Download Use the multiplication property of equality.

Chapter 2
Section 2
2.2 The Multiplication Property of Equality
Objectives
1
Use the multiplication property of equality.
2
Simplify, and then use the multiplication property of equality.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Objective 1
Use the multiplication property of
equality.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-3
Use the multiplication property of equality.
The addition property of equality is not enough to solve some equations,
such as
3x  2  17.
Since the coefficient of x is 3 rather than 1, the multiplication property of
equality is needed to change the equation to the form x = a number,
after the 2 is subtracted from both sides of the equation and we are left
with
3x  15.
Multiplication Property of Equality
If A, B, and C (C ≠ 0) represent real numbers, then the equations
A  B and AC  BC
are equivalent equations.
That is, we can multiply each side of an equation by the same nonzero
number without changing the solution.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-4
Use the multiplication property of equality. (cont’d)
This property can be used to solve 3x  15. The 3x on the left must be
changed to 1x, or x. To isolate x, we multiply each side of the equation
1
by , the reciprocal of 3, which will result in a coefficient of 1 when
3
1
3
multiplied.
3   1
3
3
Just as the addition property of equality permits subtracting the same
number from each side of an equation, the multiplication property of
equality permits dividing each side of an equation by the same
number.
DO NOT, however, divide each side by a variable, since the
variable might be equal to 0.
It is usually easier to multiply on each side if the coefficient of the variable is a
fraction, and divide on each side if the coefficient is an integer.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-5
EXAMPLE 1 Applying the Multiplication Property of Equality
Solve.
15x  75
Solution:
Check:
15(5)  75
15 x 75

15 15
75  75
x 5
The solution set is
5.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-6
EXAMPLE 2 Applying the Multiplication Property of Equality
Solve.
8x  20
Solution:
8 x 20

8
8
5
x
2
Check:
8x  20
 5
8     20
 2
20  20
 5
The solution set is    .
 2
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-7
EXAMPLE 3 Solving an Equation with Decimals
Solve.
0.7 x  5.04.
Solution:
Check:
0.7 x  5.04
0.7 x 5.04

0.7 0.7
0.7  7.2  5.04
x  7.2
5.04  5.04
The solution set is
7.2.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-8
EXAMPLE 4 Applying the Multiplication Property of Equality
Solve.
x
 6
4
Solution:
Check:
x
 4   6  4 
4
x  24
The solution set is
24.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
x
 6
4
24
 6
4
6  6
Slide 2.2-9
EXAMPLE 5 Applying the Multiplication Property of Equality
Solve.
2
 t  12
3
Solution:
Check:
 3 2
 3
    t  12   
 2 3
 2
t  18
The solution set is
18.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
2
 t  12
3
2
 (18)  12
3
12  12
Slide 2.2-10
Using the multiplication property of equality when the
coefficient of the variable is −1.
In Section 2.1, we obtained the equation  x  17. We reasoned
that since this equation says that the additive inverse (or opposite) of
x is −17, then x must equal 17.
We can also use the multiplication property of equality to obtain the
same result as detailed in the next example.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-11
EXAMPLE 6 Applying the Multiplication Property of Equality
Solve.
p  7
Solution:
Check:
1  p  7
1 1 p   7  1
1(1)  p  7
(7)  7
77
1 p  7
p  7
The solution set is
p  7
7.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-12
Objective 2
Simplify, and then use the
multiplication property of equality.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-13
EXAMPLE 7 Combing Like Terms When Solving
Solve.
4r  9r  20
Solution:
Check:
5r  20
5r 20

5 5
4r  9r  20
4(4)  9(4)  20
16  (36)  20
r  4
20  20
The solution set is
4.
Copyright © 2012, 2008, 2004 Pearson Education, Inc.
Slide 2.2-14