Transcript solutions

Chapter 2
Linear
Functions and
Equations
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2.2
Linear Equations
♦
Learn about equations and recognize a linear
equation
♦ Solve linear equations symbolically
♦ Sole linear equations graphically and
numerically
♦ Solve problems involving percentages
♦ Apply problem-solving strategies
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Equations
An equation is a statement that two
mathematical expressions are equal.
Some examples of equations are:
x  15  9x  1
xy  x  y  x
2
x 2  2x  1 2x
3
1 3  4
z50
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Solutions to Equations
To solve an equation means to find all the
values of the variable that make the equation a
true statement.
Such values are called solutions.
The set of all solutions is the solution set.
Solutions to an equation satisfy the equation.
Two equations are equivalent if they have the
same solution set.
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Types of Equations in One Variable
Contradiction – An equation for which there is
no solution.
• Example:
2x + 3 = 5 + 4x – 2x
• Simplifies to
2x + 3 = 2x + 5
• Simplifies to
3=5
• FALSE statement – there are no values
of x for which
3 = 5. The equation
has NO SOLUTION.
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Types of Equations in One Variable
Identity – An equation for which every
meaningful value of the variable is a
solution.
•
Example:
2x + 3 = 3 + 4x – 2x
• Simplifies to
2x + 3 = 2x + 3
• Simplifies to
3=3
• TRUE statement – no matter the value of
x, the statement 3 = 3 is true. The
solution is ALL REAL NUMBERS
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Types of Equations in One Variable
Conditional Equation – An equation that is satisfied
by some, but not all, values of the variable.
•
Example 1:
2x + 3 = 5 + 4x
• Simplifies to
2x – 4x = 5 – 3
• Simplifies to
2x = 2
• Solution of the equation is: x = 1
•
Example 2:
x2 = 1
• Solutions of the equation are: x = 1, x = 1
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Linear Equations in One Variable
A linear equation in one variable is an equation
that can be written in the form
ax + b = 0,
where a and b are real numbers with a ≠ 0.
If an equation is not linear, then we say it is a
nonlinear equation.
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Linear Equations in One Variable
Rules of algebra can be used to write any linear
equation in the form ax + b = 0.
A linear equation has exactly one solution
b
x .
a
Examples of linear equations:

2x  4  x
x  12  0

2 1 4x  16x


x  5  3 x 1  0
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Symbolic Solutions
Linear equations can be solved symbolically, and
the solution is always exact.
To solve a linear equation symbolically, we usually
apply the properties of equality to the given
equation and transform it into an equivalent
equation that is simpler.
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Properties of Equality
Addition Property of Equality
If a, b, and c are real numbers, then
a = b is equivalent to a + c = b + c.
Multiplication Property of Equality
If a, b, and c are real numbers with c ≠ 0, then
a = b is equivalent to ac = bc.
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Example: Solving a linear equation
symbolically
Solve the equation 3(x – 4) = 2x – 1. Check your
answer.
Solution
Apply the distributive property


3 x  4  2x  1
3x  12  2x  1
3x  2x  12  12  2x  2x  1 12
x  11
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Example: Solving a linear equation
symbolically
The solution is 11. Check the answer.


3 x  4  2x  1
?
3 11  4   2  11  1
21 21
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Example: Eliminating fractions
Solve the linear equation
t 2 1
1
 t  5  3  t 
4
3
12
Solution
To eliminate fractions, multiply each side (or
term in the equation) by the LCD, 12.
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Example: Eliminating fractions

 12 t  12 5  12 3  t




4
3
12
3 t  2 4t  60  3  t 
12 t  2
3t  6  4t  60  3  t
t  6  57  t
2t  63
63
t
2
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The solution
63
is 
.
2
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Example: Eliminating decimals
Solve the linear equation.
0.03  z  3   0.5  2z  1  0.23
Solution
To eliminate fractions, multiply each side
(or term in the equation) by the LCD, 12.
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Example: Eliminating decimals
To eliminate decimals, multiply each side (or
term in the equation) by 100.
   
