Page 137 - Algebra 1

Example
2
Solving Equations by Dividing
Solve each equation. Then check the solution.
a. 5x=20 SOLUTION
5x = 20 5x = 20
__ 55
Division Property of Equality Simplify.
x = 4
Check Substitute4forx.
5x = 20 5(4) 20
20=20 ✓
b. -12 = 3n SOLUTION
-12 = 3n -_12 = _3n
33 -4 = n
Division Property of Equality
Simplify.
Check Substitute-4forn.
-12 = 3n
-12 3(-4) -12 = -12 ✓
c. _2p=7 5
SOLUTION
_2 p = 7 5
Reading Math
_2 p c a n b e w r i t t e n a s _2 p , 55
_2 _2 _2
(5 )p, 5 · p, or 5 × p.
_2_2_2 _2
(5 ÷ 5)p=(7÷ 5) 11
_2 _5 _5 (5 · 2)p = (7 · 2)
Dividebothsidesby 5.
Divide by multiplying by the reciprocal of 5,
_ p = 35
Check Substitute _35 for p. 2
1_2 _357 (5)(2 ) 7
11
7=7 ✓
2
_2
1 1
w h i c h i s _5 . 2
122 Saxon Algebra 1