Download A Quick Algebra Review and more Schemes and Mind Maps Algebra in PDF only on Docsity!
A Quick Algebra Review
**1. Simplifying Expressions
- Solving Equations
- Problem Solving
- Inequalities
- Absolute Values
- Linear Equations
- Systems of Equations
- Laws of Exponents
- Quadratics
- Rationals
- Radicals**
Simplifying Expressions
An expression is a mathematical “phrase.” Expressions contain numbers and variables, but not an equal sign. An equa tion has an “equal” sign. For example:
Expression: Equation: 5 + 3 5 + 3 = 8 x + 3 x + 3 = 8 (x + 4)(x – 2) (x + 4)(x – 2) = 10 x² + 5x + 6 x² + 5x + 6 = 0 x – 8 x – 8 > 3
When we simplify an expression, we work until there are as few terms as possible. This process makes the expression easier to use, (that’s why it’s called “simplify”). The first thing we want to do when simplifying an expression is to combine like terms.
For example:
Now you try: x² + 5x + 3x² + x³ - 5 + 3
[You should get x³ + 4x² + 5x – 2]
Order of Operations
PEMDAS – Please Excuse My Dear Aunt Sally, remember that from Algebra class? It tells the order in which we can complete operations when solving an equation. First, complete any work inside PARENTHESIS, then evaluate EXPONENTS if there are any. Next MULTIPLY or DIVIDE numbers before ADDING or SUBTRACTING. For example:
Inside the parenthesis, look for more order of operation rules - PEMDAS. We don’t have any exponents, but we do need to multiply before we subtract, then add inside the parentheses before we multiply by negative 2 on the outside.
Simplify:
= - 2[3-(-12)]
= - 2[3+12]
= - 2[15]
= - 30
Simplify:
x² + 10x – 6 – 5x + 4
= x² + 5x – 6 + 4
= x² + 5x – 2
There are many terms to look at! Let’s start with x². There are no other terms with x² in them, so we move on. 10x and 5x are like terms, so we add their coefficients together. 10 + (-5) = 5, so we write 5x. - 6 and 4 are also like terms, so we can combine them to get - 2. Isn’t the simplified expression much nicer?
Solving Equations
An equa tion has an equa l sign. The goal of solving equations is to get the variable by itself, to SOLVE for x =. In order to do this, we must “undo” what was done to the problem initially. Follow reverse order of operations – look for addition/subtraction first, then multiplication/division, then exponents, and parenthesis. The important rule when solving an equation is to always do to one side of the equal sign what we do to the other.
For example
Your Turn: 2(x - 1) = - 3
(you should get x = -1/2)
Solve: x + 9 = - 6
x = - 15
To solve an equation we need to get our variable by itself. To “move” the 9 to the other side, we need to subtract 9 from both sides of the equal sign, since 9 was added to x in the original problem. Then we have x + 9 – 9 = - 6 – 9 so x + 0 = - 15 or just x = - 15.
Solve: 5x – 7 = 2
+7 +
5x = 9
5x = 9 5 5
x = 9/
When the equations get more complicated, just remember to “undo” what was done to the problem initially using PEMDAS rules BACKWARDS and move one thing at a time to leave the term with the variable until the end. They subtract 7; so we add 7 (to both sides). They multiply by 5; we divide by five.
Solve: 7(x + 4) = 6x + 24 distribute 7x + 28 = 6x + 24
7x = 6x - 4
x = - 4
When there are variables on both sides of the equation, add or subtract to move them to the same side, then get the term with the variable by itself. Remember, we can add together terms that are alike!
Problem Solving
Many people look at word problems and think, “I’m really bad at these!” But once we accept them, they help us solve problems in life when the equation, numbers, and variables are not given to us. They help us THINK, logically.
One of the challenging parts of solving word problems is that you to take a problem given in written English and translate it into a mathematical equation. In other words, we turn words into numbers, variables, and mathematical symbols.
