dan's beginning algebra lesson

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Simplifying Expressions

We saw in the prealgebra section that we can combine like terms, like 2n + 3n = 5n.

We also used the distributive law to expand numbers; here we distribute the 7:

7 * 13 = 7(10 + 3) = 7 * 10 + 7 * 3 = 70 + 21 = 91.

Also this works with variables; we state the law usually as a(b + c) = ab + ac.

Example:  In a typical algebra problem you're asked to expand something like 5(x + 7).

5(x + 7) = 5(x) + 5(7) = 5x + 35. This problem was. . . no problem!

 

Example:  When faced with an expression like 4x + 5(3x – 12), what do we do first?

Let's see: PEMDAS says work in parentheses first, but 3x and 12 are unlike.

Hmm, let's try the distributive law again, but just with the multiplied 5:

4x + 5(3x – 12)  =  4x + 5(3x) – 5(12)  =  4x + 15x – 60  =  19x – 60 .

That's our answer, simplified.

 

 

    Multiplying Stuff Out

    What about (4x + 5)(3x – 12) ?  Is this the same as 4x + 5(3x – 12) ?

    No, the parentheses change it. Here we can use the distributive law twice:

    (4x + 5)(3x – 12)

    = (4x + 5)(3x) – (4x + 5)(12)

    = 12x^2 + 15x – 48x – 60 (remember to change the sign on that last term)

    = 12x^2 – 33x – 60.  That worked, but it was long.

    Is this the only way? No. The best way? No. Use the "FOIL system":

    First, Outside, Inside, Last.

    (4x + 5)(3x – 12)

            F             O            I            L
    = (4x)(3x) – (4x)(12) + (5)(3x) – (5)(12)

    = 12x^2 – 48x + 15x – 60  =  12x^2 – 33x – 60.  Better!

    12 x 2 - 48 x + 15 x - 60


    Here's another example:

    (n + 3)(n – 3) = n^2 – 3n + 3n – 9 = n^2 – 9.

    Notice the "middle terms" cancel, and we're left with what's called the difference of two squares.

    In general, (a + b)(a - b) = a^2 - b^2.  Also see the basic factoring and more factoring sections.



    Basic Factoring

    We saw the distributive law work to expand things out:

    6 (2x + 7) = 6 * 2x + 6 * 7 = 12x + 42.

    The steps can be reversed to give a factorization:

    12x + 42 = 6 * 2x + 6 * 7 = 6 (2x + 7).

    This is called factoring out a common factor. In this case the common factor was 6.


    Example:  Factor the expression  8 x^3 + 20 x^2.

    Look for the common factor:

    8 x^3 = 2 * 2 * 2 * x * x * x   and   10 x^2 = 2 * 2 * 5 * x * x.

    The common part is  2 * 2 * x * x  =  4 x^2.  Write it as

    8 x^3  +  20 x^2 = (4 x^2)(2x)  +  (4 x^2)(5) = (4 x^2)(2x + 5).   That's factored!


    Another method that might work (if we're lucky) is called the grouping method. This works by

    grouping up equal size blocks of terms and factoring out a common factor from each, then

    hoping the "left-over factors" are equal. That's where the luck, or sometimes skill, comes in.


    Example:  Factor by grouping:  12x^2 – 9xy + 8x – 6y .

    There's no common factor for all terms, but we can split the pairs:

         12x^2  –   9xy      +      8x    –   6y  

    =  3x * 4x – 3x * 3y   +   2 * 4x – 2 * 3y

    =  3x(4x – 3y)   +   2(4x – 3y)  

    =  (3x + 2)(4x – 3y) .

    You can find more factoring in the Intermediate Algebra section.



    Solving Linear Equations

    An equation has to have an equals sign, as in  3x + 5 = 11 .  It's two expressions set equal to one another.

    A solution to an equation is a number that can be plugged in for the variable to make a true number statement.

    For example, putting 2 in for x above in 3x + 5 = 11 gives

    3(2) + 5 = 11 , which says 6 + 5 = 11 ; that's true! So 2 is a solution.


    But how to start with the equation, and get (not guess) the solution?

    We use some principles of equality, such as:

    • Adding the same thing (number or variable term) to both sides of an equation
    • Subtracting the same thing (number or variable term) from both sides
    • Multiplying or Dividing both sides by a non-zero quantity.

    These all keep the equation "balanced" like a scale.


    Example:  Let's solve our given equation from scratch:

     3x  +  5   =   11        try to isolate  x  so we can see what it is

            – 5      – 5       subtract 5 from each side to get constants on the right

    3x            =    6        the intermediate result

    3x / 3      =    6 / 3   divide both sides by 3 to isolate the x

         x         =      2      the solution (same as before!)  We've solved the equation.


    The thing that makes this equation linear is that the highest power of x is x^1 ; no x^2 or other powers, no variables in the denominator, and no square roots or other funny stuff.

    For "quadratic equations" go to intermediate algebra.