**dan's beginning algebra lesson**

**dan's beginning algebra lesson**

*©1997-2016 - all rights reserved - dan bach @dansmath*

**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*":

**F**irst, **O**utside, **I**nside, **L**ast.

**(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!

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*.