Transcript Ch2-Sec 2.3

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2.3 – Slide 1
Chapter 2
Equations, Inequalities, and
Applications
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2.3 – Slide 2
2.3
More on Solving Linear Equations
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2.3 – Slide 3
2.3 More on Solving Linear Equations
Objectives
1.
2.
3.
4.
Learn and use the four steps for solving a linear
equation.
Solve equations with fractions or decimals as
coefficients.
Solve equations that have no solution or infinitely
many solutions.
Write expressions for two related unknown
quantities.
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2.3 – Slide 4
2.3 More on Solving Linear Equations
Solving a Linear Equation
Solving a Linear Equation
Step 1 Simplify each side separately. Clear (eliminate) parentheses,
fractions, and decimals, using the distributive property as needed,
and combine like terms.
Step 2 Isolate the variable term on one side. Use the addition property
so that the variable term is on one side of the equation and a number
is on the other.
Step 3 Isolate the variable. Use the multiplication property to get the
equation in the form x = a number, or a number = x. (Other letters
may be used for the variable.)
Step 4 Check. Substitute the proposed solution into the original equation
to see if a true statement results.
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2.3 – Slide 5
2.3 More on Solving Linear Equations
Using the Four Steps for Solving a Linear Equation
Example 1 Solve the equation.
Step 1
Step 2
Step 3
5w + 3 – 2w – 7 = 6w + 8
3w – 4 = 6w + 8
3w – 4 + 4 = 6w + 8 + 4
3w
3w – 6w
– 3w
– 3w
–3
w
= 6w + 12
= 6w + 12 – 6w
= 12
12
=
–3
= –4
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Combine terms.
Add 4.
Combine terms.
Subtract 6w.
Combine terms.
Divide by –3.
2.3 – Slide 6
2.3 More on Solving Linear Equations
Using the Four Steps for Solving a Linear Equation
Example 1 (continued) Solve the equation.
Step 4
Check by substituting – 4 for w in the original equation.
5w + 3 – 2w – 7 = 6w + 8
5(– 4) + 3 – 2(– 4) – 7 = 6(– 4) + 8
– 20 + 3 + 8 – 7 = – 24 + 8
– 16 = – 16
?
Let w = – 4.
?
Multiply.
True
The solution to the equation is – 4.
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2.3 – Slide 7
2.3 More on Solving Linear Equations
Using the Four Steps for Solving a Linear Equation
Example 2 Solve the equation.
Step 1
Step 2
Step 3
5(h – 4) + 2
5h – 20 + 2
5h – 18
5h – 18 + 18
5h
5h – 3h
2h
=
=
=
=
=
=
=
3h
3h
3h
3h
3h
3h
14
2h = 14
2
2
h = 7
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–
–
–
–
+
+
4
4
4
4 + 18
14
14 – 3h
Distribute.
Combine terms.
Add 18.
Combine terms.
Subtract 3h.
Combine terms.
Divide by 2.
2.3 – Slide 8
2.3 More on Solving Linear Equations
Using the Four Steps for Solving a Linear Equation
Example 2 (continued) Solve the equation.
Step 4
Check by substituting 7 for h in the original equation.
5 ( h – 4 ) + 2 = 3h – 4
5 ( 7 – 4 ) + 2 = 3(7) – 4
?
Let h = 7.
5 (3) + 2 = 3(7) – 4
?
Subtract.
15 + 2 = 21 – 4
?
Multiply.
17 = 17
True
The solution to the equation is 7.
Copyright © 2010 Pearson Education, Inc. All rights reserved.
2.3 – Slide 9
2.3 More on Solving Linear Equations
Using the Four Steps for Solving a Linear Equation
Example 3 Solve the equation.
2 ( 5y + 7 ) – 16
10y + 14 – 16
10y – 2
Step 2
10y – 2 – 2
10y – 4
10y – 4 – 10y
–4
= –4
Step 3
–5
= 4
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Step 1
15y – 1 ( 10y – 2 )
15y – 10y + 2
5y + 2
5y + 2 – 2
5y
5y – 10y
–5y
–5y
–5
y
=
=
=
=
=
=
=
Distribute.
Combine terms.
Subtract 2.
Combine terms.
Subtract 10y.
Combine terms.
Divide by –5.
2.3 – Slide 10
2.3 More on Solving Linear Equations
Using the Four Steps for Solving a Linear Equation
Example 3 (continued) Solve the equation.
Step 4
Check by substituting 4 for y in the original equation.
5
15y – ( 10y – 2 ) = 2 ( 5y + 7 ) – 16
15 45
– ( 10 45
– 2 ) = 2 ( 5 45
+ 7 ) – 16 ? Let y =
12 – ( 8 – 2 ) = 2 ( 4 + 7 ) – 16
12 – 6 = 2 ( 11 ) – 16
12 – 6 = 22 – 16
6 = 6
The solution to the equation is 4 .
5
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4
5 .
? Multiply.
? Subtract, add.
? Multiply.
True
2.3 – Slide 11
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
1) We clear an equation of fractions by multiplying each side by
the least common denominator (LCD) of all the fractions in the
equation. It is a good idea to do this in Step 1 to avoid messy
computations.
2) When clearing an equation containing decimals, choose the
smallest exponent on 10 needed to eliminate the decimals.
