Multiplying both members by 5×7×4=140, we have, 20x20 +644-28x-980-140-35x;

Multiplying both members of the equation by 70, the least common multiple of the denominators, we have, 70x-42x+42+280-700-35x-60x+80+56x-56.

NOTE 3.-In the resolution of equations, it is often convenient to depart from the rule given, and take advantage of every circumstance that presents itself for simplifying the quantities which enter into the equation, and also the equation itself; observing always to perform the same operation upon both members, that the equality may not be destroyed.

Since a is a factor in each of the terms of this equation, it can be simplified by dividing both members by a, and the equation becomes,

c2 + x2

CX

Expunging b, which is common to both members of the equation, and transposing a2 to the first member, we have, x2-a2bx-ab, or (x-a)b.

Dividing both members of this equation by x-a, we obtain,

Dividing both members by x, the equation becomes,

17. Given 5ax + 2b + 4bx=2x+5c to find x.

6

= 20

to find x.

2

X= 18.

4a b

x= 2a-3c

27. Given 9ax3- 15abx1 = 6ax3 +12ax2 to find x,

SOLUTION OF PROBLEMS PRODUCING SIMPLE EQUATIONS,

(34.) In every problem there is something given and known, and something unknown to be determined; and it is by means of the relations of the known quantities to the unknown, that we are enabled to arrive at the solution.

The solution of a problem, by means of algebra, consists of two parts; the first is to express in an equation the conditions of the problem; and the second, to reduce that equa tion, so as to find the value of the unknown quantity.

For expressing, in an equation, the conditions of a problem, the following rule may be observed,

RULE,

Express, by means of a letter, the unknown quantity, and perform upon it all the arithmetical operations which would be necessary to verify the result, were the problem already solved. This will form an equation, which, reduced according to the foregoing rules, will give the value of the unknown quantity.

EXAMPLES.

1. To find a number, such that the sum of its third and fourth parts, shall be equal to its half, plus 13.

If we suppose the number already found, and we wished to verify it, we should add its third and fourth parts together, and see if the sum was equal to its half, increased by 13.

Putting x for the number, we proceed in the same manner:

4

2

is its third part, its fourth part, and 1⁄2 its half. We should then have the following equation:

Clearing of fractions, 4x+3x=6x+156,

Transposing,

and reducing,

or,

Whence,

9191, which shows the result to be correct.

2. The sum of the two numbers is 30, the difference of their squares is 120; required the numbers.

Let x equal the least of the two numbers; then since their sum is 30, the greater must be 30-x. The square of 30-x is 900 60x+x2. Subtracting from this, the square of x, the remainder, by the conditions of the problem, must be 120, We have, therefore, the following equation

The least number is 13, and the greater 30- 13 = 17,

It is sometimes convenient to substitute letters in the place

of the known quantities.