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SECTION VIII.

EQUATIONS OF THE FIRST DEGREE CONTAINING TWO UNKNOWN QUANTITIES.

(106.) IN the examples which have been hitherto given, each problem has contained but one unknown quantity; or, if there have been more, they have been so related to each other, that all have been expressed by means of the same letter. This, however, can not always be done, and we are now to consider how equations of this kind are resolved.

(107.) If we have two equations with two unknown quantities, we must endeavor to deduce from them a single equation containing only one unknown quantity. We must, therefore, make one of the unknown quantities disappear; or, as it is termed, we must eliminate it. There are three different methods of elimination which may be practiced.

The first is by substitution;

66

second is by comparison;

66 third is by addition and subtraction.

(108.)

ELIMINATION BY SUBSTITUTION.

Ex. 1. Let it be proposed to solve the system of equations

QUEST.-When we have two equations with two unknown quanti. ties, how must we proceed? What is meant by elimination? How many methods of elimination are practiced?

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From the second equation we find the value of x in terms of y, which gives

x=y+6.

Substituting the expression y+6 for x in the first equation, it becomes

y+6+y=12;

from which we find that y=3; and since we have already seen that x=y+6, we find that x=3+6=9.

To verify these values, substitute them for x and y in the original equations, and we shall obtain

9+3=12,

9-3 6.

Ex. 2. Again: take the equations

2x+3y=13

3y = 13. }

5x+4y=22.

From the first equation we find

13-2x

y= 3

Substituting this value of y in the second equation,

it becomes

5x+4x

13-2x
3

=22,

an equation containing only x, which, when solved,

gives

x=2;

and this value of x, substituted in either of the origi. nal equations, gives

y=3.

(109.) The method thus exemplified is expressed

in the following

RULE.

Find an expression for the value of one of the unknown quantities in either of the equations; then substitute this value in the place of its equal in the other equation.

Ex. 3. Find the values of x and y in the equations 3x+y=13 and 2x-y=2.

Ans. x=3, y=4. Ex. 4. Find the values of x and y in the equations 3x-2y=11, and 2x+5y=39.

Ans. x=7, y=5.

Ex. 5. Find the values of x and y in the equations 2x+7y=65, and 6x-2y=34.

Ans. x=8, y=7.

Ex. 6. Find the values of x and y in the equations 4x+6y=84, and 8x-3y=18.

Ans. x=6, y=10.

Ex. 7. Find the values of x and y in the equations 5x-9y=14, and 8x-6y=56.

Ans. x=10, y=4.

Ex. 8. Find the values of x and y in the equations

7x-15y=75, and 10x+3y=156.

Ans. x=15, y=2.

Ex. 9. Find the values of x and y in the equations 8x+7y=208, and 5x+6y=156.

Ans. x=12, y=16.

Ex. 10. Find the values of x and y in the equations 10x+8y=202, and 9x-4y=25.

Ans. x=9, y=14. Ex. 11. A market-woman sold to one boy 4 apples

QUEST.-Give the rule for elimination by substitution.

and 3 pears for 17 cents, and to another 7 apples and She sold them in each case at

5 pears for 29 cents.

the same rate. What was the price of each?

Ans. The apples 2, and the pears 3 cents each. Ex. 12. There are two brothers, A and B. If twice A's age be added to three times B's age, the sum will be 54; but if twice B's age be subtracted from five times A's age, the remainder will be 40. What are their ages?

Ans. A is 12 years and B 10 years.

Ex. 13. Two farmers, A and B, driving their sheep to market, says A to B, if you will give me half of your sheep, I shall have 75. Says B to A, if you will give me one third of your sheep, I shall have 75. How many had each?

Ans. A had 45 and B had 60. Ex. 14. A grocer had two kinds of wine, marked A and B. He mixed together 3 gallons of A and 2 gallons of B, and sold the mixture at $1.35 per gallon. He also mixed 4 gallons of A and 6 of B, and sold the mixture at $1.40 per gallon. What was the value of each kind of wine?

Ans. A was worth $1.25 per gallon.
B was worth $1.50 per gallon.

Ex. 15. Find two numbers such that their sum shall be 42, and the greater shall be equal to five times the Ans. 7 and 35.

ess.

(110.)

ELIMINATION BY COMPARISON.

Ex. 1. To illustrate this method, take the equations

x+y=12,
x-y= 6.

Derive from each equation an expression for z in terms of y, and we shall have

x=12-y,
x= 6+y.

These two values of x must be equal to each other, and, by comparing them, we shall obtain

12-y=6+y,

an equation involving only one unknown quantity ; whence we obtain

y=3.

Substituting this value of y in the expression x=6 +y, and we find x=9.

Ex. 2. Again: take the equations

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Putting these values of y equal to each other, we

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an equation containing only x; whence, by clearing of fractions and transposing, we obtain

x=2.

Substituting this value of x in either of the preceding expressions for y, we find

y=3.

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