 So let's take a look at solving equations by the addition method and you might say well Why do we need to learn a second method of solving equations? Don't we already know how to solve multilinear systems of equations and the answer is sure So here's a system of equations the x plus 4y equals 11 to x minus 7y equals 14 If the only tool you have is a hammer then every problem is a nail So I have these substitution methods so I can solve one of these equations for one of the variables and I'll end up with this horrible expression involving fractions Which will then have to substitute into the other equation and get another horrible expression involving fractions and it's painful all around So if the only tool you have is a hammer then every problem is a nail Even if the problem is trying to open a jar of pickles. So if you want to open a jar of pickles with a hammer I Wouldn't recommend it but as an alternative. We have a different way of solving the system of equations What we can do is I can add the two equations 3x is the same as 11 2x minus 7 y is the same as 14 if I add 14 it's the same as adding 2x minus 7 y and I can add Equations together and I can eliminate one of the variables that way well in this particular case if I add 14 Adding 14 is the same as adding 2x minus 7 y. That's not going to eliminate one of the variables So I do have to do a little bit of preparation And in this case the first thing I might notice is that the y terms in my two equations have Opposite signs so if I can get the coefficients equal The fact that they have opposite signs means that they will cancel each other out So I can do this very easily by multiplying by the coefficient of the other equation so here I have 3x plus 4y equals 11 and In this first equation my y coefficient in the other equation is going to be 7 So if I multiply both sides by 7 Then what I'm going to end up with as my coefficient of y is going to be 7 times 4 is going to be 28 Well, if I then multiply this second equation by the coefficient in the other equation, so that's going to be 4 Then what I'll end up with in this equation 4 times 7 subtracted. That's 4 times 7. That's 28 That's minus 28 and I'll have a plus 28 up here I'll have a minus 28 down here and when I add those two will cancel each other out So I'll expand my expressions And I'll expand my other expression 4 times 2x minus 7y and Now I'm ready to add the two equations. So again 56 is the same thing as 8x minus 28 y so on the right hand side if I add 56 I can add something that's equal to 56 and not change the equality. So I'll add 77 plus 56 is 133 21x plus 28 y plus 8x minus 28 y The 28 y minus 28 y drops out and I have 29 x and I have this equation, which I can then solve for x 133 over 29 and If I'm so keen on the substitution method well once I have one of the variables I can substitute into the other equation into either equation really to solve for the other variable So I'll drop that in I'll do some arithmetic of fractions and I'll get a solution and Well, I can do this, but it involves working a lot with fractions So here's an alternative if it worked before It'll work again. I can try to get the coefficients of x equal but opposite then add So here the coefficients of y were equal and opposite. So when I added I got an equation in x That was easy to solve if I can get the coefficients of x equal but opposite. I can add They'll drop out and I'll get an equation in y that's easy to solve So again, here's my original system of equations and I'm going to multiply by the Coefficient in the other equation. So this first equation I want to get rid of x. I'll multiply by 2 in the second equation I want to get a Minus so here I have 6x. I want to get a minus 6x So I'm going to multiply by the coefficient in the other equation That's 3 but then I'll make that a negative 3 so multiplying this first equation by 2 That's my coefficient multiplying the second equation by negative 3. That's my coefficient with a negative and I will expand those terms out and now I can add and That gets me 29 y equals negative 20 y equals 20 negative 20 over 29 And so here I'm able to solve for y without having to go through a lot of fraction arithmetic