6. Extracting the square root, x-1=6, and x=7. XVI. Let x=the number of yards which he bo't; then, 1. By the conditions of the problem, 6000 5400 XC x-15 -10 2. Clear. of fractions, 6000x-90000=5400x-10x2 +150x 3. Transposing and uniting, 10x2+450x=90000 4. Dividing by 10, +45x=9000 5. Completing the square, (Art.311.b.) 4x2+180x+2025= 36000+2025 6. Extracting the square root, 2x+45=195 7. Transposing and uniting, 2x=150, and x=75=No. y'ds. The price per yard is found by dividing 54 dollars by 75 yards, which 80 cents the price per yard. XVII. Let x=the number of days which both traveled before they met; then x-3=B's daily progress, and (x-3) Xx the distance that B traveled, and 9x=the distance that A traveled; therefore, 1. By the given conditions, (x-3)×x+9x=247 2. Expanding and uniting, x2+6x=247 3. Comp. the sq. (Art. 305.) x2+6x+9=247+9 4. Extracting the square root, x+3=16, and x=13 Therefore, 13×9=117=the distance A travels, and 247 -117=130=B's distance. XVIII. Let x=No. yards in the finer, and x+2=No. yards in the coarser piece; then, 1. By the conditions of the problem, 320 360 -4 x+2. x 2. Clearing of fractions, 320x=360x+720-4x2 - 8x 3. Transposing and uniting, 4x2 - 32x=720 4. Dividing by 4, x2-8x=180 5. Completing the square, (Art. 305.) x2-8x+16=180+16 6. Extracting the square root, x-4=14, and x=18=the number of yards in the finer piece; and 18+2=20 No. yards in the coarser; and the prices were 20, and 16 shillings per yard. XIX, Let x=the number of gallons of Teneriffe; then the Teneriffe per gallon, and the number of gallons being x, 2 2 x 2 -4x=the cost of the whole Teneriffe, and 27x+ -4x- = 2 the cost of the whole of the Madeira and Teneriffe, 54 +x =the number of gallons of both. 540+10x=what the whole was sold for. 540+10x+576=what was given, 1. Clearing of fractions, 54x+x2-8x=1080+20x+1152 2. Transposing and uniting, x2+26x=2232 3. Completing the square, (Art.305.), x2+26x+169=2232 +169 4. Extracting the square root, x+13=49, and x= 36=the Teneriffe; 36÷2=18=the price of the Madeira per gal. XX. Let x=the required number; then 1. By the conditions of the problem, 240−x2+20 2. Clearing of fractions, 240—x2+20=4x x 4 3. Placing 2 under the radical sign, ✓160-4x2 +20=4x 4. Transposing 20, 160-4x=-20+4x 5. Squaring both sides, 160-4x2-400—160x+16x2 6. Transposing and uniting, 20x2-160x=-240 7. Dividing by 20, x2-8x=-12 8. Comp. the square, (Art. 305.) x2-8x+16=-12+16 9. Extracting the square root, x-4=2, and x=6. XXI. Let x=the persons age: then, 1. By the conditions of the problem, √x+2=12 2. Transposing √x=12 x 2 3. Squaring both sides, x= 576-48x+x2 4 4. Clearing of fractions, 4x=576–48x+x2 5. Transposing and uniting, -x2+52x=576 6. Changing signs, x2 -52x=-576 7. Completing square, (Art. 305.) x2-56x+676=676-576 8. Extracting the square root, x-26-10, and x-16. XXII. Let x=number of gallons in smaller cask then, 58 x 1. By the given conditions, 2x+5 2 3 2. Clearing of fractions, 174=2x2+5x-12x-30 5. Comp. the square, (311, b.) 16x2-56x+49=1632+49 6. Extracting the square root, 4x—7=41, and x= =12=No. gals. in the smaller cask. 12+5=17=No. galls. in the larger cask. 58÷29=2=the price per gallon. XXIII. Let x=the coins of copper; then 24-x=the coins of silver. 1. By the problem, aX(24-x)=24x-x2, value of copper (24-x)×x=24x-x2, value of the silver 2. 66 66 3. Uniting the two values, 2x2-48x=-216 4. C. sq. (Art. 311, b.) 16x2-384x+2304=-1728+2304 5. Extracting the square root, 4x-28=44, and x= =the copper coins, and 24-18=6=the silver coins. XXIV. Let x=the number of oxen: then, 1. By the conditions of the problem, 80 80 X x+4 +1 2. Clearing of fractions, 80x+320=80x+x2+4x 3. Transposing and uniting, -x2 — 4x——320 4. Changing signs, x2+4x=320 =18. 5. Com. the square, (Art. 311. b.) x2+4x+4=320+4 6. Extracting the square root, x+2=18, and x=16. 18 PROBLEMS CONTAINING TWO OR MORE UNKNOWN QUANTITIES. I. Let x=the greater of the two numbers, and y=the less. 1. By the first condition of the problem, x+y=24 2. By the second condition, x=5y 3. Transposing y in the first equation, x=24-y 4. Making the second and third equal, 5y=24—y 5. Transposing y, 6y=24 6. Dividing by 6, y=4, and x=20. II. Let x=the greater quantity, and y=the less. 1. By the first condition, x+y=h 66 2. By the second x2-y2=d 3. Transposing y2 in the second equation, x2=d+y3 4. By evolution, (Art. 297,) x=d+ya 5. Transposing y in the first equation, x=h—y 6. Making the fourth and fifth equal, √d+y2=h—y 7. Squaring each side of the equation, d+y=h3-Qhy+y3 8. Cancelling y2, d=h2 — 2hy 9. Transposing d and 2hy, 2hy-ha-d 10. Dividing by 2h, y=. h2-d 2h III. 1. By the first conditions of the problem, ax+by=h 2. By the second condition, x+y=d 3. Transposing by, in the first equation, ax=h-by 4. Dividing by a, x-h-by x=' a 5. Transposing y in the second equation, x=d-y 6. Making the fourth and fifth equal, d-y= 7. Clearing of fractions, ad-ay=h-by 8. Transposing ad and by, by-ay-h-ad 9. Dividing by b—a, y= h-ad b-a h-by a IV. Let x=1 the distance which the privateer sails from the commencement of the chase, until she overtakes the ship; and y=the distance which the ship sails, from the commencement of the chase until she is overtaken then, 1. By the conditions of the problem x=y+20 3. Converting the proportion into an equation, 8y=7x 5. Substituting for y in the 1st equation, x=. 7x 8 6. Clearing of fractions, 8x=7x+160 7. Transposing and uniting, x=160. = V. Let the age of A; and y =the age of B: then, 1. By the problem, x-7=the age of A, 7 years ago y-7=the age of B, 7 years ago x+7=the age of A, 7 years hence y+7=the age of B, 7 years hence 2. 3. 4. 66 5. By the first condition, x-7=3X(y-7)=3y-21 8. Substituting 3y-14 for x in the 6th, 3y-14+7=2y+14 9. Transposing and uniting, y=21=the age of B. VI. Let x=greater, and y=less of the two numbers: then, 1. By the first condition of the problem, x: y::3:2 3. