Chapter 9 - XYZ Custom Plus

532
Chapter 9 Solving Equations and Inequalities
2. Solve for a: 1 } 5
a 1 3 5 7
Example 2
Solve for a: 1 3 a 1 2 5 7
Solution We begin by adding 22 to both sides to get 1 3
multiply by 3 to solve for a.
1
} 3
a 1 2 5 7
a by itself. We then
1
} 3
a 1 2 1 (−2) 5 7 1 (−2) Add 22 to both sides
1
} 3
a 5 5 Addition
3 ? 1 } 3
a 5 3 ? 5 Multiply both sides by 3
a 5 15
Multiplication
We can check our solution to see that it is correct:
the equation
When a 5 15
1
} 3
a 1 2 5 7
becomes
1
} 3
(15) 1 2 5 7
5 1 2 5 7
7 5 7 A true statement
3. Solve for y: 3 } 5
y 5 6
Example 3
Solve for y: 2 } 3
y 5 12
Solution
In this case we multiply each side of the equation by the reciprocal
of 2 } 3
, which is 3 } 2
.
2
} 3
y 5 12
3
− 2 1 2 } 3
y 2
5 3 − 2
(12)
y 5 18
The solution checks because } 2 } of 18 is 12.
3
Note The reciprocal of a negative number is also a negative number. Remember,
reciprocals are two numbers that have a product of 1. Since 1 is a positive number,
any two numbers we multiply to get 1 must both have the same sign. Here
are some negative numbers and their reciprocals:
The reciprocal of 22 is 2 } 1 . 2
The reciprocal of 27 is 2 1 } 7
.
The reciprocal of 2 1 } 3
is 23.
The reciprocal of 2 3 } 4
is 2 4 } 3
.
Answers
2. 20 3. 10
The reciprocal of 2 9 } 5
is 2 5 } 9
.