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# Systems of equations with elimination: 2x-y=14 & -6x+3y=-42

CCSS.Math:

## Video transcript

solve for x and y and we have the system of equations right here we have 2x minus y is equal to 14 and negative 6x plus 3y is equal to negative 42 so we could try to solve this by elimination and maybe we want to let's see if we can eliminate let's see if we can eliminate our Y variables first we have a 3y here and we have a negative y up here and we can they want to lymon eight if you just add negative y plus 3y that won't eliminate but if we could turn this negative Y into a negative 3y then it would cancel out with a 3y and the best way to turn a negative y to a negative 3y is to multiply this entire top equation by three so let's do that let's multiply let me get some space over to the left let's multiply this entire top equation by three so I'm going to multiply it by three so I multiply 2x by three I get 6x I multiply negative Y by three I get negative 3y and then I multiply 14 by 3 is 42 right 3 times 10 is 30 plus 12 it's 42 and then we can add both of these equations and something interesting should already maybe be showing up on your radar let's add both of these equations let's add the left-hand side negative 6x plus 6x well those cancel out we get 0 then we have 3y minus 3y those cancel out we get another 0 and then finally you get negative 42 plus 42 well that's 0 so we end up with just 0 is equal to 0 which is clearly true but it's not putting any constraints on the X or Y and that's because whenever you have a situation like this where you just get something that's obviously true 0 equals 0 or 1 equals 1 or 5 equals 5 what we're dealing is a situation is where both of our constraints both of our equations are actually the same equation so this right here is a dependent system it is a dependent it is a dependent system and you see it right over here if you take that first equation you multiply it by 3 you got 6x minus 3y is equal to 42 if we then multiply it by negative 1 if we then multiplied it by negative 1 you would get this you would get the exact same equation as the second equation you would get negative 6x plus 3y is equal to negative 42 or another way to think about it if you want to go from the first equation to the second equation you just multiply both sides of the equation x times negative 3 so both of these constraints are actually the same constraints they're just kind of a scaled-up multiple of each other so if you were to graph them and I might as well grab them for you right here this first equation right here let me do it over here it's 2x minus y is equal to 14 could subtract 2x from both sides and you would get let me just subtract it subtract 2x subtract 2x on the left-hand side you're left with just negative Y on the right hand side you have negative 2x plus 14 multiply both sides by negative 1 and you get Y is equal to 2x minus 14 so this first equation over here if I were to draw my axes so that is my y-axis and then that is my x-axis this graph right over here it's y-intercept is negative 14 so this is 0 negative 14 0 negative 14 it has a slope of 2 so it's going to look something like this it's going to look something like that and then the second equation if you were to graph it because this is a dependent system it's the exact same line if you were to put this in slope intercept form and graph it you would get the exact same thing you would go right on top of it so there's actually an infinite number of solutions these two lines are the same line so they intersect everywhere on each of the respective lines they're the same line and that's why and when you get something like this 0 equals 0 or 1 equals 1 that's the tell-tale sign that you're dealing with a dependent system if you've got something like 0 equals 1 then you would have no solution so this would right here would be an inconsistent inconsistence this would be an inconsistent system and this would be the situation so let me make it very clear this is a situation we're dealing with the same same lines the lines are the same lines this is a situation where you have parallel lines parallel parallel lines and so they never intersect and then obviously you have the the easiest situation I think this is one most of us familiar with where you have something like X is equal to 1 or Y is equal to 2 or anything like this doesn't have to be 1 or 2 and then this is clearly a situation where you have an independent system where you have two different lines that intersect in exactly one place anyway this there's an infinite number of solutions any X and y that satisfy the first equation will also satisfy the second equation because there's fundamentally the same constraint