10.3 more on solving linear equations

Slide - 1Copyright © 2018, 2014, 2010 Pearson Education Inc.A L W A Y S L E A R N I N G
2
Equations,
Inequalities, and
Applications
10

1. Learn and use the four steps for solving a
linear equation.
2. Solve equations that have no solution or
infinitely many solutions.
3. Solve equations with fractions or decimals
as coefficients.
4. Write expressions for two related unknown
quantities.
Objectives
10.3 More on Solving Linear Equations

Solving a Linear Equation in One Variable
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
of equality so that all terms with variables are on one side of
the equation and all constants (numbers) are on the other side.
Step 3 Isolate the variable. Use the multiplication property of equality
to obtain an equation that has just the variable with coefficient 1
on one side.
Step 4 Check. Substitute the value found into the original equation. If a
true statement results, write the solution set. If not, rework the
problem
Solving a Linear Equation

–6x + 5 = 17
–6x + 5 – 5 = 17 – 5
No step 1:
Step 2 Subtract 5.
–6x = 12 Combine terms.
Divide by –6.
w = –2
Step 3
Example Solve the equation.
6 1
6
2
6


 
w
Step 4
2
6 5 17
6( ) 5 17
17 17

 
 


x

5w + 3 – 2w – 7 = 6w + 8
3w – 4 = 6w + 8 Combine terms.
3w – 4 + 4 = 6w + 8 + 4
Step 1
Step 2 Add 4.
3w = 6w + 12 Combine terms.
3w – 6w = 6w + 12 – 6w Subtract 6w.
Combine terms.–3w = 12
Divide by –3.
w = –4
Step 3
3 1
3
2
3


 
w

5w + 3 – 2w – 7 = 6w + 8
? Let w = –4.
Step 4
5(–4) + 3 – 2(–4) – 7 = 6(–4) + 8
Check by substituting –4 for w in the original equation.
–20 + 3 + 8 – 7 = –24 + 8 ? Multiply.
–16 = –16 True
The solution to the equation is –4.
Example (cont) Solve the equation.

5 ( h – 4 ) + 2 = 3h – 4
5h – 20 + 2 = 3h – 4 Distribute.
5h – 18 = 3h – 4 Combine terms.
5h – 18 + 18 = 3h – 4 + 18 Add 18.
5h = 3h + 14 Combine terms.
Subtract 3h.5h – 3h = 3h + 14 – 3h
Combine terms.2h = 14
h = 7
Divide by 2.
2
2 2
14

h

Check by substituting 7 for h in the original equation.Step 4
5 (h – 4) + 2 = 3h – 4
5 (7 – 4) + 2 = 3(7) – 4
5 (3) + 2 = 3(7) – 4
15 + 2 = 21 – 4
17 = 17
? Let h = 7.
? Subtract.
True
? Multiply.
The solution to the equation is 7.

15y – ( 10y – 2 ) = 2 ( 5y + 7 ) – 16
15y – 10y + 2 = 10y + 14 – 16 Distribute.
5y + 2 = 10y – 2
Step 1
Step 2
Combine terms.
5y + 2 – 2 = 10y – 2 – 2
Subtract 2.
5y = 10y – 4
Combine terms.
Subtract 10y.
5y – 10y = 10y – 4 – 10y
Combine terms.
–5y = – 4
Step 3 Divide by –5.
1
5
5
5
4 

 
y
4
5
y

Step 4 Check by substituting for y in the original equation.4
5
15y – ( 10y – 2 ) = 2 ( 5y + 7 ) – 16
? Multiply.
True
? Subtract, add.
? Let y = .
4
5
12 – ( 8 – 2 ) = 2 ( 4 + 7 ) – 16
12 – 6 = 2 ( 11 ) – 16
15 – ( 10 – 2 ) = 2 ( 5 + 7 ) – 164
5
4
5
4
5
? Multiply.12 – 6 = 22 – 16
6 = 6
The solution to the equation is
4
.
5

Solve Equations with No Solution or Infinitely Many Solutions
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 Subtract 24.
8n = 8n Combine terms.
Subtract 8n.8n – 8n = 8n – 8n
True0 = 0
Solution Set: {all real numbers}.

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
Do not write {0} as the solution set of the equation. While 0 is a
solution, there are infinitely many other solutions.

6x – ( 4 – 3x ) = 8 + 3 ( 3x – 9 )
6x – 4 + 3x = 8 + 9x – 27 Distribute.
9x – 4 = –19 + 9x Combine terms.
9x – 4 – 9x = –19 + 9x – 9x Subtract 9x.
–4 = –19 False
There is no solution. Solution set:
1
0

Solve 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 .
CAUTION
Do not write { } to represent the empty set.
0
0

Solve Equations with Fraction or Decimal Coefficients
Multiply by LCD, 8.
5
8
m 3
4
m 1
2
m– 10 = +
5
8
m 3
4
m 1
2
m– 10 = +8 8
5
8
m 3
4
m 1
2
m– 10 = +8 88 8 Distribute.
5m – 80 = 6m + 4m Multiply.
Now use the four steps to solve this
equivalent equation.

Combine terms.
Subtract 5m.
5m – 80 = 6m + 4m
Divide by 5.
Step 1
Step 2
Step 3
5m – 80 = 10m
5m – 80 – 5m = 10m – 5m
– 80 = 5m Combine terms.
– 80 5m=
5 5
– 16 = m

Check by substituting –16 for m in the original equation.Step 4
? Let m = –16.
? Multiply.
True
The solution to the equation is –16.
5
8
m 3
4
m 1
2
m– 10 = +
5
8
(–16) 3
4
(–16) 1
2
(–16)– 10 = +
–10 – 10 = –12 – 8
–20 = –20

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
how many places to the right?
Answer: 4 places.

0.2v – 0.03 ( 11 + v ) = – 0.06 ( 31 )
Multiply by 100.20v – 3 ( 11 + v ) = – 6 ( 31 )
Distribute.20v – 33 – 3v = – 186
17v – 33 = – 186 Combine terms.
Step 1

17v – 33 = – 186 From Step 1
17v – 33 + 33 = – 186 + 33 Add 33.
17v = – 153 Combine terms.
Divide by 17.
v = – 9
17 17
17v – 153
=
Step 2
Step 3
Check to confirm that
– 9 is the solution.

Write Expressions for Two Related Unknown Quantities
Example
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

Writing Expressions for Two Related Unknown Quantities
Example
Two numbers have a product of 24. If one of the numbers is
represented by x, find an expression for the other number.
x represents one of the numbers
To find the other number, we would divide.
24
x

10.3 more on solving linear equations

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