There are three important steps to “translating” a word problem into an equation we can work with:
Let’s look at an example:
The fence around my rectangular back yard is 48 feet long. My yard is 3ft longer than twice the width. What is the width of my yard? What is the length?
First, we have to make sure we understand the problem. So what’s going on here? Drawing a picture often helps with this step.
length
width
- Understand the problem
- Define the variables
- Write an equation
Yard
We know that the problem is describing a person’s rectangular yard. We also know that one side is the width and the other side is the length. The perimeter of (distance around) the yard is 48ft. To arrive at that perimeter, we add length + length + width + width, or use the formula 2 l + 2 w = p ( l = length, w = width, p = perimeter)
Check to see that it works
2(7) + 3 14 + 3
17ft = length
Your turn:
I have a box. The length of the box is 12 in. The height of the box is 5 in. The box has a total volume of 360 in.^2 What is the width of the box? Note: The formula for volume is V = lwh , where v = volume, l = length, w = width, and h = height.
(You should get w = 6in.)
Inequalities
We’ve all been taught little tricks to remember the inequality sign. For example; when given x < 10, we know that x is less than ten because x has the LITTLE side of the sign and 10 has the BIG side of the sign.
Solving inequalities is similar to solving equations; what you do to one side of an inequality, we must do to the other. If we are given x + 7 > 13 and asked to solve, we would undo the addition on the left side by subtracting 7 from both sides. We would then be left with x > 6 , which is our answer.
Suppose we were given ¼ x < 2. To undo the division, we would multiply both sides by 4. The result would be x < 8.
But if we had – ¼ x < 2 , we would multiply by both sided of the inequality by -4 and the rule is that when multiplying (or dividing) by a negative number , we must always flip the sign of an inequality. So we would get x > 2.
Substitute our newly found width and simplify using order of operations
So now we know that the yard is 7ft wide and 17ft long.
Let’s practice:
It’s always good to check our answer. To do this, plug values in for x. Let’s try 1 , since it is greater than 0 and an easier number to work with. 1 – 5 is -
- -4 times -2 is -8. Is -8 < 10? You bet!
Your turn: 4x + 2 > 10
[You should get x > 2]
Absolute Values
The absolute value of a number is its distance from zero on the number line. Since we can’t have negative distance, absolute values are always positive.
│5│= 5
│-7│= 7
│-8 + 2│= 6
When solving absolute value equations, we must consider that the number inside could have been negative before you applied the absolute value!
For example: │x + 2│= 3
x + 2 and x + 2
3 -
When we take the absolute value, both are equal to 3.
So we end up with TWO SOLUTIONS for this equation.
Solve:
x – 5 > - 5
x > 0
Just as with an equation, we start here by dividing both sides by - 2. Since we are dividing by a negative, we must flip the inequality sign.
Next, we add five to both sides and we are left with x > 0.
3x – 2y = 4 -3x -3x
y = 3x - 2 2 Now it is easy to compare the equation we have to the slope-intercept form and identify the slope and the y-intercept of this line. y = 3x - 2 2
y = mx + b, so the slope, m= 3/2 and the y-intercept, b=-2.
Your turn: Find the slope and y-intercept of the linear equation x – 3y = - 6
[You should get m = 1/3 and b = 2]
Writing the Equation of a Line in Slope Intercept Form
Let’s try this example:
What is the equation of a line that has a slope of - 2/3 and the point (9, 2)?
Since we know that lines can be written in the form y =mx + b, use this equation as a template and substitute the information that you have into the equation to find the y-intercept.
y = mx + b 2 = - 2/3 ( 9 ) + b 2 = -2(3) + b 2 = -6 + b 2 + 6 = b 8 = b
Substitute - 2/3 in for the slope m, and (9, 2) in for the point (x, y) in the equation.
To solve we subtract 3x from both sides of the equation. Be careful to note that the 2y remaining is negative. Then divide both sides by -2.
