Page 642 - Algebra 1

Example
4
Application: Archery
A 60-centimeter indoor archery target has several rings around a circular center. If the average diameter of a ring is d, and an arrow landing in that ring scores p points, the inner and outer diameters of the ring are given by the absolute-value equation
⎪d + 6p - 63⎥ = 3. Find the inner and outer diameters of the 8-point ring.
8 pt.
inner diameter outer diameter
SOLUTION
⎪d + 6p - 63⎥ = 3 ⎪d + 6 · 8 - 63⎥ = 3 ⎪d + 48 - 63⎥ = 3 ⎪d - 15⎥ = 3
Substitute 8 for p. Multiply. Subtract.
Caution
The absolute-value bars act as grouping symbols. Be sure to use the order of operations to simplify any expression within them.
Write the absolute-value equation as two equations and solve.
d - 15 = 3 or d - 15 = -3
d = 18 Add 15 to both sides. d = 12
The inner diameter of the ring is 12 centimeters and the outer diameter of
the ring is 18 centimeters.
Solve each equation. Then graph the solution.
(Ex 1) _ ⎪x⎥
a. 7 +10=18 b. 3⎪x⎥-11=10
Solve each equation.
Lesson Practice
c.
(Ex 2)
d.
(Ex 2)
e.
(Ex 3)
f.
(Ex 3)
g.
(Ex 3)
h.
(Ex 4)
_4⎪x⎥ _+23=11
9
⎪x⎥ + 3
-2=1 ⎪7x⎥+2=37
2 5⎪x+1⎥-2=23
9 _x - 1 = 45 ⎪2 ⎥
Investments A factory produces items that cost $5 to make. The factory would like to invest $100 plus or minus $10 in the first batch. Use the equation ⎪5x - 100⎥ = 10 to find the least and greatest number of items the factory can produce.
Lesson 94 627