Solving Equations
After explaining the basic concepts in solving equations, I show students a simple equation to solve such as 2x + 4 = 8. I tell students that the goal is to get x by itself on one side of the equation. So the first step is to subtract 4 from both sides, and we are left with 2x = 4. Most are able to follow this procedure. Then when I ask them what should be the next step, some students want to subtract the 2 from the 2x. So I explain that 2x means 2 multiplied by x, so to get x by itself we need to divide by 2. This satisfies some students but others remain confused.
I have achieved good results by turning this equation into a word problem: Two cantaloupes = $4. I ask “How much is one cantaloupe?” and they all answer $2. Then I ask “How did you figure this out?” This helps them realize that they had to divide both sides by 2 in order to get the answer.
Collecting Like Terms
I use a similar strategy when I am explaining how to add like terms. For example, if we have an expression such as 6x2 + 2x4 - 2x2 + 7 + 3x2 - 3 and the goal is to combine like terms, I first ask three students to name their favorite fruit. If I get answers such as apples, oranges and bananas, I first rewrite this expression as 6 apples + 2 oranges – 2 apples + 7 bananas + 3 oranges - 3 bananas. Then I ask them to evaluate how many fruits of each kind are left. Most can figure this out as 4 apples + 5 oranges + 4 bananas.
I use a similar strategy when I am explaining how to add like terms. For example, if we have an expression such as 6x2 + 2x4 - 2x2 + 7 + 3x2 - 3 and the goal is to combine like terms, I first ask three students to name their favorite fruit. If I get answers such as apples, oranges and bananas, I first rewrite this expression as 6 apples + 2 oranges – 2 apples + 7 bananas + 3 oranges - 3 bananas. Then I ask them to evaluate how many fruits of each kind are left. Most can figure this out as 4 apples + 5 oranges + 4 bananas.
Then I tell students that in math we also have different kinds of fruits but they have mathematical names such as x4, x2 and just plain numbers. They are then usually able to make the connection and get the right answer.
Factoring Polynomials
Most algebra textbooks introduce factoring of polynomials before describing the purpose of factoring. I prefer to start by showing my students a graph of a quadratic equation which is a parabola, and explaining that the solutions of this quadratic are the x intercepts (they are already familiar with the concept of intercepts from linear equations). Then I indicate that there are three ways of solving this type of equation: 1) The quadratic formula, 2) completing the square and 3) factoring (when possible). Then the reason for factoring becomes more understandable in view of the ultimate goal.
Most algebra textbooks introduce factoring of polynomials before describing the purpose of factoring. I prefer to start by showing my students a graph of a quadratic equation which is a parabola, and explaining that the solutions of this quadratic are the x intercepts (they are already familiar with the concept of intercepts from linear equations). Then I indicate that there are three ways of solving this type of equation: 1) The quadratic formula, 2) completing the square and 3) factoring (when possible). Then the reason for factoring becomes more understandable in view of the ultimate goal.
Systems of Equations
Before showing the two methods of solving a system of two equations (substitution and elimination), I first write this equation on the board: x + y = 4.
Before showing the two methods of solving a system of two equations (substitution and elimination), I first write this equation on the board: x + y = 4.
Then I ask my students “What is x and what is y?” After the common answers of 2 and 2, 3 and 1 and 4 and 0, most of them soon realize that an equation in two variables does not have a unique solution, because there are an infinite number of possible answers. Then I write two equations like this:
x + y = 4
x - y = 2
x - y = 2
I ask “What is x and what is y?” Most are able to figure out that x is 3 and y is 1. This makes the explanation that we need two equations to solve for x and y more understandable.
Finding the LCM (Least Common Multiple) of Rational Expressions
This topic is often quite challenging for students. What I try to do is clearly demonstrate the analogy with finding the LCM of numbers, which they have studied in an earlier chapter and know fairly well by the time we get to rational expressions.
This topic is often quite challenging for students. What I try to do is clearly demonstrate the analogy with finding the LCM of numbers, which they have studied in an earlier chapter and know fairly well by the time we get to rational expressions.
In order to find the LCM of 12, 15 and 18, break each number down into its prime factors:
12: 3,2,2
15: 3,5
18: 3,3,2
15: 3,5
18: 3,3,2
So the LCM is 3 x 3 x 2 x 2 x 5 = 180
To find the LCM of x2 - 4, x2 + 4x + 4, 2x - 4, break each expression into its prime factors (after defining a prime factor for an algebraic expression):
To find the LCM of x2 - 4, x2 + 4x + 4, 2x - 4, break each expression into its prime factors (after defining a prime factor for an algebraic expression):
x2 - 4: (x + 2) (x - 2)
x2 + 4x + 4: (x + 2) (x + 2)
2x - 4: 2(x - 2)
Using the same procedure we used for numbers, we pick each prime factor to the maximum number of times it appears in any of the three expressions:
So the LCM is 2(x + 2) (x +2) (x - 2)
x2 + 4x + 4: (x + 2) (x + 2)
2x - 4: 2(x - 2)
Using the same procedure we used for numbers, we pick each prime factor to the maximum number of times it appears in any of the three expressions:
So the LCM is 2(x + 2) (x +2) (x - 2)
No comments:
Post a Comment