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Find the values of z in each of these equations, and we have

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Put the first value of z = the second; and also, to the third, and we have two equations.

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Find the value of x in each of these equations, and put the two values of x equal to each other; and we shall have an equation containing only y, from which the value of y may be found.

8. Given

x+y+2=60
3x=y+z
2y=x+z

to find the values of x, y, and z."

9. A man sold a horse, a cow, and a sheep for 100 dollars. The cow and sheep came to as much as the horse; and the price of the horse is equal to five times the difference between the price of the cow and the sheep. What is the price of each?

We will give a third method of reducing three equations with three letters.

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To find the values of x, y, and z.

Multiplying the first equation by 2, and subtracting the second, gives

(4)

10y-11z=26

Multiplying the first by 3, and subtracting the third from the product, gives

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Multiplying equation (5) by 2, and subtracting it from equation (4,) we have

5 z=20

2 = 4, &c.

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13. There are three numbers. The sum of the first and second is 10. The sum of the first and third is 12; and the sum of the second and third 14, What are the numbers?

14. A man bought at one time, a sheep and 2 hogs for 14 dollars; at another time, a sheep and 2 cows for 26 dollars; and at another, 2 hogs and a cow for 24 dollars. He bought them at the same rate each time. How much did he give for one of each kind?

SECTION XXX.

Equations of Four or Five Letters,

We have given three different methods of finding the values of the letters when we have 3 equations containing 3 letters. It is obvious that the same methods may be employed when we have 4 equations containing 4 letters, or 5 equations and 5 letters; or in general whenever we have as many equations as letters.

Given the following equations to find the values of the letters which they contain.

1.

2.

3.

u+x=3
u+y= 4
u+z = 5
y + z =
u+x+y=18°
u+x+x=20
u+y+z = 22
x + y + z=24

2u+2x+2y= 18
2u+3x+4y: = 29
3x+4y+5z = 50
4u+6y+2z = 42

4.

u+v+x+y=10
u+v+x+x=11
u+v+ y + z = 12
u+x+y+z = 13
u + x + y + z:

= 14

5. There are 4 numbers. The sum of the first and second is 8. The sum of the first and third is 10. The sum of the first and fourth is 12; and the sum of the second and fourth is 14. What are the numbers?

6. A man has 4 sons. The sum of the ages of the first, second and third is 12 years. The sum of the ages of the first, second and fourth is 14 years. The sum of the ages of the first, third and fourth is 16 years. The sum of the ages of the second, third and fourth is 18 years. What are their ages?

7. There are four numbers. The first added to the second gives 20. The first added to the third gives 25. The first added to the fourth gives 30. the fourth gives 40. What are the numbers?:

The second added to

8. Divide 72 into 4 such parts, that the first increased by 5 may be equal to the second diminished by 5, and also to the third multiplied by 5, and the fourth divided by 5.

9. There are 4 numbers. The first added to of the second gives 15. The first added to the third gives 20. The first added to of the fourth gives 15; and the second added to of the third gives 11. What are the numbers?

10. There are 5 numbers. The first, second, third and fourth make 16. The first, second, third and fifth make 18. The first, third, fourth and fifth make 21. The second, third, fourth and fifth make 22; and the first, second, fourth and fifth make 19. What are the numbers?

11. There are five numbers. The first, second, third and fourth make 100. The first, second, third and fifth make 110. The first, third, fourth and fifth 130. The first, second, fourth and fifth make 120. The second, third, fourth and fifth make 140. What are the numbers?

SECTION XXXI.

Explanation of the signs + and -.

1. What number must you add to 30, that the sum may

20 ?

Let

x= the number,

Then 30+x= = 20

be

Subtract 30 from each member. Subtracting 30 from the first member, leaves x. When we attempt to subtract 30 from the second member, we take away the whole of the second member, and there remains still 10 more to take away; and we write it 0-10, or simply — 10..

Hence X== -10

10, then, is the number to be added to 30, that the sum may be 20; but 10 must be subtracted from 30 to make 20; hence, adding 10 is, in reality, subtracting 10.

Was the preceding question stated right? How should it have been stated?.

2. What number must be added to 60, that the sum may be 50?

3. What number must be added to 45, that the sum may be 30?

4. What number must be subtracted from 10, that the re-mainder may be 15?:

Let

Then 10

x= the number,.

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Adding x to both numbers, we have ›

10=15+ x

Subtracting 15 from both members, and we have

x=- 5

5, then, is the number to be subtracted from 10, that the remainder may be 15; but 5 must be added to 10 to make 15. Hence subtracting — 5 is the same as adding 5.

5. What number must be subtracted from 4, that the remainder may be 6?

It will be seen that all the preceding questions are improperly stated; and by following the statement of the questions, we get the answer a negative number, which shows that it is to be taken in an opposite sense from what is stated in the

question: that is, it is added when it should have been subtracted, or subtracted when it should have been added; and that in fact, the letter entered the equation with the wrong sign.

There are two numbers, 30 and 20. How much must be added to the 30 that the product of the two may be 500? Is this question correctly stated? How would you correct it? Solve the two following questions, and correct the statements of them.

6. There are two numbers, five times one added to six times the other gives 3; and four times one added to twice the other gives 8. What are the numbers?

7. A man worked 7 days, and had his son with him 3 days; and received for wages 22 shillings. He afterwards worked 5 days, and had his son with him one day, and received for wages 18 shillings. What are his daily wages, and the daily wages of his son?

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-a; and subtracting the

6 is the same as adding 6, hence the remainder is

10. Multiply a·

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b by c-d.

.b

It is evident that the answer is c times a-b, wanting d times ab; but c times a-b is ca- - cb, and d times a is a d bd; hence the remainder is a d cacb, which gives ca-cb-ad+bd.,

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11. Multiply a 6 by b .6.

12. Multiply b· by c

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bd subtracted from

-3; and y

6 by c-m.

13. Multiply o -a by o

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5 by 0-6.

It will be seen from the preceding questions that we can obtain the result of the analysis by observing the following rule,

Rule for multiplication. If two numbers have the same sign, if they are both plus, or both minus, their product must have the sign plus; if they have different signs, if one be plus and the other minus, their product must have the sign minus See the illustration on following page.

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