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Current time:0:00Total duration:9:21

Video transcript

in the last video we saw what a system of equations is and in this video I'm going to show you one algebraic technique for solving systems of equations where you don't have to graph the two lines and try to figure out exactly where the intersect this will give you an exact algebraic answer and future videos we'll see more methods of doing this so let's say you had two equations one is X plus 2y is equal to 9 and the other equation is 3x plus 5y is equal to 20 now if we did what we did in the last video we could graph each of these these are lines you could put them in either a slope intercept form or point-slope form they're in standard form right now and then you could graph each of these lines figure out where they intersect and that would be a solution to that but it's sometimes hard to find two just by looking figure out exactly where they intersect so let's figure out a way to algebraically do this and what I'm going to do is the substitution method I'm going to use one of the equations to solve for one of the variables and then I'm going to substitute back in for that variable over here so let me show you what I'm talking about so let me solve for x using this top equation so the top equation says X plus 2y is equal to 9 I want to solve for X so let's subtract 2y from both sides of this equation so I'm left with X is equal to 9 minus 2y this is what this first equation is telling me I just rearranged it a little bit the first equation is saying that so in order to satisfy both of these equations X has to satisfy this constraint right here so I can substitute this back in for X we say we're saying this top equation says X has to be equal to this well if X has to be equal to that let's substitute this in for X so the second equation will become 3 times X and instead of an X I'll write this thing 9 minus 2y 3 times 9 minus 2y plus 5y is e equal to 20 that's why it's called the substitution method I just substituted for X and the reason why that's useful is now I have one equation with one unknown and I can solve for y so let's do that 3 times 9 is 27 3 times negative 2 is negative 6y plus 5y is equal to 20 add the negative 6y plus the 5y add those two terms you have 27 see this will be minus y is equal to 20 let's subtract 27 from both sides and you get let me write it out here so let's subtract 27 from both sides the left hand side the 27 s cancel each other out you're left with negative y is equal to 20 minus 27 is negative 7 and then we can multiply both sides of this equation by negative 1 multiply both sides by negative 1 we get Y is equal to 7 so we thought we found the Y value of the point of intersection of these two lines Y is equal to 7 let me write it over here y is equal to 7 so I don't have to keep scrolling down and back up Y is equal to 7 well if we know Y we can now solve for X X is equal to 9 minus 2y so let's do that X is equal to 9 minus 2 times y 2 times 7 or X is equal to 9 minus 14 or X is equal to negative 5 so we've just using substitution we've able to found a pair of X&Y points that satisfy these equations the point X is equal to negative 5 y is equal to 7 satisfy both of these and you can try it out you could try it out negative 5 plus 2 times 7 that's negative 5 plus 14 that is indeed 9 you do this equation 3 times negative 5 is negative 15 plus 5 times y plus 5 times 7 so negative 15 plus 35 is indeed 20 so this satisfies both equations if you Graff both of these equations they would intersect at the point negative five comma seven now let's use let's use our newly found skill to do an actual word problem let's say that they tell us that the sum sum of two numbers the sum of two numbers is 70 and they defer they differ or maybe we could say their difference they differ by 11 by 11 what are the numbers what are the numbers so let's do this word problem so let's define some variables let's let X be the larger number the larger number and let Y be the smaller number I'm just arbitrarily creating these variables one of them is larger than the other they differ by 11 be the smaller number now this first statement the sum of the two numbers is 70 that tells us that X plus y must be equal to 70 X plus y must be equal to 70 that second statement the second statement that they defer by 11 that means the larger number minus the smaller number must be 11 that tells us that X minus y must be equal to 11 so there we have we have two equations and two unknowns we have a system of two equations we can now solve it using the substitution method so let's solve for x on this equation right here so if you add Y to both sides of this equation if you add Y to both sides of this equation what do you get on the left hand side you just get an X because these cancel out and then on the right hand side you get X is equal to 11 plus y or y plus 11 so we get X is equal to 11 plus y using the second equation and then we can substitute it back into this top equation so instead of writing X plus y is equal to 70 we can substitute this in for X we've already used the second the second equation the magenta one now we have to use the top constraint so if we substitute this in we get Y plus 11 remember this is what X was or that we're substituting that in for X plus y is equal to 70 this is this is X and that constraint was given to us by the second equation or by this second statement I just substituted this X with y plus 11 and I was able to do that because that's the constraint the second equation gave us so now let's just solve for y we get CY plus eleven plus y is equal to 70 that's 2y plus 11 is equal to 70 and then if we subtract if we subtract 11 from both sides we get 2y is equal to what is that 59 you subtract 10 from 7 you get 60 so it's going to be 59 so Y is equal to 59 over 2 or another way to write it you could write that as 59 over 2 is the same thing as let's see 25 20 29 point 5 y is equal to 29 point 5 now what is X going to be equal to well we already figured out X is equal to Y plus 11 so X is going to be equal to 29 point 5 that's what Y is we just figure that out plus 11 which is equal to C add 10 you get thirty nine point five you add another one you get 40 point five and we're done if you wanted to find the intersection of these two lines it would intersect at the point forty point five comma twenty nine point five and you could have done you could have used this equation to solve for X and then substitute in this one you could have used this equation to solve for y and then substitute this one you could use this equation to solve for y and then substitute in that equation the important thing is is you use both constraints now let's just verify that this actually works out what's the sum of these two numbers forty point five plus twenty nine point five that indeed is seventy and the difference between the two is indeed eleven they're exactly eleven apart anyway hopefully you found that useful