16. Extract. square root, y+1=±5, y=4, and 24÷4=6=x. Example 8. It is required to prove that a:x::√2a-y: ✔y on the supposition that (a+x)2(a−x)2::x+y: x-y 1. By expanding, a2+2ax+xa2-2ax+x2:x+y: x-y 2. Adding and subtracting, (Art. 388.) 2a2+2x2:4ax:: 2x: 2y 3. Dividing the terms in each coup. by 2, a2 +x2 :2ax::x:y 4. Transferring x, (Art. 374. Cor.) a2+x22a::x:y 5. Inverting the means, a2+x2x2::2a: y 6. Subt. consequents from antecedents, a2 x2:2a-yy 7. Evolving the terms in each couplet, ax::√2a-y: √y Example 9. It is required to prove that dx=cy, if x is to y in the triplicate ratio of ab, and a;b::Vc+x: Vd+y. 1. By the 1st supposition, a3: b3::x:y 2. By the 2d supposition, ab::Vc+x: Vd+y 3. Involv. to the 3d power, (Art. 399.) a3: b3::c+x:d+y 4. By equality of ratios in the 1st and 2d, c+x:d+y::x:y 5. Inverting the means, c+x:x::d+yy 6. Subt. consequents from antecedents, cx::dy 7. Multiplying extremes and means, dx=cy. Example 10. Let x= the greater number, and y= the less. 1. By the first condition, x2 y2(x-y)2::4:1 3. Expanding in the first, x2-y2 ; x2 -2xy+y2::4; 1 4. Subt. conseq. from antecedents, 2xy-2y2 : (x − y)2::3; 1 5. Div. homologous terms by y, 2x-2y: (x-y)2:3÷y:1 6. Div. analogous terms by x-y, 2:x-y::3÷y:1 7. Conv. propor. into an equation, 2=(3x-3y)÷y 8. Mult. each side by y, 2y=3x-3y, 5y=3x, and x= 5y 9. Substituting a's value in the 2d, 5y2÷3=135 10. Multiplying each side by 3, 5y2=405 11. Dividing by 5, and evolving, y2=81, and y=9 12. Substituting y's value in the 2d, xx9=135, and x=15. Example 11. Let x=the greater number, and y= the less. 1. By the proposed conditions, −y:x+y::2:3 3. By compounding the 1st and 2d, (x-y)x(x+y): (x+y) Xxy::6 15 4. Dividing analogous terms, x-y: xy::2:5 5. Conv. proportion into an equation, 5x-5y=2xy 11. Substituting y's value in the 8th, x=5x2=10. Example 12. Let x=the greater No. and 24-x-the less. 1. By the proposed conditions, xx (24-x): x2+(24-x)2 ::3: 10 2. Expanding, 24x-x2: x2+576-48x+x2::3: 10 3. Multiplying the antecedents by 2, 48x-2x2: 2x2-48x +576::6: 10 1 1. Adding and subtracting, 576 : 4x2-96x+576::16:4 5. Extracting the square root, 24: 2x-24::4: 2 3. Div. antecedents by 4,and conseq. by 2, 6 : x- -12::1 : 7. Conv. proportion into an equation, x-12=6, and x=18 3. 24-18=6y=the less. Therefore the Nos. are 18 & 6. Example 13. Let x=the rum, and y=the brandy. 1. By the proposed conditions, x-yy::100: x 2. x-yx:4y 3. By compounding ratios, (x-y)2 xy::400 xy 4. Dividing the consequents by xy, (x-y)21:400:1 5. Extracting the square root, x-y:1::20:1 6. Converting the proportion into an equation, x-y=20 7. Substi. 20 for x-y in the 1st equation, 20: y::100: x 8. Dividing the antecedents by 20, 1y::5: x 9. Converting the proportion into an equation, x=5y 10. Substi. 5y for x in the 6th, 5y-y=20, y=5, &x=20 Example 14. Let 3x and 2x-the numbers. 1. By the proposed conditions, 3x+62x-6::3:1 2. Adding and subtracting, x+12 5x::2:4 3. Conv. the proportion into an equation, 4x+48=10x 4. Tr. and uniting, 48=6x, and x=8=the less. =the greater. 3×8=24 Second method. Let x=the greater No. and y=the less. 1. By the first proposed conditions, xy::3: 2. By the second, 66 x+6y-6:3:1 3y 3. Conv. 1st prop. into an equation, 2x=3y, and x= 2 5. Conv. prop. into an equation, 3+6=3y—18, & y=16 2 6. Substituting y's value in the 3d, 2x=48, and x=24 Example 15. Let x and y=the numbers. 1. By the first condition, xy=320 2. By the second 66 x3 y3(x-y)3::61:1 3. Exp. 1st consequent,x3-y3· x3-3x2y+3xy2 —y3::61:1 4. Subtracting, 3x2y-3xy2(x-y)3::60:1 5. Div. terms in 1st couplet by x-y, 3xy: (x—y)2::60;1 According to the example, the product of the two required numbers xy=320; and the first antecedent in the 5th equation is 3xy=3X320=960: therefore, 6. Substituting 960 for 3xy, 960 : (x−y)2::60 : 1 9. Converting the proportion into an equation, 8=2x-2y 320 320 14. Substi. y's value in the 1st, x=. =20. y 16 Therefore the values of x and y are 20 and 16. Example 16. Let x=the greater number and y=the less. 1. By the proposed conditions, ay::16:9 2. 66 66 x: 24:24 y 3. By compounding, 2: 24y::334: 9y 4. Changing the order of the means, x2: 384::24y: 9y 5. Div. the terms in the 2d couplet by y, x2: 384::24 : 9 6. Conv. the proportion into an equation, 9x2=9216 7. Dividing by 9, x2=1024 8. Extracting the square root, x=32 9. Substituting a's value in the 1st, 32 y::16:9 10. Conv. the propor. into an equation, 16y=288, and y=18. Therefore the values of x and y are 32 and 18. ARITHMETICAL PROGRESSION. Three problems under Article 426. I. Let a=1 the 1st term, z=the last term, n=the number of terms, and d=the common difference: then by the 1st for. mula, z=a+(n−1)d. Substituting numbers, 7+(9−1)×3 31 the last term. II. Putting a, z, n, and d, for the 1st term, the last term, the number of terms, and the common difference, as above; by the 2d formula, a=z-(n-1)d. Substituting numbers, 60-(12-1)x5=5=the first term. III. To find 6 arithmetical means between 1 and 43. One problem and seven examples, under Art. 429. Problem. What is the sum of the natural series of num. bers, 1, 2, 3, 4, 5, &c. up to 1000? By the formula under Art. 428, s= Xn. Substituting the numbers which a+z 4. Clearing of fractions, 2s=120+760, and s=440. Example 2. In going from the box to the first stone, and in returning, two yards were traveled over; and in going from the box to the second stone, and in returning, four yards were traveled over: Therefore 2=the first term, and 2 the common difference. 4. Clearing of fractions, s=101 x 100 10100 yards 5. 10100 yards=5 miles and 1300 yards. Example 3. The first term =3, the common difference= 3, the number of terms=150, and the last term=50(Art.426.) By substituting numbers for letters in the formula, the sum of 150 terms of the series: = +50 X150=3 22650 =3775. Example 4. The first term =5, the sum of the series= 1455, and the number of terms 30. By the 3d formula, Substituting numbers, the com 2s-2an (Art. 429.) d= n2 - -n Example 5. By the 4th formula, (Art. 429,) Example 6. The first term =1, the common difference= , and the number of terms=32, By formula first, s= |