3 z  3  50 2z  1 0.23
0.03 z  3  0.5 2z  1  0.23
3z  9  100z  50  23
97z  59  23
97z  82
82
z
97
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The solution
82
is 
.
97
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Intersection-of-Graphs Method
The intersection-of-graphs method can be used
to solve an equation graphically.
STEP 1:
STEP 2:
STEP 3:
Set y1 equal to the left side of the
equation, and set y2 equal to the right
side of the equation.
Graph y1 and y2.
Locate any points of intersection. The
x-coordinates of these points
correspond to solutions to the equation.
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Example: Solving an equation
graphically and symbolically
Solve 2x  1 1 x  2 graphically and symbolically.
2
Solution
Graph y1  2x  1
1
and y 2  x  2
2
Intersect at (2, 3),
solution is 2.
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Example: Solving an equation
graphically and symbolically
1
2x  1 x  2
2
1
2x  x  3
2
2 3
2
 x  3
3 2
3
3
x3
2
x2
Solution is 2, agrees with graphical.
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Example: Solving an equation
numerically


1
Solve 3 2x    x  0 numerically to the
3
nearest tenth.
Solution
Enter Y1 
32x    X / 3
Make a table for y1,
incrementing by 1.
This will show the solution
is located in the interval 1
< x < 2.
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Example: Solving an equation
numerically
Make a table for y1, start
at 1, increment by 0.1.
Solution lies in
1.4 < x < 1.5
Make a table for y1, start at
1.4, increment by 0.01.
Solution lies in
1.43 < x < 1.44
Solution is 1.4 to the
nearest tenth
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Example: Solving for a variable
The area of a trapezoid with bases a and b and
1
height h is given by A  h a  b .
2
Solve this equation for b.
Solution
Multiply each side by 2
2A  h  a  b 
Divide each side by h
2A
ab
h
2A
ab
h
Subtract a from each side
isolates b
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Solving Application Problems
STEP 1: Read the problem and make sure you
understand it. Assign a variable to
what you are being asked. If
necessary, write other quantities in
terms of the variable.
STEP 2: Write an equation that relates the
quantities described in the problem.
You may need to sketch a diagram and
refer to known formulas.
STEP 3: Solve the equation and determine the
solution.
STEP 4: Look back and check your solution.
Does it seem reasonable?
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Example: Solving an application
involving motion
In 1 hour an athlete traveled 10.1 miles by running
first at 8 miles per hour and then at 11 miles per
hour. How long did the athlete run at each speed?
Solution
STEP 1:
Let x represent the time in hours
running at 8 mph, then 1 – x represents
the time spent running at 11 mph.
x: Time spent running at 11 miles per hour
1 – x: Time spent running at 9 miles per hour
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Example: Solving an application
involving motion
STEP 2:
d = rt; total distance is 10.1
d  r1t1  r2t2
10.1  8 x  111  x 
STEP 3:
Solve symbolically
10.1  8 x  111  x 
3x  0.9
x  0.3
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Example: Solving an application
involving motion
STEP 3:
The athlete runs 0.3 hour (18 min) at 8
miles per hour and 0.7 hour (42 min) at
11 miles per hour.
STEP 4:
check the solution
 
 
8 0.3  11 0.7  10.1
This sounds reasonable. The runner’s average
speed was 10.1 miles per hour so the runner must
have run longer at 11 miles per hour than at 8
miles per hour.
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Example: Mixing acid in chemistry
Pure water is being added to 153 milliliters of a
30% solution of hydrochloric acid. How much water
should be added to dilute the solution to a 13%
mixture?
Solution
STEP 1:
Let x be the amount of pure water
added to the 153 ml of 30% acid to
make a 13% solution
x: Amount of pure water to be added
x + 153: Final volume of 13% solution
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Example: Mixing acid in chemistry
STEP 2:
Pure water contains no acid, so the
amount of acid before the water is
added equals the amount of acid after
water is added. Pure acid before is
30% of 153, pure acid after is 13% of x
+ 153.
The equation is:
0.13  x  153   0.30 153 
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Example: Mixing acid in chemistry
STEP 3:
Solve: divide each side by 0.13
0.13  x  153   0.30 153 
x  153 
x
 
0.30 153
0.13
  153
0.30 153
0.13
x  200.08
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Example: Mixing acid in chemistry
STEP 3:
We should add about 200 milliliters of
water.
STEP 4:
Initially the solution contains 0.30(153) =
45.9 milliliters of pure acid. If we add
200 milliliters of water to the 153
milliliters, the final solution is 353
milliliters, which includes 45.9 milliliters
of pure acid.
45.9
 0.13 or ≈ 13%
Concentration 
353
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