# Beginning Algebra

September 12, 2016
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We saw in the prealgebra section that we can combine like terms, like 4n + 2n = 6n.

We also used the distributive law to expand numbers; here we distribute the 6:

6 * 17 = 6(10 + 7) = 6 * 10 + 6 * 7 = 60 + 42 = 102.

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

In a typical algebra problem you're asked to expand something like 3(x + 5).

3(x + 5) = 3(x) + 3(5) = 3x + 15.

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

Let's see:

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 .

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

= 12x

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

(4x + 5)(3x – 12)

F O I L

= (4x)(3x) – (4x)(12) + (5)(3x) – (5)(12)

= 12x

12x

(n + 3)(n – 3) = n

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

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.

Factor the expression 8x

Look for the common factor:

8x

The common part is 2 * 2 * x * x = 4x

8x

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.

Factor by grouping: 12x

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

12x

= 3x * 4x – 3x * 3y + 2 * 4x – 2 * 3y

= 3x(4x – 3y) + 2(4x – 3y)

= (3x + 2)(4x – 3y) .

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

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.

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

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## Simplifying Algebraic Expressions

We saw in the prealgebra section that we can combine like terms, like 4n + 2n = 6n.

We also used the distributive law to expand numbers; here we distribute the 6:

6 * 17 = 6(10 + 7) = 6 * 10 + 6 * 7 = 60 + 42 = 102.

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 3(x + 5).

3(x + 5) = 3(x) + 3(5) = 3x + 15.

**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.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 .

## Multiplying Algebraic Expressions

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!12x

^{2}- 48 x + 15 x - 60**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 8x

^{3}+ 20x^{2}.Look for the common factor:

8x

^{3}= 2 * 2 * 2 * x * x * x and 10x^{2}= 2 * 2 * 5 * x * x.The common part is 2 * 2 * x * x = 4x

^{2}. Write it as8x

^{3}+ 20x^{2}= (4x^{2})(2x) + (4x^{2})(5) = (4x^{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) .

## 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.

**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.
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