Page 18 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 1] USEFUL MATH 3

Second, a quantity that multiplies one side of an equation may be shifted so as to divide the other side, and vice
versa. Thus if ab = c, then a = c/b; and if a/d = e, then a = de.
When each side of an equation consists of a fraction, all we need to do to remove the fractions is to cross
multiply:
a c
= → ad = bc
b d
What was originally the denominator (lower part) of each fraction now multiplies the numerator (upper part) of
the other side of the equation.

SOLVED PROBLEM 1.5

Solve 5x − 10 = 20 for x.
We want to have just x on the left-hand side of the equation. The first step is to shift the −10 to the right-hand
side, where it becomes +10:

5x − 10 = 20
5x = 20 + 10 = 30
Now we shift the 5 so that it divides the right-hand side:
5x = 30
30
x = = 6
5
The solution is x = 6.

SOLVED PROBLEM 1.6
Examples of cross multiplication:
x y y 5
= =
2 7 8 x
7x = 2y xy = (5)(8) = 40
3x 4y y 3x + 2
= =
5 3 5 4
3(3x) = 5(4y) 4y = 5(3x + 2) = 15x + 10

9x = 20y

SOLVED PROBLEM 1.7

Solve the equation
5 3
=
a + 2 a − 2
for the value of a.

We proceed as follows:
Cross multiply 5(a − 2) = 3(a + 2)
Multiply out both sides 5a − 10 = 3a + 6
Shift −10 and 3a 5a − 3a = 6 + 10
Carry out indicated addition and subtraction 2a = 16
Divide both sides by 2 a = 8