Lesson 2: Algebra |
1 Introduction to Algebra: Expressions, Terms and Factors
Lesson 2: Algebra | |||
: Expressions, Terms and Factors | < Lessons | | ||
![]() Literal numbers can be used to represent either variables (the value of the letter can change) or constants (the value of the letter never changes). We have all used algebra at some point in our lives and were probably exposed to it at an early age. ExampleFor example, we all learned the area of a rectangle is equal to the width (w) times the height (h), while in algebraic terms this equation looks like the following:
So once we know the width and the height we can solve the equation. A term is an algebraic expression involving letter and/or numbers (called factors) which are multiplied together. Example
Example
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2 Adding and Subtracting Algebraic Terms
Lesson 2: Algebra | |||
Algebraic Terms | |||
In algebra we can only add and subtract terms with the same
variables raised to the same power. These terms are called like
terms.
Example
Remember we can only add or subtract like terms. Now lets try an example: ExampleSimplify the following:
Notice that the only like terms in this problem are 8x and -3x. we cannot do anything with the 6y or 9a terms. So we group together the terms we can subtract, and just leave the rest. ExampleLet’s try another one:
Hint: go back to the order of operations if you are not sure what to do first.
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3 Multiplication of Algebraic Expressions
Lesson 2: Algebra | |||
of Algebraic Expressions | |||
To explore the multiplication of algebraic expressions lets work
through a few examples.
Example
ExampleLet’s try another one:
ExampleLet’s try one more:
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4 Division of Algebraic Expressions
Lesson 2: Algebra | |||
of Algebraic Expressions | |||
Now that you have mastered multiplication of algebraic functions
lets go over a few examples of dividing algebraic functions.
Example
ExampleLet’s try another example:
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5 Dividing by a Fraction
Lesson 2: Algebra | ||||||||||||||
Remember the reciprocal of a number x is 1/x, for example the
reciprocal of 8 is 1/8. To divide by a fraction you simply multiply
by the reciprocal of the fraction.
ExampleLet’s do an example:
ExampleLets try another example, I will show you two different ways you can solve this problem and you can decide for yourself which one is easier.
Solution 1: multiplying by the reciprocal:
Solution 2: multiplying the top and bottom
Notice that the two answers are the same. |
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6 Solving Algebraic Equations
Lesson 2: Algebra | |||||||||
The goal of algebraic equations is to find the number or set
of numbers which makes the equation true. Sometimes the answer can
be simple, other times it can be very complex, so we need to have a
process that allows us to find and answer which is correct.
This process is aiming to get the letter on the left hand side of the equation by itself. To do this we must balance whatever we do to one side of the equation with the other. So for example if you add 4 to one side of an equation you must also add four to the other. ExampleLet’s look at some examples:
ExampleLet’s try this one:
ExampleLet’ try one more!
In some cases it might be possible that there are no possible solutions to the equation or the solutions do not make since. There for it is important to always check your solutions. |
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7 Algebraic Formulas
Lesson 2: Algebra | |||
It is very common in many scientific fields to have a formula that
we will have to rearrange to solve for a different variable. In such
cases we solve the equation exactly how we did in the last section.
ExampleLet’s look at a few examples.
So why might you want to do this? Well lets say you knew the area and heights for many rectangles and wanted to get the width of each, well by manipulating the formula we can now quickly solve the equation for the variable we are most interested in. ExampleLet’s try another example:
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8 Word problems
Lesson 2: Algebra | |||
The reason we learn mathematics is we can solve problems, not just
move numbers and letters around in an equation. But most real world
problems are stated using verbal communication, so what we need to
be able to do is translate words into mathematical statements.
Here is a general outline of how you can tackle word problems:
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9 Example word problem 1
Lesson 2: Algebra | |||
Solve the Following:Two streams have a combined flow rate of 32.4 cubic feet per second. We know that one stream has a flow rate that is 5 cubic feet per second higher than the other. What is the flow rate of the two streams? Solution 1:
We also know that together the two streams will release the following:
When we solve for x we get a flow rate of 13.5 ft3/sec and since we know that the second stream has a flow rate which is 5 ft3/sec faster we can add 5 to the first flow rate of 13.5 to get 18.5 ft3/sec. Now let’s check our work. If we add together the tow flow rate we should get a combined flow of 32 ft3/sec. Since 13.5 + 18.5 add up to 32 we know we have done this problem correctly. Solution 2: Lets start by assuming that X is equal to the larger flow rate and Y is equal to the smaller flow rate.
Now let’s set up our equations like this:
If we simplify these two equations we see that we get:
Since Y is 5 ft3/sec lower then Y = 13.5 |
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10 Review Questions
Lesson 2: Algebra | |||
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