Download 10.2 More on Solving Linear Equations

10
Real Numbers, Equations,
and Inequalities
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 1
10.2 More on Solving Linear Equations
Objectives
1. Solve more difficult linear equations.
2. Solve equations that have no solution or
infinitely many solutions.
3. Solve equations by first clearing fractions
and decimals.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 2
Solve More Difficult Linear Equations
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 3
Solve More Difficult Linear Equations
Example 1
Solve this equation and check the solution: –2 + 3(2x + 7) = 7 + 3x
Step 1 Use the distributive property. –2 + 3(2x + 7) = 7 + 3x
Step 2 Combine like terms.
–2 + 6x + 21 = 7 + 3x
6x + 19 = 7 + 3x
Step 3 Subtract 19 from both sides.
–19 –19
6x + 0 = –12 + 3x
6x = –12 + 3x
Subtract 3x from both sides.
–3x –3x
3x = –12 + 0
3 x 12
Step 4 Divide both sides by 3.

3
3
x  4
The solution is –4.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 4
Solve More Difficult Linear Equations
Step 5 Check the solution by replacing each x with –4 in the original
equation.
–2 + 3(2x + 7) = 7 + 3x
–2 + 3[2(–4) + 7] = 7 + 3(–4)
–2 + 3[(–8) + 7] = 7 + –12
–2 + 3(–1) = –5
–2 + –3 = –5
–5 = –5 ✓ Balances
When x is replaced with –4, the equation balances, so 4 is the
correct solution (not –5).
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 5
Solve More Difficult Linear Equations
Example 2
Solve this equation: 8a – (3 + 2a) = 3a + 1
Start by removing the parentheses on the left side of the
equation. In this situation, the – sign acts like a factor of –1,
changing the sign of every term inside the parentheses.
8a  (3  2a)  3a  1
8a  1(3  2a)  3a  1
8a  3  2a  3a  1
6a  3  3a  1
6a  3  3  3a  1  3
6a  3a  4
6a  3a  3a  4  3a
3a  4
3a 4

3 3
4
a
3
The solution is 4/3. You can check
the solution by going back to the
original equation and replacing
each a with 4/3.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 6
Solve More Difficult Linear Equations
Example 3
Solve: 4(8 – 3t) = 32 – 8(t + 2)
4(8  3t )  32  8(t  2)
32  12t  32  8t  16
32  12t  32  32  8t  16  32
12t  16  8t
12t  8t  16  8t  8t
4t  16
4t 16

4
4
t4
The solution is 4.
The check is left to you.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 7
Solving Equations With Infinitely Many Solutions
Example 4
Solve: 5x – 15 = 5(x – 3)
5 x  15  5( x  3)
5 x  15  5 x  15
5 x  15  15  5 x  15  15
5x  5x
5x  5x  5x  5x
00
When both sides of an equation are exactly the same, the equation
is called an identity. An identity is true for all replacements of the
variables.
The solution for 5x – 15 = 5(x – 3) is all real numbers.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 8
Solving Equations With No Solution
Example 5
Solve: 2w – 7(w + 1) = –5w + 4
2w  7( w  1)  5w  4
2w  7 w  7  5w  4
5w  7  5w  4
5w  7  5w  5w  4  5w
7  4
As in Example 4, the variable has “disappeared.” This time,
however, we are left with a false statement. Whenever this happens
in solving an equation, it is a signal that the equation has no
solution. So, you write “no solution.”
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 9
Solve Equations with Fraction or Decimal Coefficients
Example 6 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
+
8
Multiply by LCD, 8.
1 m
2
5m – 80 = 6m + 4m
Distribute.
Multiply.
Now use the four steps to solve this
equivalent equation.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 10
Solve Equations with Fraction or Decimal Coefficients
Example 6 (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
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Combine terms.
Subtract 5m.
Combine terms.
Divide by 5.
Slide 11
Solve Equations with Fraction or Decimal Coefficients
Example 6 (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.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 12
Solve 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
how many places to the right?
Answer: 4 places.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 13
Solve Equations with Fraction or Decimal Coefficients
Example 7 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
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Multiply by 100.
Distribute.
Combine terms.
Slide 14
Solve Equations with Fraction or Decimal Coefficients
Example 7 (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
Check to confirm that
– 9 is the solution.
Copyright © 2014, 2010, 2007 Pearson Education, Inc.
Slide 15