Now you have everything you need to write the equation of the line with slope -2/3 through the point (9,2). Put the slope and y-intercept back into the equation and you have y = -2/3x + 8.
Sometimes, the slope isn’t given. Let’s try another example:
Find the equation of a line that passes through the points (3, 1) and (6, 2).
The first thing we need to do is find the slope of the line through these two points. Remember that:
slope = m =^2 2 1
y y x x
m =^2 6 3
m =^1 3
Now we can substitute the slope and one of the points (either one will work) into the equation y =mx + b and solve for b.
We’ll use the point (3, 1) for (x, y) y = mx + b 1 = 1 /3 ( 3 ) + b 1 = 1 + b 1 - 1 = b 0 = b Now we have everything we need to write the equation of the line with points (3, 1) and (6, 2). We found that the slope m =1/3 then we found the y-intercept. Putting them into the equation you have y = 1/3x + 0
Your turn: Find the equation of the line that goes through the points (5, 4) and (1, - 4).
[You should get y = 2x - 6]
No one is trying to confuse you with all of these subscripts…Its just that we have two points (x, y) so we call the first one (x 1 , y 1 ) and the second (x 2 , y 2 ) Substitute (3, 1) in for one point and (6, 2) in for the other. Be careful to go the same direction when you subtract and WATCH YOUR SIGNS!
is called a system of equations. The solution to a system of equations must satisfy both equations. There are several different methods to accomplish this. We will review two: graphing and substitution.
Graphing:
What does the graph for a system of equations look like? Well, one equation makes a line, so two equations make….two lines! For the above example, the graph would look like this:
Since each point on a line is a solution to the equation – the point that satisfies both equations is the point where they intersect. The solution for this system of equations is the point (1, 2).
Your turn: Solve the system of equations by graphing: y = -1/2 x – 2 y = -7/2 x + 4
[You should get (2, -3)]
Note: Often we need to solve the equations for y-intercept form before graphing them.
y = 2x y = - x + 3
Substitution:
We can also solve systems of equations by substitution. Let’s look at another example: -2x + y = - x – 2y = - To use the substitution method, we first need to solve one equation for either x or y. Looking at the equations, the first one could easily be solved for y without dividing, or the second equation could be solved for x. Either choice works fine. Let’s start with the first equation.
-2x + y = - +2x +2x y = 2x -
So our two equations now look like this: y = 2x - x – 2y = -
Now we know y is equal to 2x – 1. We can SUBSTITUTE this equality for y in the second equation.
y = 2x - 1
x – 2 y = - 4
x - 2(2x – 1 ) = - 4 x - 4x + 2 = - -2 -
-3x = - -3 -
x = 2
y = 2( 2 ) - 1 y = 4 – 1 y = 3
We know that y = 2x - 1 , so anywhere we see y in the other equation, we substitute 2x - 1.
Don’t forget to distribute your negative sign…
Now we know half the solution, x = 2
The point where the lines will intersect is (2, _)
But since x = 2 , we can substitute 2 in for x in either of the equations and solve for y to find the other value of our point (2, 3)
Your turn: x^5 (x^2 )-
[you should get x^11 ]
Multiplying Polynomials – FOIL
When multiplying polynomials, we need to make sure to distribute each of the terms to every other term. There are several ways of doing this; we will focus on the most common one.
Do you remember FOIL? FOIL is an acronym we use to help us multiply polynomials. FOIL stands for F irst, O uter, I nner, L ast. It depicts the order in which we multiply. Let’s try an example:
Simplify:
(2x + 1)(x – 2)
Simplify:
(- 6 p^2 q)(- 4 p^4 q^5 )
= (- 6 )(- 4 )p2+4q1+
=24p^6 q^6
Simplify:
5 x -^2 – 3 + 2x^0
= 2 5 x
=^2
5 (^) x - 1
Check for order of operations first. We can multiply coefficients together then use the laws of exponents to combine bases that are alike.
