Download Evaluating algebraic expressions:

Evaluating algebraic expressions:
PEMDAS
Parenthesis
Exponents
Multiplication
Division
Add
Subtract
Inverse Operations:
Original
Inverse
+
-
x,

(opposite)
2
sin
Sin-1
cos
cos-1
tan
tan-1
Solving equations:
Single variable
1. Remove any parentheses and combine like terms, on each side of the equation.
2. Clear any fractions by multiplying all terms on both sides by the denominator
3. Group like terms on each side of the equal sign.
4. Isolate the variable on one side of the equation by using inverse operations
5. Divide or multiply both sides of the equation by the coefficient of the variable to solve
6. Check the solution by substituting your answer into the original equation wherever you
have a variable. If the solution is correct, both sides of the equation will be equal.
Solve for the given variable:
1) 5c - 4 - 2c + 1 = 8c + 2
2) 8(m + 5) = 16
Literal Equations
Can be solved the same way as all other equations. Just remember you can
only combine like terms and it is ok to have an answer with multiple parts.
Solve for x:
3) xy - d = m
x
z y
4)
y
Solving Quadratic Equations
Any equations with an x2
Always keep your squared variable positive
1-Set your equation equal to 0
2-Identify the type of equation you have
3-Pick a method of factoring
1. x2 – 5x = 24
2. x2 + 5x = 0
Solving Systems of Equations Algebraically - Using Substitution
The substitution method is used to eliminate one of the variables by replacement when
solving a system of equations.
1. Solve one of the equations for either "x =" or "y =".
This example solves the second equation for "y =".
3y - 2x = 11
y = 9 - 2x
2. Replace the "y" value in the first equation by what "y" now equals. Grab the "y" value
and plug it into the other equation.
3(9 - 2x) - 2x = 11
3. Solve this new equation for "x".
(27 - 6x) - 2x = 11
27 - 6x - 2x = 11
27 - 8x = 11
-8x = -16
x=2
4. Place this new "x" value into either of the ORIGINAL equations in order to solve for "y".
Pick the easier one to work with!
y = 9 - 2(2)
y=9-4
y=5
Solve: x = y - 5
2x – y = 7
Solving Systems of Equations Algebraically - Using Addition
or Subtraction
Think of the adding or subtracting method as simply "eliminating" one of the variables to
make your life easier.
1. Solve this system of equations:
x - 2y = 14
x + 3y = 9
a. First, be sure that the variables are "lined up" under one another. In this problem, they
are already "lined up".
x - 2y = 14
x + 3y = 9
b. Decide which variable ("x" or "y") will be easier to eliminate. In order to eliminate a
variable, the numbers in front of them (the coefficients) must be the same or negatives of
one another. Looks like "x" is the easier variable to eliminate in this problem.
x - 2y = 14
x + 3y = 9
c. Now, subtract to eliminate the "x" variable.
(Remember: when you subtract signed numbers, you change the signs and add)
x - 2y = 14
-x - 3y = - 9
- 5y = 5
d. Solve this simple equation.
-5y = 5
y = -1
e. Plug "y = -1" into either of the ORIGINAL equations to get the value for "x".
x - 2y = 14
x - 2(-1) = 14
x + 2 = 14
x = 12
Ex: The sum of two numbers is 19. Their difference is 9. Find each number.
Solving a Linear Quadratic System Algebraically
Through substitution
We'll use the following example.
Solve algebraically:
y = x2 - x - 6
y = 2x - 2
First we solve for one
y = 2x - 2
Since this is already
of the variables in the
done for us in this
linear equation.
example, we can go to
the next step.
2
Next we substitute for
y=x -x-6
that variable in the
2x-2 = x2 - x - 6
Subtract 2 from both
quadratic equation, and
2x = x2 - x - 4
sides.
solve the resulting
2
Subtract 2x from both
equation.
0 = x - 3x - 4
sides.
0 =(x-4)(x+1)
Factor.
x-4=0
x+1=0
Set each factor =0 and
x=4
x = -1
solve.
We now have two values for x, but we still need to find the corresponding values
for y.
We find the yvalues by
substituting each
value
of x into the linear
y = 2x - 2
y = 2(4) - 2
y=8-2
y=6
(4,6)
equation.
{(4,6),(-1,-4)}
y = 2x - 2
y = 2(-1) - 2
y = -2 - 2
y = -4
(-1,-4)