Transcript PowerPoint Lesson

Basic Algebra
Grade: 9th
Gustavo Miranda
LRC 320
Vocabulary
Expressions:
• Don’t contain an “=” sign
• Contain variables
• Values are substituted into variables
Variables:
• Represented as letters, i.e. “x”
Equations:
• Also contain variables
• Want to solve for specified variable
• Contain an “=” sign
Evaluating Basic Expressions
• When evaluating, we want to plug in our given
value into the variable and simplify by doing
the indicated operation.
Basic Expression:
Ex:
x + 5; when x = 2
Evaluating Advanced Expressions
• For advanced expressions, we want to
combine like terms so that the expression
becomes simpler and easier to evaluate.
Advanced Expression:
Ex:
Want this expression to be
simplified to:
x + 12 – 10x - 30;
When x = 3
-9x – 18; when x = 3
Examples for Evaluating Expressions
Ex: Evaluate the expression.
x – 5; when x = 10
Ex: Evaluate the expression.
2x + 5 – 3x + 10; when x = 2
Solution:
Solution:
x – 5 (plug in given value for x)
= 10 – 5
=
So, the solution to the
expression is 5.
2x + 5 – 3x + 10 (Simplify terms with x)
= -1x + 5 + 10 (Now simplify constants)
= -1x + 15 (Plug in value for x)
= -1(2) + 15 (Simplify)
= -2 + 15
= 13, solution is 13
Solving Basic Equations
• With equations, they are expressions that are
equal to a number.
• The goal with equations is to solve for a variable
that contains a number for its solution.
Ex: (Basic equation)
x + 3 = 10
Solving Advanced Equations
• With advanced expressions we wanted to simplify first; the
same process is done with equations with more terms.
Ex: Solve for x.
x + 3 – 3x = 15
Want a simplified equation.
-2x + 3 = 15
Examples for Solving Equations
Ex: Solve for x.
x + 5 = 10
Ex: Solve for x.
x - 5 - 3x - 1 = 9
Solution:
Solution:
x + 5 = 10 (Subtract 5 from both sides of equation)
- 5 -5
x – 5 – 3x – 1 = 9 (combine like terms)
-2x – 6 = 9 (Add 6 to both sides of equation)
+ 6 +6
So, x = 5
-2x = 15
-2
(Now divide -2 to both sides of equation)
-2
So, x = -15/2
Practice Exercises
Evaluate the following expressions:
1) 2x + 3; when x = 2
2) x – 3; when x = 10
3) 3x – 5x +1; when x = 3
Solve the following equations for x.
a) x + 5 = 2
b) 2x – 3 + x = 6
c) 3x – 10 = 5
Solutions:
1) 7 2) 7
3) -5
a) x = -3
b) x = 3
c) x = 5