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Therefore x=4, z=5, and the fraction=1.

XXXVI. Let x=the number of apples which he bought,

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and z the number of pears; then =the price of the ap

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4

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1. By the conditions of the problem, +

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+ =30 4 5

3. Clearing the first of fractions, 15x+82-1560 4. Clearing the second of fractions, 5x+4x=600 5. Multiplying by 3, 15x+12z=1800

6. Subtracting the 3d from the 5th, 4z=240, and z=60 7. By substituting in the 4th, 5x+240=600, and x=72. Therefore x=72=the apples, and z=60=the pears.

er.

SOLUTIONS BY SUBSTITUTION, Art. 337.

Let x the less number required; then a-x=the greatAlso let a 30, and b=120.

1. By the first condition, x+a-x—a

2. By the second

66

(a2-2ax+x2)—x2 —b

3. Uniting terms in the 2d, a2 -2ax-b

4. Dividing by a, a- -2x=

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The same letter marked with different accents, is made the co-efficient of the same unknown quantity in different equations. The particular advantage of this method is, that

the expressions here obtained may be considered as general solutions, which give the values of the unknown quantities in other equations of a similar nature,

Required the values of x and y, from the following equations; letting a=10, a'=40, b=6, b'=4, c=100, & c'=200. 1. Given the equation, ax+by=c

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3. Multiplying the 1st by b', ab'x+bb'y=cb'

4. Multiplying the 2d by b, ba'x+bb'y=bc'

5. Subtracting the 4th from the 3d, ab'x-ba'x-cb'—bc'

6. Dividing by ab' —ba', x=

cb'-bc'

ab'-ba'

7. Multiplying the 1st by a', a'ax+a'by=a'c

8. Multiplying the 2d by a, a'ax+ab'y=ac'

9. Subtracting the 7th from the 8th, ab'y-a'by-ac'-a'c

10. Dividing by ab'-a'b, y=

ac' - a'c

ab'-a'b

11. Restoring a's, a's, b's, b''s, c's and c''s values in 6th, x= 100×4-6X200

10X4-6X40

12. Restoring a's, a's, b's, b's, c's and c''s values in 10th, y= 10x200-40×100

10×4-40×6

13. Clearing of fractions in 11th, 40x-240x=400-1200 14. Clearing of fractions in 12th, 40y-240y=2000-4000 15. Transposing and uniting in the 13th, 200x=-800 16. Transposing and uniting in the 14th, 200y=-2000 17. Changing signs in the 15th, 200x=800 18. Changing signs in the 16th, 200y=2000 19. Dividing by 200 in the 17th, x=4 20. Dividing by 200 in the 18th, y=10.

This solution is not necessarily extended beyond the 12th step. For x=(100×4-6x200)-(10×4-6X4)=4 Also, y=(10x200-40×100)÷(10×4-40×6)=10.

39.6

EXAMPLES OF GEOMETRICAL RATIO. Art. 362.

385

3159

Example 1. 11:9, and 44:35=Y, and 44=31, and 328. But (329)-(315)=: Therefore 119 is greater than 44 35 by 315

Example 2. a+3:2=(Art. 359. Cor.) 6a+18; a; and

α

6

2a+7=(Art. 359. Cor.) 6a+21:a. But 6a+21 is

3

greater than 6a+18 by 3. Therefore the ratio of 2a+7 is greater than the ratio of a+3 : 2 by 3.

a

3

6

Example 3. By Art. 349. 65-13=5= the consequent. Example 4. By Art. 349. 18x7=126= the antecedent.

Example 5. By Art. 352. the ratio compounded of the ratios 3 7, of 2a 5b, and of 7x+1: 3y-2=3×2a+(7x +1):7×5b×(3y—2)=42ax+6a :105–70b.

Example 6. By Art. 352. the ratio compounded of x+y: b, of x-ya+b, and of a+b;h=(x+y)x(xy)x(a+b) :b+(a+b)xh=ax2-by2+bx2-by2 abh +b2h=x2-y2

:bh.

Example 7. The ratios 5x+7: 2x−3, and x+2:2+3 when compounded=5x2 +17x+14: x23x+2+6x-9. Because 5x2+17x+14 the antecedent is greater than x2 3x-2 +6x-9 the consequent, the ratio is one (Art. 350.) of greater inequality.

Example 8. x+y: a, x—y; b, & b; abx2-aby2

a

xa —y2 —bx—2 by2 •

a

=bx2 —by2 : bx2 —by2=1(Art. 350.) Therefore,

the ratio compounded of x+y: a, and x-y: b, and b: x2y2 is a ratio of equality.

