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# 用乘法預備聯立方程式做加減消去法 (英): 用乘法預備聯立方程式做加減消去法

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- Let's solve a few more systems of equations using
- elimination, but in these it won't be kind of a one-step
- elimination.
- We're going to have to massage the equations a little bit in
- order to prepare them for elimination.
- So let's say that we have an equation, 5x minus 10y is
- equal to 15.
- And we have another equation, 3x minus 2y is equal to 3.
- And I said we want to do this using elimination.
- Once again, we could use substitution, we could graph
- both of these lines and figure out where they intersect.
- But we're going to use elimination.
- But the first thing you might say, hey, Sal, you know, with
- elimination, you were subtracting the left-hand side
- of one equation from another, or adding the two, and then
- adding the two right-hand sides.
- And I could do that, because it was essentially adding the
- same thing to both sides of the equation.
- But here, it's not obvious that that
- would be of any help.
- If we added these two left-hand sides, you would get
- 8x minus 12y.
- That wouldn't eliminate any variables.
- And on the right-hand side, you would just
- be left with a number.
- And if you subtracted, that wouldn't
- eliminate any variables.
- So how is elimination going to help here?
- And the answer is, we can multiply both of these
- equations in such a way that maybe we can get one of these
- terms to cancel out with one of the others.
- And you could really pick which term you
- want to cancel out.
- Let's say we want to cancel out the y terms. So I'll just
- rewrite this 5x minus 10y here.
- 5x minus 10y is equal to 15.
- Now, is there anything that I can multiply this green
- equation by so that this negative 2y term becomes a
- term that will cancel out with the negative 10y?
- So I essentially want to make this negative 2y into a
- positive 10y.
- Right?
- Because if this is a positive 10y, it'll cancel out when I
- add the left-hand sides of this equation.
- So what can I multiply this equation by?
- Well, if I multiply it by negative 5, negative 5 times
- negative 2 right here would be positive 10.
- So let's do that.
- Let's multiply this equation times negative 5.
- So you multiply the left-hand side by negative 5, and
- multiply the right-hand side by negative 5.
- And what do you get?
- Remember, we're not fundamentally
- changing the equation.
- We're not changing the information in the equation.
- We're doing the same thing to both sides of it.
- So the left-hand side of the equation becomes negative 5
- times 3x is negative 15x.
- And then negative 5 times negative 2y is plus 10y, is
- equal to 3 times negative 5 is negative 15.
- And now, we're ready to do our elimination.
- If we add this to the left-hand side of the yellow
- equation, and we add the negative 15 to the right-hand
- side of the yellow equation, we are adding the same thing
- to both sides of the equation.
- Because this is equal to that.
- So let's do that.
- So 5x minus 15y-- we have this little negative sign there, we
- don't want to lose that-- that's negative 10x.
- The y's cancel out.
- Negative 10y plus 10y, that's 0y.
- That was the whole point behind multiplying this by
- negative 5.
- Is going to be equal to-- 15 minus 15 is 0.
- So negative 10x is equal to 0.
- Divide both sides by negative 10, and you get
- x is equal to 0.
- And now we can substitute back into either of these equations
- to figure out what y must be equal to.
- Let's substitute into the top equation.
- So we get 5 times 0, minus 10y, is equal to 15.
- Or negative 10y is equal to 15.
- Let me write that.
- Negative 10y is equal to 15.
- Divide both sides by negative 10.
- And we are left with y is equal to
- 15/10, is negative 3/2.
- So if you were to graph it, the point of intersection
- would be the point 0, negative 3/2.
- And you can verify that it also satisfies this equation.
- The original equation over here was 3x minus
- 2y is equal to 3.
- 3 times 0, which is 0, minus 2 times negative 3/2 is, this is
- 0, this is positive 3.
- Right?
- These cancel out, these become positive.
- Plus positive 3 is equal to 3.
- So this does indeed satisfy both equations.
- Let's do another one of these where we have to multiply, and
- to massage the equations, and then we can eliminate one of
- the variables.
- Let's do another one.
- Let's say we have 5x plus 7y is equal to 15.
- And we have 7-- let me do another color-- 7x minus 3y is
- equal to 5.
- Now once again, if you just added or subtracted both the
- left-hand sides, you're not going to
- eliminate any variables.
- These aren't in any way kind of have the same coefficient
- or the negative of their coefficient.
- So let's pick a variable to eliminate.
- Let's say we want to eliminate the x's this time.
- And you could literally pick on one of the
- variables or another.
