side of the equation so I'm going to choose to get the X's on the left so let's bring that 7x on to the left and we can do that by subtracting 7x from both sides minus 7x this is minus 7x the right hand side these two 7 X's will cancel out and on the left hand side we have minus 7x plus X well that's minus 6 X plus 2 is equal to and on the right all we have left is 7 now we just have to get rid of this 2 and we can just do that by subtracting 2 from both sides and we're left with minus 6 X is equal to 5 now it's a level 1 problem we just have to multiply both sides times the reciprocal of the coefficient on the left hand side and the coefficient is negative 6 so we multiply both sides of this equation by negative 1 over 6 negative 1 over 6 the right the left hand side negative 1 over 6 times negative 6 well that just equals 1 so we just X is equal to 5 times negative one-sixth well that's negative 5 over 6 and we're done and if you wanted to check it you could just take that x equals negative 5 6 and put it back in the original equation to confirm that it worked let's do another one and I'm making these up on the fly so I apologize let me let me think 3 times X plus 5 is equal to 8 times X plus 2 well we do the same thing here although now we have two expressions that we want to get out of the denominators you want to get this X plus 5 out and we want to get this X plus 2 out so let's do the X plus 5 first well just like we did before we multiply both sides of this equation by X plus 5 you can say X plus 5 over 1 times X plus 5 over 1 on the left hand side they get cancelled out so we're left with 3 is equal to 8 times X plus 5 all of that over X plus 2 on the now on that on the top just to simplify it we once again just multiply that 8 times your whole expression so it's 8x plus 40 over X plus 2 now we want to get rid of this X plus 2 and we do it the same way we can multiply both sides of this equation by X plus 2 over 1 X plus 2 we could just say we're multiplying both sides by X plus 2 that one is a little unnecessary so the left-hand side becomes 3x plus 6 remember always distribute the 3 times because you're multiplying it times a whole expression X plus 2 and on the right hand side well this X plus 2 and this X plus 2 will cancel out and we're left with 8x plus 40 and this is now a level-3 problem well if we subtract 8x from both sides 8x plus I think I'm running out of space minus 8x well on the right hand side the 8x is canceled out on the left hand side we have minus 5x plus 6 is equal to on the right hand side all we have left is 40 now we can subtract 6 from both sides of this equation let me just write it out here minus 6 plus minus 6 now I'm going to hope I don't lose you guys by trying to go up here but if we subtract minus 6 from both sides on the left hand side we're just left with minus 5x equals and on the right hand side we have 34 now that's a level 1 problem we just multiply both sides times negative 1/5 times negative 1/5 left hand side we have X and then the right hand side we have negative 34 over 5 unless I made some careless mistakes I think that's right and I think if you if you understood what we just did here you're ready to tackle some level 4 linear equations have fun