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# 用代入消去法解聯立方程式 (英) : 用代入消去法解聯立方程式

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- 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 they intersect.
- This will give you an exact algebraic answer.
- And in 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 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, to 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'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 this 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 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-- let's 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.
- And 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, and we get y is equal to 7.
- So we found the y value of the point of intersection of these
- two lines. y is equal to 7.
- Let me write over here, 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 been able
- to find a pair of x and 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.
- 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 were to graph both of these equations, they would
- intersect at the point negative 5 comma 7.
- Now let's use our newly found skill to do
- an actual word problem.
- Let's say that they tell us that the sum of
- two numbers is 70.
- And they differ-- or maybe we could say their difference--
- they differ by 11.
- 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, 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.
- Now, this first statement, the sum of the two numbers is 70.
- That tells us that x plus y must be equal to 70.
- That second statement, that they differ 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 it.
- 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, 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 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, we're substituting that in for x--
- plus y is equal to 70.
- This is x.
- And that constraint was given to us by this 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 y plus 11, plus y is equal to 70.
- That's 2y plus 11 is equal to 70.
- And then if we subtract 11 from both sides, we get 2y is
- equal to-- what is that?
- 59?
- You subtract 10 from 70, 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-- 29.5.
- y is equal to 29.5.
- 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 29.5-- that's what y is, we
- just figured that out-- plus 11, which is equal to-- so you
- add 10, you get 39.5.
- You add another 1, you get 40.5.
- And we're done.
- If you wanted to find the intersection of these two
- lines, it would intersect at the point 40.5 comma 29.5.
- And you could have used this equation to solve for x and
- then substituted in this one.
- You could have used this equation to solve for y and
- then substituted in this one.
- You could use this equation to solve for y and then
- substitute into 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?
- 40.5 plus 29.5, that indeed is 70.
- And the difference between the two is indeed 11.
- They're exactly 11 apart.
- Anyway, hopefully you found that useful.

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