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2.3 – Slide 12
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
Example 4 Solve the equation.
5 m – 10 =
8
8
8
5 m
8
3 m + 1 m
4
2
5 m – 10
8
=
8
3 m + 1 m
4
2
– 8 10
=
8
3 m
4
+
5m – 80 = 6m + 4m
8
Multiply by LCD, 8.
1 m
2
Distribute.
Multiply.
Now use the four steps to solve this
equivalent equation.
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2.3 – Slide 13
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
Example 4 (continued) Solve the equation.
5m – 80 = 6m + 4m
Step 1
5m – 80 = 10m
Step 2
5m – 80 – 5m = 10m – 5m
– 80 = 5m
Step 3
– 80 = 5m
5
5
– 16 = m
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Combine terms.
Subtract 5m.
Combine terms.
Divide by 5.
2.3 – Slide 14
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
Example 4 (continued) Solve the equation.
Step 4
Check by substituting –16 for m in the original equation.
5 m – 10 =
8
5 (–16)
– 10 =
8
3 m + 1 m
4
2
3 (–16)
1 (–16)
+
4
2
–10 – 10 = –12 – 8
–20 = –20
?
Let m = –16.
?
Multiply.
True
The solution to the equation is –16.
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2.3 – Slide 15
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
Note
Multiplying by 10 is the same as moving the decimal point one
place to the right.
Example:
1.5 ( 10 ) = 15.
Multiplying by 100 is the same as moving the decimal point two
places to the right.
Example:
5.24 ( 100 ) = 524.
Multiplying by 10,000 is the same as moving the decimal point
Answer: 4 places.
how many places to the right?
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2.3 – Slide 16
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
Example 5 Solve the equation.
0.2v – 0.03 ( 11 + v ) = – 0.06 ( 31 )
20v – 3 ( 11 + v ) = – 6 ( 31 )
Step 1
20v – 33 – 3v = – 186
17v – 33 = – 186
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Multiply by 100.
Distribute.
Combine terms.
2.3 – Slide 17
2.3 More on Solving Linear Equations
Solving Equations with Fraction or Decimal Coefficients
Example 5 (continued) Solve the equation.
17v – 33 = – 186
Step 2
Step 3
From Step 1
17v – 33 + 33 = – 186 + 33
Add 33.
17v = – 153
Combine terms.
17v
– 153
=
17
17
Divide by 17.
v = –9
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Check to confirm that
– 9 is the solution.
2.3 – Slide 18
2.3 More on Solving Linear Equations
Solving Equations with No or Infinitely Many Solutions
Example 6 Solve the equation.
4 ( 2n + 6 ) = 2 ( 3n + 12 ) + 2n
8n + 24 = 6n + 24 + 2n
Distribute.
8n + 24 = 8n + 24
Combine terms.
8n + 24 – 24 = 8n + 24 – 24
8n = 8n
8n – 8n = 8n – 8n
0 = 0
Subtract 24.
Combine terms.
Subtract 8n.
True
Solution Set: {all real numbers}.
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2.3 – Slide 19
2.3 More on Solving Linear Equations
Solving Equations with No or Infinitely Many Solutions
An equation with both sides exactly the same, like 0 = 0, is called an
identity. An identity is true for all replacements of the variables.
We write the solution set as {all real numbers}.
CAUTION
In example 6, do not write {0} as the solution set of the equation.
While 0 is a solution, there are infinitely many other solutions.
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2.3 – Slide 20
2.3 More on Solving Linear Equations
Solving Equations with No or Infinitely Many Solutions
Example 7 Solve the equation.
6x – 1 ( 4 – 3x )
6x – 4 + 3x
9x – 4
9x – 4 – 9x
–4
=
=
=
=
=
8 +
8 +
–19
–19
–19
3 ( 3x – 9 )
9x – 27
+ 9x
+ 9x – 9x
Distribute.
Combine terms.
Subtract 9x.
False
There is no solution. Solution set: 0
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2.3 – Slide 21
2.3 More on Solving Linear Equations
Solving Equations with No or Infinitely Many Solutions
Again, the variable has disappeared, but this time a false statement
(– 4 = – 19) results. This is the signal that the equation, called a
contradiction, has no solution. Its solution set is the empty set, or
null set, symbolized 0 .
CAUTION
Do not write {0 } to represent the empty set.
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2.3 – Slide 22
2.3 More on Solving Linear Equations
Writing Expressions for Two Related Unknown Quantities
Example 8
Two numbers have a sum of 32. If one of the numbers is
represented by c, write an expression for the other number.
Given:
c represents one number.
The sum of the two numbers is 32.
Solution:
32 – c represents the other number.
Check:
One number + the other number = 32
c
+
(32 – c)
= 32
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2.3 – Slide 23
2.3 More on Solving Linear Equations
Writing Expressions for Two Related Unknown Quantities
Example 9
Andrew arrived at work with x dollars. How much money will
he have if he finds $15 in his desk? How much money will he
have if he spends $10 on his lunch?
Given:
x represents the amount of money Andrew has currently.
He found $15 in his drawer. He then spent $10 on lunch.
Solution for Part A:
x + 15
Solution for Part B:
x + 15 – 10
or
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x + 5
2.3 – Slide 24