Solving Problems Using Systems Of Equations

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Let's explore a few more methods for solving systems of equations.

Let's say I have the equation, 3x plus 4y is equal to 2.5.

But is there anything that we could add or subtract to both sides of this equation that might eliminate one of the variables? The left-hand side-- you're just left with a 4y, because these two guys cancel out-- is equal to-- this is 5 minus 21 over 2. So that's negative 16 over 2, which is the same thing-- well, I'll write it out as negative 16 over 2. Peter also buys 3 candy bars, but could only afford 1 additional Fruit Roll-Up. And we're going to solve this using elimination.

And then we would have one equation in one variable, and we can solve for it. So this is going to be 21 over 2 plus 4y is equal to 5/2. Or we could write that-- let's continue up here-- 4y-- I'm just continuing this train of thought up here-- 4y is equal to negative 8. You could solve this using any of the techniques we've seen so far-- substitution, elimination, even graphing, although it's kind of hard to eyeball things with the graphing. Remember, with elimination, you're going to add-- let's focus on this top equation right here. And that is going to be equal to $2.84 minus $1.79.

There is a whole field of mathematics devoted to the study of linear equations called linear algebra.

Convenient systems usually seem very tough to solve at first.

You would get Ax plus By, plus D is equal to C plus D. Anything you do to one side of the equation, you have to do to the other side. This second equation is telling me that explicitly. Peter also buys 3 candy bars, but can only afford 1 additional Fruit Roll-Up. What is the cost of each candy bar and each Fruit Roll-Up?

But you're saying, hey, Sal, wait, on the left-hand side, you're adding 5x minus 4y to the equation.

So I could, for example, I could add D to both sides of the equation. That's equivalent to-- let's see, this is 17.5 plus 8. So here it says, Nadia and Peter visit the candy store.

Because D is equal to D, so I won't be changing the equation. Nadia buys 3 candy bars and 4 Fruit Roll-Ups for .84. Let's let x equal cost of candy bar-- I was going to do a c and a f for Fruit Roll-Up, but I'll just stick with x and y-- cost of candy bar.


Comments Solving Problems Using Systems Of Equations

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