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Solving linear systems by substitution (old)

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 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.