# Solving Problems Using Systems Of Equations We'll go back to our same example to illustrate this.  If we subtract 0.35 from both sides, what do we get?

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.

And you're probably saying, Sal, hold on, how can you just add two equations like that? When I looked at these two equations, I said, oh, I have a 4y, I have a negative 4y. And that's going to be equal to 2.5 plus 25.5 is 28. So let's verify that it also satisfies this bottom equation.

And remember, when you're doing any equation, if I have any equation of the form-- well, really, any equation-- Ax plus By is equal to C, if I want to do something to this equation, I just have to add the same thing to both sides of the equation. If we were to add the left-hand side, 3x plus 5x is 8x. If you just add these two together, they are going to cancel out. 5 times 7/2 is 35 over 2 minus 4 times negative 2, so minus negative 8. Now let's see if we can use our newly found skills to tackle a word problem, our newly found skills in elimination. Nadia buys 3 candy bars, so the cost of 3 candy bars is going to be 3x.

## One thought on “Solving Problems Using Systems Of Equations”

1. In activities where students are synthesizing information, such as in a reading or science activity, teachers should be asking them to create or invent new ideas or to compare and contrast what they are seeing.