- It doesn't matter.
- You can say let's eliminate the y's first. But I'm going
- to choose to eliminate the x's first. And so what I need to
- do is massage one or both of these equations in a way that
- these guys have the same coefficients, or their
- coefficients are the negatives of each other, so that when I
- add the left-hand sides, they're going to eliminate
- each other.
- Now, there's nothing obvious-- I can multiply this by a
- fraction to make it equal to negative 5.
- Or I can multiply this by a fraction to make it equal to
- negative 7.
- But even a more fun thing to do is I can try to get both of
- them to be their least common multiple.
- I could get both of these to 35.
- And the way I can do it is by multiplying by each other.
- So I can multiply this top equation by 7.
- And I'm picking 7 so that this becomes a 35.
- And I can multiply this bottom equation by negative 5.
- And the reason why I'm doing that is so this becomes a
- negative 35.
- Remember, my point is I want to eliminate the x's.
- So if I make this a 35, and if I make this a negative 35,
- then I'm going to be all set.
- I can add the left-hand and the right-hand
- sides of the equations.
- So this top equation, when you multiply it by 7, it becomes--
- let me scroll up a little bit-- we multiply it by 7, it
- becomes 35x plus 49y is equal to-- let's see, this is 70
- plus 35 is equal to 105.
- Right?
- 15 and 70, plus 35, is equal to 105.
- That's what the top equation becomes.
- This bottom equation becomes negative 5 times 7x, is
- negative 35x, negative 5 times negative 3y is plus 15y.
- The negatives cancel out.
- And then 5-- this isn't a minus 5-- this is times
- negative 5.
- 5 times negative 5 is equal to negative 25.
- Now, we can start with this top equation and add the same
- thing to both sides, where that same thing is negative
- 25, which is also equal to this expression.
- So let's add the left-hand sides and
- the right-hand sides.
- Because we're really adding the same thing to both sides
- of the equation.
- So the left-hand side, the x's cancel out.
- 35x minus 35x.
- That was the whole point.
- They cancel out, and on the y's, you get 49y plus 15y,
- that is 64y.
- 64y is equal to 105 minus 25 is equal to 80.
- Divide both sides by 64, and you get y is equal to 80/64.
- And let's see, if you divide the numerator and the
- denominator by 8-- actually you could probably do 16.
- 16 would be better.
- But let's do 8 first, just because we
- know our 8 times tables.
- So that becomes 10/8, and then you can divide this by 2, and
- you get 5/4.
- If you divided just straight up by 16, you would've gone
- straight to 5/4.
- So y is equal to 5/4.
- Let's figure out what x is.
- So we can substitute either into one of these equations,
- or into one of the original equations.
- Let's substitute into the second of the original
- equations, where we had 7x minus 3y is equal to 5.
- That was the original version of the second equation that we
- later transformed into this.
- So we get 7x minus 3 times y, times 5/4, is equal to 5.
- Or 7x minus 15/4 is equal to 5.
- Let's add 15/4-- Oh, sorry, I didn't do that right.
- This would be 7x minus 3 times 4-- Oh, sorry, that was right.
- What am I doing?
- 3 times is 15/4.
- Is equal to 5.
- Let's add 15/4 to both sides.
- And what do we get?
- The left-hand side just becomes a 7x.
- These guys cancel out.
- And that's going to be equal to 5, is the
- same thing as 20/4.
- 20/4 plus 15/4.
- Or we get that-- let me scroll down a little bit-- 7x is
- equal to 35/4.
- We can multiply both sides by 1/7, or we could divide both
- sides by 7, same thing.
- Let's multiply both sides by 1/7.
- The same thing as dividing by 7.
- So these cancel out and you're left with x is equal to--
- Here, if you divide 35 by 7, you get 5.
- You divide 7 by 7, you get 1.
- So x is equal to 5/4 as well.
- So the point of intersection of this right here is both x
- and y are going to be equal to 5/4.
- So if you looked at it as a graph, it'd be 5/4 comma 5/4.
- And let's verify that this satisfies the top equation.
- And if you take 5 times 5/4, plus 7 times
- 5/4, what do you get?
- It should be equal to 15.
- So this is equal to 25/4, plus-- what is this?
- This is plus 35/4.
- Which is equal to 60/4, which is indeed equal to 15.
- So it does definitely satisfy that top equation.
- And you could check out this bottom equation for yourself,
- but it should, because we actually used this bottom
- equation to figure out that x is equal to 5/4.

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