1 - 8 A n I n t r o d u c t i o n t o E q u a t i o n s
Quick Review
An equation c a n b e t r u e o r fa ls e , o r i t c a n b e a n open
sentence with a variable. A solution of an equation
is th e v a lu e ( o r v a lu e s ) o f th e v a r ia b le t h a t m a k e s th e
e q u a t io n t r u e.Ex a m p l e
Is c = 6 a solution of the equation 25 = 3c - 2?
25 = 3c - 2
25 — 3 • 6 — 2 Substitute 6 for c.
25 + 16 Simplify.
No, c = 6 is n o t a so lu tio n o f the equation 25 = 3c — 2.Ex e r c i s e s
Tell whether the given number is a solution of each
equation.- 17 = 37 + 4/;/= -5 88. -3 a 2 = 27; a = 3
- 3b-9 = 21;b = -1 0 90. -2b + 4 = 3; b = \
Use a table to find or estimate the solution of each
equation.
91.x+(-2) = 8 92. 3m-13 = 24
93.41-2 = 9 94. 6 b - 3 = 17
1 - 9 P a t t e r n s , E q u a t i o n s , a n d G r a p h s
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Qu ick Review
Y o u c a n r e p r e s e n t th e r e la t io n s h ip b e tw e e n t w o v a r y in g
quantities in different ways, including tables, equations, and
graphs. A solution of an equation with two variables is an
ordered pair (x, y) that makes the equation true.Ex a m p l e
Bo makes $15 more per week than Sue. How can you
represent this with an equation and a table?
First w rite an equation. Let b = B o ’s e a r n in g s a n d
s = S u e 's e a r n in g s. B o m a k e s $ 1 5 m o r e t h a n S u e , so
b = s + 15. Y ou can use th e e q u a tio n to m ake a ta b le fo r
s = 2 5 , 5 0 , 7 5 , a n d 100.Su e ' s Ea r n i n g s ( s) (^255075100)
Bo' s Earnings (b) 40 65 90 115
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Ex e r c i s e s
Tell whether the given ordered pair is a solution of each
equation.
- 3 x + 5 = y ; (1 , 8 )
- y = — 2 (x + 3); ( - 6 , 0 )
- y= (x — 1.2)(—3); (0,1.2)
- 10 - 5x = y; (-4 ,1 0 )
- Describe the pattern in the table using words,
an equation, and a graph. Extend the pattern for
x = 5, 6 , a n d 7.
on
1 1 572 Ch ap t er 1 Ch ap t er Revi ew