dan's beginning algebra lesson
©1997-2016 - all rights reserved - dan bach @dansmath
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!
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.
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) .
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.