Section 2.1 - Shelton State Community College

Download Report

Transcript Section 2.1 - Shelton State Community College

Section 2.1
Linear Equations in One Variable
Introduction
• A linear equation can be written in the
form ax = b*
• where a, b, and c are real numbers and
• a is not equal to 0.
• Recall that an equation will have an equal
sign, while an expression will not.
• We solve equations, we simplify
expressions.
Expression or Equation?
 3x  2  4  x  4
4( x  3)  2( x  1)  10
Solutions
• The solution to every linear equation is a
real number that satisfies the equation.
• Our strategy for solving is based on the
idea of isolating the variable.
Steps to Solving a Linear Equation
1. Got fractions? Multiply by the LCD.
2. Got decimals? Multiply by a power of 10.
3. Apply the distributive property on each side to
remove parentheses.
4. Combine like terms on each side.
5. Move variable terms to the left side, nonvariable terms to the right side. “Change sides,
change signs.” Simplify.*
6. Divide both sides by the coefficient of the
variable term.
Examples
5 x  4  21
2(3  2 x)  x  4
4( x  2)  2( x  3)  6
4[2 x  (3  x)  5]  (2  7 x)
More Examples
2[( x  1)  4]  5  [(6 x  7)  9 x]
7
 x6
8
3x  2 x  4

2
7
5
0.08x  0.12(260 x)  0.48x
Weird Stuff Happens at Step #5
• Sometimes, when you’re moving variable
terms to the left and non-variable terms
the right, all the variables cancel out on
both sides of the equation.
• When this happens, you have two
possibilities:
1.Contradiction
2.Identity
Contradictions
• All the variables cancel out
• The resulting statement, which is usually
0 = some number, is FALSE
• The original linear equation has NO
SOLUTION, which can be represented
symbolically by Ø.
Identities
• All the variables cancel out
• The resulting equation, which is usually
0 = 0 or
some number = the same number, is TRUE
• The solution to the original equation is ALL
REAL NUMBERS
Still More Examples
 6 x  2 x  11  2(2 x  3)  4
4(2 x  7)  2 x  25  3(2 x  1)
4[6  (1  2 x)]  10x  2(10  3x)  8 x