When we multiply powers, we add the exponents.
Note: If the base has a power of 1 you may want to write it in so you don’t forget to add the power.
2x^2 – 2x
2x^2 – 2x + x – 2
2x^2 – x – 2
Your turn: (2x + 3)(x – 1)
[You should get 2x^2 + x – 3 ]
Quadratics
Quadratic equations are equations that have a variable to the second power, like x^2 + x = 6. Since x^2 and x are not like terms they can not be combined. We need a new way for finding solutions to quadratic equations.
Solving by factoring:
To solve an equation by factoring, one side of the equation must be equal to zero. In the equation 3x^2 = x. We would need to subtract x from both sides, so we would have 3x^2 – x = 0.
Our goal with factoring is to find two terms that multiply together to give us zero. Since x is a factor in both 3x^2 and x , we can factor out an x from the equation and rewrite it, x (3x – 1) = 0
Our first step is to multiply the first terms together (red). Then we multiply the outer terms (blue).
Next, we multiply the inner terms (green), followed by the last terms (pink).
Now, we combine like terms and our answer is simplified!
- 2, we get -1, but if we add (-2) + 3, we get 1, which is exactly what we are looking for. We now know that one set of parentheses must have a -2, and the other must have a three.
(x – 2)(x + 3) = 0
Again, we know that two things multiplied together will only equal zero if one of them is zero. So we set x – 2 and x + 3 equal to zero and solve for x. x – 2 = 0 x + 3= 0 x = 2 x = -
Our solutions to the quadratic equation are x = -3 and 2.
Let’s try another example:
2 x^2 – 7x – 4 = 0
(2x )(x )
(2x + 1)(x – 4)
2x + 1 = 0 x – 4 = 0
- 1 -1 - 4 - 4
2x = - 1 x = 4 2 2
x = - 1/
Your turn: 2 x^2 + 7x = - 4
[You should get x = -3 and -1/2]
This one’s a bit harder because of the 2 in front of the x^2. When setting up our parentheses, we have to think about how our F irst terms can multiply together to give us 2. The only way to do this is by 2x and x.
Here, it really becomes a puzzle. We have to find two numbers that multiply together to give us four. Because of the 2 in front of the x^2 , we know that double one of these numbers plus the other must give us - 7. Luckily 4 has only two sets of factors: 1 & 4 and 2 & 2. One of these must be negative and one positive, since the 4 is negative. If you double - 4, you get - 8. If you add 1 to - 8, you get - 7, which is exactly what we are looking for. Since the 4 must be multiplied by the 2 in the first set of parentheses, it has to go in the other set of parenthesis.
Then solve both for 0 to find our solutions.
Solving by the Quadratic Formula:
x =
(^2 ) 2
b b ac a
Factoring can be the easiest way to solve a quadratic equation when the coefficient of x^2 is one. But as the equations get harder there is another strategy we can use. The quadratic formula solves equations of the form: ax^2 + bx + c = 0 ( the equation must be solved for zero). where; a is the coefficient (“number” of x^2 ’s we have) of x^2 b is the coefficient (“number” of x’s we have) of x, and c is the constant term of our equation when solved for zero.
In the equation, x^2 + x – 6 = 0, a= 1, b= 1 and c= -6_._ By substituting these values of a, b, and c into the quadratic formula we are able to solve for x.
x =
(^2 ) 2
b b ac a
1 (1)^2 (4)(1)( 6) (2)(1)
x
1 1 ( 24) 2
x
1 25 2
x
1 5 1 5 2 2
x x
6 4 2 2
x x
x = -3, 2
Make sure that when b is squared that it is in parenthesis. If it isn’t, and the value is negative, we’ll get the wrong solution.
Be careful with the signs inside the radical. Remember, negative times a negative equals a positive.
Notice that there are TWO solutions.
