Next Lesson: Isolating a Variable

In this lesson, students will learn how to solve for a variable that may be in multiple terms, as well as trapped inside of parentheses. I realized I needed to make this a specific objective in Algebra 2 when my Calculus students in previous years didn't know how to isolate dy/dx after taking an implicit derivative. They have also had trouble in science: for example, the Gas Laws in Chemistry throw them for a loop when they have to solve for a variable that's in the denominator. Hopefully, getting in some good practice with this solving technique now will pay off when it is reinforced later on.

We will start by looking at x + xy = 1, and finding the value of y for different values of x. Students will likely conclude that it would be easier to isolate y first. We'll then look at some specific formulas (Gas Laws, temperature conversion) and how to isolate for any given variable. The key to isolating a variable:
1) use the distributive property (if the variable is trapped inside parentheses).
2) get all the terms with the variable on one side of the equation.
3) get all the terms without the variable on the other side.
4) if necessary, factor the variable out, and then divide what remains.

The type of problem I want them to be able to do: Isolate y in (3 + 7x)y – 6(xy + 5) = 15

Following the direct instruction portion, students will have independent practice time with a related homework. These will be posted on ILoveMath.

Update:
The lesson went ok, but I realized I made the same mistake as last year... I need to teach students how to deal with the division form of the distributive property first, so they can understand why certain steps are invalid. For example, given the equation xy = 2y + 7, students will tend to "divide both sides by y" to get x = 2 + 7. I didn't think of this because none of the problems I gave require a division step like that - but I forgot to take into account that they would mistakenly do it. Woops. I think that will come up in the next lesson.