the end of two years Blends to A enough to pay off his debts, and has 32 (a) dollars to spare. What is the income

of each?

Ans. 280 dollars, or (35a). 38. What number is that of which, and added

together make 73 (a)?

Ans. 84. General Ans.

84a
73

39. A person after spending 100 dollars more than of his income, had remaining 35 dollars more than of it. Required his income. Ans. $450.

40. A person after spending (a) dollars more than of his income, had remaining (b) dollars more than of it.

Required his income.

Ans. 21(a+b) dollars.

11

41. What two numbers are those in proportion of 2 to 3, and if 4 be added to each of them, the sums will be in proportion of 5 to 7? Ans. 16 and 24.

SIMPLE EQUATIONS.

CHAPTER II.

(Art. 45.) We have given a sufficient number of examples, and introduced the reader sufficiently far into the science previous to giving instructions for the solution of questions containing two or more unknown quantities.

There are many simple problems which one may meet with in algebra which cannot be solved by the use of a single unknown quantity, and there are also some which may be solved by a single unknown letter, that may become much more simple by using two or more unknown quantities When two unknown quantities are used, two independent equations must exist, in which the value of the unknown letters must be the same in each. When three unknown quantities are used there must exist three independent equations, in which the value of any one of the unknown letters is the same in each.

In short there must be as many independent equations as unknown quantities used in the question.

For more definite illustration let us suppose the following question:

A merchant sends me a bill of 16 dollars for 3 pair of shoes and 2 pair of boots; afterwards he sends another bill of 23 dollars for 4 pair of shoes and 3 pair of boots, charging at the same rate. What was his price for a pair of shoes, and what for a pair of boots?

This can be resolved by one unknown quantity, but it is far more simple to use two.

Let x= the price of a pair shoes,

And y= the price of a pair of boots.

Then by the question

And

3x+2y=16

4x+3y=23.

These two equations are independent; that is, one cannot be converted into the other by multiplication or division, notwithstanding the value of x and of y are the same in both equations.

Having intimated that this problem can be resolved with one unknown quantity, we will explain in what manner, before we proceed to a general solution of equations containing two unknown quantities.

Let x= the price of a pair of shoes.

Then 3x= the price of three pair of shoes.
And 16-3x= the price of two pair of boots.

16-3x

Consequently

the price of one pair of boots.

2

Now 4 pair of shoes which cost 4x, and 3 pair of boots 48-9x 2

which cost

dollars.

being added together must equal 23

That is, 4x+24-x=23.

Or,

1-x=0. Therefore x=2 dollars, the price of a pair of shoes. Substitute the value of x in the

16-3x

expression

2

pair of boots.

and we find 5 dollars for the price of a

Now let us resume the equations,

3x+2y=16 (4)
4x+3y=23 (B)

FIRST METHOD OF ELIMINATION.

(Art. 46.) Transpose the terms containing y to the right hand sides of the equations, and divide by the coefficients

Put the two expressions for x equal to each other. (Ax. 7.)

An equation which readily gives y=5, which taken as the value of yin either equation (C) or (D) will give x=2. This method of elimination just explained is called the method by comparison.

SECOND METHOD OF ELIMINATION.

(Art. 47.) To explain another method of solution, let us

The value of x from equation (A) is x=(16-2y).

Substitute this value for x in equation (B), and we have

4×(16-2y)+3y=23, an equation containing only y. Reducing it, we find y=5 the same as before.

This method of elimination is called the method by sub

stitution, and consists in finding the value of one unknown quantity from one equation to put that value in the other which will cause one unknown quantity to disappear.

THIRD METHOD OF ELIMINATION.

(Art. 48.) Resume again 3x+2y=16

4x+3y=23

(A)

(B)

If the coefficients of either a or y were alike in both

equations, that term will disappear by subtraction.

If the signs are unlike and the coefficients equal, the quantity will disappear by addition.

To make the coefficients of x equal, multiply each equation by the coefficient of x in the other.

To make the coefficients of y equal, multiply each equa

tion by the coefficient of y in the other.

Multiply equation (A) by 4 and 12x+8y=64

Multiply equation (B) by 3 and 12x+9y=69

Difference

y=5 as before.

To continue this investigation, let us take the equations

Multiply equation (A) by 2, and equation (B) by 3, and

Equations in which the coefficients of y are equal, and

the signs unlike. In this case add, and the y's will destroy

This method of elimination is called the method by addition and subtraction.

FOURTH METHOD OF ELIMINATION.

(Art. 48.) Take the equations 2x+3y=23. (A)

Multiply one of the equations, for example (A), by some

indeterminate quantity, say m.

• Remainder, (C) (2m-5)x+(3m+2)y=23m-10 As m is an indeterminate quantity, we can assume it of any value to suit our pleasure, and whatever the assumption may be, the equation is still true.

Let us assume it of such a value as shall make the coefficient of y, (3m+2)=0.

The whole term will then be 0 times y, which is 0, and equation (C) becomes

But 3m+2=0. Therefore m= .

Which substitute for m in equation (D), and we have

x=

-23×3-10-23×2-30-76
-2x-5-2×2-15 -19

4.

This is a French method introduced by Bezout, but it is too indirect and metaphysical to be much practised, or in fact much known.

Of the other three methods, sometimes one is preferable and sometimes another, according to the relation of the coefficients and the positions in which they stand.

No one should be prejudiced against either method, and in practice we use either one, or modifications of them, as the case may require. The forms may be disregarded when the principles are kept in view.