α

Example 9. The ratio compounded of 7: 5, of 42 92 and of 3323-3024 3240 (By reduction) 14: 15.

Example 10. The ratio compounded of 3: 7, of x3 : y3 ; and of 49: 9=21x3 ; 21y3=x3 ;y3.

EXAMPLES OF PROPORTION.

To nine of the sixteen examples which follow, abridged solutions are furnished in Day's Algebra, from which some of the solutions here given may differ. Those which are essentially the same, are more full, and have the several steps more particularly explained.

Example 1. Let x-6=the greater, then 55-a-the less. 1. By the conditions proposed, a: 44-x::9; 2 2. By adding terms, (Art. 389.) x : 44::9; 11 3. Multiplying extremes and means, 11x=396 4. Dividing by 11, x = 36

5. From the supposition x-6=30= the greater, and 4930 19 the less.

Example 2. Let x= the number required.

1. By the proposed conditions x+1:x+5::x+5x+13 2. Subtr. antecedents from consequents, (Art. 389.)x+1: 4x+58

3. Multiplying extremes and means, 8x+8=4x+20

4. Transposing and uniting, x=3= the required number.

Example 3. Let x= half their sum, and y= half their dif ference.

1. By the proposed conditions, x+y:x−y::2x:42

2.

66

66

66

x+yx-y::2y:6

3. By equality of ratios, (Art. 384) 2x: 42::2y: 6.
4. Dividing the terms in each couplet by 2, a 21:y 3
5. Changing the order of the means, xy::21:3

6. Multiplying extremes and means, 3x=21y, and x=7y
7. Mult. the extremes and means in 2d, 6x+6y=2xy-2y2
8. Substituting x's value, 42y+6y=14y2-2y2

9. Dividing by y, 42+6=14y-2y

10. Uniting, 48=12y, and y=4

x=7y=7×4=28= half the sum. Therefore 28x2=56= the sum, and y×2=4×2=8= the difference. Therefore x+y=28+4=32= the greater, and x-y=28-4=24= the less.

Example 4. Let 2x= the difference of the parts; then as the sum dif. gr. and the sum - dif. = the less.

=

1. By the conditions (9+x)2 (9−x)2::25; 16 2. Extracting the square root (Art. 391.) 9+x: 9-x::5;4

3. Multiplying extremes and means, 36+4x=45 — 5x

4. Transposing and uniting, 9x=9, and x=1

=

5. By the supposition 2 the difference, 10= the greater, and 8 the less.

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Example 5. Let x=the greater part; then 14-x-the less.

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2. Mult. analagous terms, (382.) x2 :(14−x)2::16:9 3. Extracting the square root, (Art. 391.) x 14-x::4:3 4. By adding terms, (Art. 388.) x;14::4;7

5. Dividing the consequent by 7, a 2::4:1

6. Mult. extremes and means, x=8= greater, and 14—8= 6 the less.

Example 6. Let x=the greater part; then 20-x=the less. 1. By the proposed conditions, a 20-x::32:12

2. By adding terms (Art. 388.) ≈:20::9: 10

3. Multiplying extremes and means, 10x=180, and x=18. The parts then are 18 and 2; and a mean proportional between two quantities is found (Art. 376.) by extracting. the square root of their product. Hence 2x18=6= the mean proportional between 2 and 18.

Example 7. Let x= the greater number, and y= the less. 1. By the proposed conditions, xy=24

2. 66

66

66

66

x3 y3(x-y)3:19:1 3. By expanding (x—y)3, x3—y3 : x3-3x2y+3xy2 —y3::

19:1

4. Subt. consequents from antecedents, 3x2y-3xy2 :(x—y)3

::18:1

:1

5. Dividing by x-y, (Art. 382. 5.) 3xy : (x—y)2::18 ;
6. Substi. for xy in 1st equation, 3×24 (x-y)2:18:1
7. Multiplying extremes and means, (x—y)2 × 18=72
8. Dividing by 18, (x—y)2=4

9. Extracting the square root, x-y=2
10. By the first equation, ay=24

11. Transposing y in the ninth, x=2+y

12. Dividing by y in the tenth, x=24÷y

13. Making the 11th and 12th equal, 2+y=24÷y. 14. Clearing of fractions, y2+2y=24

15. Completing the square, (Art. 305.) y2 +2y+1=24+1

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