Math 6 Honors Periods 1, 6 & 7 (Thursday)

Combined Operations 8-5

In order to solve an equation of the form

ax + b = c or ax –b = c or b – ax = c

where a, b, c are given numbers and x is the variable, we must use more than one transformation

Solve the equation 3n - 5 = 10 + 6

Simplify the numerical expression

3n - 5 = 10 + 6
3n – 5 = 16

add 5 to both sides

3n – 5 + 5 = 16 + 5
or
3n – 5 = 16
+ 5 = +5

3n = 21

divide both sides by 3 (or multiply each side by the reciprocal of 3)

3n/3 = 21/3

n = 7

General procedures for solving equations

Simplify each side of the equation
If there are still indicated additions or subtractions, use the inverse operation to undo them
If there are indicated multiplications or division involving the variable, use the inverse operations to undo them

The books says you must always perform the same operation on both sides of the equation. I say, “do to one side what you have done to the other side.”

Solve the equation
(3/2)n + 7 = 22

subtract 7 from both sides

(3/2)n + 7 - 7 = 22 - 7

(3/2)n =15


multiply both sides by 2/3, the reciprocal of 3/2

(2/3)(3/2)n = 15(2/3)

n = 10

Solve the equation
40 – (5/3)n = 15

add (5/3)n to both sides

40 – (5/3)n + (5/3)n = 15 + (5/3)n


40 = 15 + (5/3)n
subtract 15 from both sides

40 – 15 = 15-15 + (5/3)n

25 = (5/3)n multiply both sides by 3/5

(3/5)(25) = (5/3)n (3/5)

15 = n