Transcript Linear Equations in One Variable

Linear Equations in One Variable
Objective: To find solutions of linear
equations.
Linear Equations in One Variable
An equation in x is a statement that two
algebraic expressions are equal. For example,
3x – 5 = 7 is an equation.
Solutions of Equations
• To solve an equation in x means to find all values of x
for which the equation is true. Such values are called
solutions.
Solutions of Equations
• To solve an equation in x means to find all values of x
for which the equation is true. Such values are called
solutions.
• For instance, x = 4 is the solution of the equation
3x – 5 = 7 since replacing x with 4 makes a true
statement.
Identity vs. Conditional Equation
• Identity- An equation that is true for every real
number in the domain of the variable.
Identity vs. Conditional Equation
• Identity- An equation that is true for every real
number in the domain of the variable.
2
x
 9  ( x  3)( x  3)
• For example,
is an identity since it is always true.
Identity vs. Conditional Equation
• Conditional Equation- An equation that is true for
just some (or even none) of the real numbers in the
domain of the variable.
Identity vs. Conditional Equation
• Conditional Equation- An equation that is true for
just some (or even none) of the real numbers in the
domain of the variable.
• For example, x 2  9  0
is conditional because x = 3 and x = -3 are the only
solutions.
Definition of a Linear Equation
• A linear equation in one variable x is an equation
that can be written in the standard form ax + b = 0,
where a and b are real numbers and a cannot
equal 0.
Example 1a
• Solve the following linear equation.
3x  6  0
Example 1a
• Solve the following linear equation.
3x  6  0
3x  6
x2
Example 1b
• You Try
• Solve the following linear equation.
5x  4  3x  8
Example 1b
• You Try
• Solve the following linear equation.
5x  4  3x  8
2 x  12
x  6
Example 2
• Solve the following linear equations.
6( x  1)  4  3(7 x  1)
Example 2
• Solve the following linear equations.
6( x  1)  4  3(7 x  1)
6x  6  4  21x  3
6x  2  21x  3
Example 2
• Solve the following linear equations.
6( x  1)  4  3(7 x  1)
6x  6  4  21x  3
6x  2  21x  3
15x  5
1
x
3
Linear Equations in other forms
• Some equations involve fractions. Our goal is to get
rid of the fraction by multiplying by the common
denominator.
x 3x

2
3 4
Linear Equations in other forms
• Some equations involve fractions. Our goal is to get
rid of the fraction by multiplying by the common
denominator.
• The common denominator is 12. Multiply everything
by 12.
x
3x
12  12  2 12
3
4
Linear Equations in other forms
• Some equations involve fractions. Our goal is to get
rid of the fraction by multiplying by the common
denominator.
• The common denominator is 12. Multiply everything
by 12.
x
3x
12  12  2 12
3
4
4x  9x  24
13x  24
24
x
13
Linear Equations in other forms
• You Try.
• Solve the following equation.
2x x
 4
3 5
Linear Equations in other forms
• You Try.
• Solve the following equation.
2x x
 4
3 5
2x
x
15  15  4 15
3
5
10x  3x  60
13x  60
60
x
13
Extraneous Solutions
• When multiplying or dividing an equation by a
variable expression, it is possible to introduce an
extraneous solution.
• An extraneous solution is one that you get by solving
the equation but does not satisfy the original
equation.
Example 4
• Solve the following.
1
3
6x

 2
x2 x2 x 4
Example 4
• Solve the following.
1
3
6x

 2
x2 x2 x 4
1
3
6x
( x  2)( x  2)

( x  2)( x  2)  2
( x 2  4)
x2 x2
x 4
Example 4
• Solve the following.
1
3
6x

 2
x2 x2 x 4
1
3
6x
( x  2)( x  2)

( x  2)( x  2)  2
( x 2  4)
x2 x2
x 4
( x  2)  3( x  2)  6 x
Example 4
• Solve the following.
1
3
6x

 2
x2 x2 x 4
1
3
6x
( x  2)( x  2)

( x  2)( x  2)  2
( x 2  4)
x2 x2
x 4
( x  2)  3( x  2)  6 x
x  2  3x  6  6 x
x  2  3x  6
4 x  8
x  2
Example 4
• Solve the following.
• If we try to replace each x value with x = -2, we will
get a zero in the denominator of a fraction, which we
cannot have. There are no solutions.
1
3
6x

 2
x2 x2 x 4
4 x  8
x  2
Example 4
• You Try
• Solve the following.
15
6
4  3
x
x
Example 4
• You Try
• Solve the following.
15
6
4  3
x
x
15
6
x  4 x  x  3x
x
x
15  4x  6  3x
 7 x  9
x  9/7
Intercepts
• To find the x-intercepts, set y equal to zero and solve
for x.
Intercepts
• To find the x-intercepts, set y equal to zero and solve
for x.
• To find the y-intercepts, set x equal to zero and solve
for y.
Intercepts
• To find the x-intercepts, set y equal to zero and solve
for x.
• To find the y-intercepts, set x equal to zero and solve
for y.
• Find the x and y-intercepts for the following
equation.
y  4x 1
Intercepts
• To find the x-intercepts, set y equal to zero and solve
for x.
• To find the y-intercepts, set x equal to zero and solve
for y.
• Find the x and y-intercepts for the following
equation.
y  4x 1
• x-intercept (y = 0)
0  4x  1
x  1 / 4
(1 / 4,0)
Intercepts
• To find the x-intercepts, set y equal to zero and solve
for x.
• To find the y-intercepts, set x equal to zero and solve
for y.
• Find the x and y-intercepts for the following
equation.
y  4x 1
• x-intercept (y = 0) 0  4x  1 x  1 / 4
• y-intercept (x = 0) y  4(0)  1
y 1
(1 / 4,0)
(0,1)
Intercepts
• You Try
• Find the x and y-intercepts for the following
equation. 2 y  x  1
Intercepts
• You Try
• Find the x and y-intercepts for the following
equation. 2 y  x  1
• x-intercept (y = 0)
• y-intercept (x = 0)
0  x 1
x 1
(1,0)
2 y  (0)  1 y  1 / 2 (0,1 / 2)
Class work
• Pages 94-95
• 23, 25, 29, 31, 34, 35, 46, 47
Homework
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Pages 94-95
3-36, multiples of 3
45-53 odd
71,73,75