The circumference of a circle is 2 times the radius times π:

c = 2 ′′ r

The capacity of a circular tank is the area of base times height. Letting v stand for volume or capacity, we have

The capacity of a cubical box or bin is the cube of the length, that is

V= = 13

It is clear from these illustrations that algebraic equations are time-savers. We shall now discuss methods of solving such equations. In most practical work all the quantities in the equation are known except one, and the object is to find the value of the one unknown quantity. For example, we may know the height and radius of a circular tank, and wish to find the capacity. The equation for capacity of a circular tank gives us the rule for solving the problem. We put the numbers in the equation in place of the letters and then do the work as the equation indicates. If the radius of our tank is 5 feet and the height 8 feet, then for this particular tank

v3.1416 X 52 X 8

It is necessary to know certain rules that must always be observed in solving equations.

First rule: Whatever is done to one side of an equation must be done to the other side.

If this rule is violated, the equation is no longer true. This rule can be illustrated best with numbers. Take the equation

2+3=5

We can add the same number to both sides of the equation. Let us add 4, then we have

2+3+45 + 4,

which is true. But if we add 4 to one side only, we get 2+3+4,

which does not equal 5.

Subtracting 3 from both sides, we get

2 2

=

Multiplying both sides by 2, we get

4+6=10

Take the equation

a=b

Multiply both sides by c, then

ac= bc

Divide both sides by c, then

a b

==

C C

Square both sides, then

a2 = b2

Transposing is subtracting the same number from both sides of an equation.

Take the equation

a+b= c

Subtracting b from both sides, we have

a=c-b

So when b appears on the right side of the equation it has the minus sign.

Illustrating this with numbers, let

When the 3 appears on the right side, it has the minus sign. This gives us our second rule, viz.:

Any term in an equation may be transposed by changing its sign.

Clearing an Equation of Fractions.

Third rule: To clear an equation of fractions, multiply both sides of the equation by a number that will contain the denominators of all the fractions.

For example, take the equation

The quantity bd contains both denominators. Multiplying both sides by b d, we have

since a bd divided by b equals a d, and cbd divided by d equals cb.

We shall now apply these rules in solving some simple equations:

If 2x 10, what does x equal? Dividing both sides by 2, we get x

=

5.

10, what does a equal? Subtracting

If a +3 3 from both sides, we get a 7.

If b

3 times 4

=

3 X 22, what is the value of b? 22

=

12; b = 12.

4, and

Examples to be solved:

1. a +5 15, what does a equal?

=

2. — 12, what does a equal?— Answer, a — 36.

=

4. c = 2 xг. If r — 3, π — 3.1416, what does c equal? - Answer, c

5. v = 13. swer, v = 512.

18.8496.

If 1 = 8, what does v equal?——- An

6. ar. If r = 3, what does a equal? - Answer, a= 28.2744.

7. vr h. If r 4, h3, what is the value of v?- Answer, v 150.7968.

8. For total pressure on piston, we have the equation P — « r2 p, in which p is pressure per square inch. 50, what does P equal? Answer,

If r

=

6, P

P = 5,654.88.

=

9. In the equation v = r2 h, if v = 251.328, r = 4, what does h equal?- Answer, h = 5.

251.328 π †2

Lever, Safety Valve, Torque.

Rules for lever problems can be understood best by considering first one of the simplest levers, the crowbar. The same rules can be applied to any other lever as to the crowbar.

Suppose a man weighing 150 pounds is lifting a weight with a crow bar (Fig. 12). The support on which the bar rests is called the fulcrum. Suppose the man is pushing down on the end of the bar 4 feet from the fulcrum, and the weight is 1 foot from the fulcrum. Then the man's weight will balance a weight of 600 pounds. The rule is simple. The weight of the