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Hence, when we have two equations, containing two unknown quantities, we have the following

RULE,

FOR ELIMINATION BY COMPARISON.

Find an expression for the value of the same unknown quantity in each of the given equations, and place these values equal to each other; there will thus be formed a new equation, containing only one unknown quantity.

Find the value of each of the unknown quantities in the following equations, by the preceding rule.

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ELIMINATION BY ADDITION AND SUBTRACTION.

ART. 160.-Elimination by addition and subtraction, consists in multiplying or dividing two equations, so as to render the coëfficient of one of the unknown quantities, the same in both; and then, by adding or subtracting, to cause the term containing it to disappear.

To explain this method, we will take the same equations used to illustrate elimination by substitution and comparison.

x+2y=17 (1.) 2x+3y=28 (2.)

If we multiply equation (1) by 2, so as to make the coëfficient of x the same as in the second equation, we have

2x+4y=34 (3.)

2x+3y=28, equation (2) brought down. y=6

Since the coefficient of x has the same sign in these equations, if we subtract, the terms containing x will cancel each other, and the resulting equation will contain only y, the value of which may then readily be found. After this, by substituting the value of y, as before, the value of x is easily obtained.

To illustrate the method of eliminating, when the coefficients of the unknown quantity to be eliminated, have contrary signs in the two equations, suppose we have the following, in which it is required to eliminate y.

3x-5y=6 (1.)

4x+3y=37 (2.)

It is obvious, that if we multiply equation (1) by 3 and (2) by 5, that the coefficients of y will be the same.

9x-15y= 18 20x+15y=185

Thus,

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Substituting this value of x in equation (2), we have

28+3y=37
3y= 9
y= 3

From this we see, that after making the coëfficients of the quantity to be eliminated, the same in both equations, if the signs are alike, we must subtract; but if they are unlike, we must add them. Hence, when we have two equations, containing two unknown quantities, we have the following

RULE,

FOR ELIMINATION BY ADDITION AND SUBTRACTION.

Multiply, or divide the equations, if necessary, so that one of the unknown quantities will have the same coefficient in both. Then take the difference, or the sum of the equations, according as the signs of the equal terms are alike or unlike, and the resulting equation will contain only one unknown quantity.

REMARK. When the coefficients of the unknown quantities to be eliminated are prime to each other, they may be equalized, by multiplying each

REVIEW.-159. What is the rule for elimination by comparison? 160. In what does elimination by addition and subtraction consist? What is the rule for elimination by addition and subtraction?

equation by the coefficient of the unknown quantity in the other. When the coefficients are not prime, find their least common multiple, and multiply each equation by the quotient obtained by dividing the least common multiple by the coefficient of the unknown quantity to be eliminated in the other equation.

If the equations have fractional coefficients, they ought to be cleared before applying the rule.

Find the value of the unknown quantities in each of the following equations, by the preceding rule.

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ART. 161.-The questions contained in Art. 156, were all capable of being solved by using one unknown quantity; although, several of the examples contained two, and in some cases more, unknown quantities. In those questions, however, there was such a connection existing between the several quantities, that it was easy to express each one in terms of the other. But it frequently happens, that in a problem containing two unknown quantities, there may be no direct relation existing between them, by means of which either of them may be found in terms of the other. In such a case, it becomes necessary to use a separate symbol for each unknown quantity, and then to find the equations containing these symbols, on the same principle as where there was but one unknown quantity; that is, in brief, regard the symbols as the answer to the question, and then proceed in the same manner as it would be necessary to do, to prove the answer. After the equations are obtained, the values of the unknown quantities may be found, by either of the three different modes of elimination.

We shall first give two examples, which can be solved by using either one or two unknown quantities.

In general, no more symbols should be used, than are really necessary; unless, by using them, the solution is rendered more simple.

1. Given, the sum of two numbers equal to 25, and their difference equal to 9, to find the numbers.

Solution, by using one unknown quantity.

Let x= the less number; then x+9= the greater.

And x+x+9=25.

2x=16

x=8, the less number; and x+9=17, the greater.

Solution, by using two unknown quantities.

Let x the greater, and y- the less.

Then x+y=25 (1.)

And x-y 9 (2.)

2x=34, by adding the two equations together.

x=17, the greater number.

2y=16, by subtracting equation (2) from equation (1). y= 8, the less number.

2. The sum of two numbers is 44, and they are to each other as 5 to 6; required the numbers.

Solution, by using one unknown quantity.

Let 5x the less number; then 6x= the greater.
And 5x+6x=44.

11x=44

x=4

5x=20, the less number.

6x=24, the greater number.

Solution, by using two unknown quantities.
Let x= the less number, and y= the greater.
Then x+y=44 (1.)

And xy::56

or, 6x=5y (2.) by multiplying means and extremes.
6x+6y=264 (3.) by multiplying equation (1) by 6.
6y=264-5y, by subtracting equation (2) from (3).
11y=264

y=24 and x=44—y=20.

Several of the following questions may also be solved by using only one unknown quantity.

3. There is a certain number consisting of two places of figures; the sum of the figures is equal to 6, and, if from the double of the

REVIEW.-161. In solving questions, when does it become necessary to use a separate symbol for each unknown quantity? How are the equations formed, from which the values of the unknown quantities are to be obtained?

number 6 be subtracted, the remainder is a number whose digits are those of the former in an inverted order; required the number?

In solving questions of this kind, the pupil must be reminded, that any number consisting of two places of figures, is equal to 10 times the figure in the ten's place, plus the figure in the unit's place. Thus, 23 is equal to 10X2+3. In a similar manner, 325 is equal to 100×3+10×2+5.

Let x= the digit in the place of tens, and y= that in the place of units.

Then 10x+y= the number.

And 10y+x the number, with the digits inverted.

Then x+y=6 (1.)

And 2(10x+y)—-6=10y+x (2.)

or, 20x+2y-6=10y+x.

19x=8y+6

8x=-8y+48, from equation (1), by multiplying by 8

and transposing.

27x=54 by adding.

x=2

y=6-2-4. Ans. 24.

4. What two numbers are those, to which if 5 be added, the sums will be to each other as 5 to 6; but, if 5 be subtracted from each, the remainders will be to each other as 3 to 4?

By the conditions of the question, we have the following proportions:

x+5y+5::5:6

x-5: y-5::3:4.

Since, in every proportion, the product of the means is equal to the product of the extremes, we have the two equations

6(x+5)=5(y+5)
4(x-5)=3(y-5)

From these equations, the values of x and y are readily found to be 20 and 25.

REMARK. Instead of saying, that the two sums will be to each other as 5 to 6, it will be the same to say, that the quotient of the second divided by the first, is equal to §, since six divided by 5, expresses the ratio of 5 to 6. This would give the following equations:

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which may be readily obtained from those given above.

NOTE. In solving the following questions, after finding the equations, the values of the unknown quantities may be found by either of the three methods of elimination.

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