Solving for the Unknown - Basic Mathematics for Water and Wastewater Operators - page 123

■ Example 8.28
Problem:

= ××
××

160 240834 . .

061 .

1970 x 834 .

Solution:

x = ××

160 240834 . . 3202 56

10 022

.

×× =

= 032 .

1970 061834 . . , 178

■ Example 8.29
Problem:

x 2 × 0.865 = 3176

Note: When solving for x 2 the procedure is the same as solving for x (but there is
one extra step at the end). Determine if x 2 is in the numerator, simplify numbers,
get x 2 by itself, and solve the equation. Finally, take the square root of both sides.

Solution:

2

x × 0 865 . = 3176

3176
0 865
3672

2

x =

.

2

x =

2

x =

=

3672
60 x .. 6

■ Example 8.30
Problem:

0.785 × D 2 = 5038

Solution:

5038
0 785
6418

2

D =

.

2

D =

2

D = 6418

D = 80 1 .