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Topic D: Systems of linear equations and their solutions

Video transcript

solve for x and wine we have two equations with two unknowns negative 3y plus 4x is equal to 11 and y plus 2x is equal to 13 and what we can do is try to solve this using using elimination maybe we can add these two equations together to get some variables to cancel out but if we just add Y and negative 3y those won't cancel out and if we just add a 2x and 4x those won't cancel out but maybe we can scale one of these equations so that we will get some cancellation if this Y could somehow become a 3y then 3y and negative 3i would cancel out and so the easiest way to turn this Y into a 3y so just multiply this entire equation so we don't actually change the information in it to multiply this entire equation by 3 so let's multiply both sides by 3 if we multiply the left-hand side by 3 we get 3y 3y plus 6x to multiply every term by 3 3y plus 6x is equal to 3 times 13 is 39 so this equation in this equation are the exact same equation I've just multiplied both sides by 3 to go from here to here so I multiplied by 3 and of course we have this original equation up here that negative 3y plus 4x is equal to 11 and now if we add the left hand side and the right hand sides of this equation something interesting happens the negative 3y and the positive 3y cancel out that was the whole point behind multiplying the second guy over here by 3 so let's add them so if we add these guys cancel out we have 4x plus 6x is 10 X and that is equal to 11 plus 39 which is 50 and then we can divide both sides by 10 divide both sides by 10 we get X is equal to 50 divided by 10 is 5 and then we can go substitute back to solve for y if X is 5 we could use the second equation right over here we get y plus 2 times 5 I'll do it in that same color 2 times 5 is equal to 13 or Y plus 10 is equal to 13 we can subtract 10 from both sides and we will get why these cancel out y is equal to 3 so our solution that we get for x and y is that X is equal to 5 and Y is equal to 3 and we can verify that it works in both equations in this top equation let me do it a new color negative 3 times 3 plus 4 times 5 is this is negative 9 plus 20 which is indeed equal to 11 so both of these satisfy the first equation and then if we take the second equation 3 plus 2 times 5 that's 3 plus 10 that does indeed equal 13 so it definitely satisfies both and I want to be clear you could have solved this with substitution or you could have you could have changed which equation you scale so that we get different cancellations when we add or different eliminations so another way that we could have done this we could have kept the first equation the same negative 3y plus 4x is equal to 11 but maybe we want this 2x to cancel with this 4x and the only way we could do that is if this negative this 2x could become a negative 4x and the way to get from a to X to a negative 4x is to multiply it by negative 2 so let's multiply this entire equation by negative 2 so it become negative 2y minus 4x is equal to negative 26 and if we then add the two equations the X's will cancel out you have negative 3y minus 2y which is negative 5y is equal to 11 minus 26 that's negative 15 now you can divide both sides by negative 5 and you get the negatives cancel out Y is equal to 3 so in this situation we eliminated the X's first to solve for y and then you could go back substitute and solve for X you'll get X is equal to 5 the first time we did it we eliminated the Y's we scaled this equation right here so that the Y's would cancel out and so we solved for the X first and then substitute it back either way works you get X is equal to 5 y is equal to 3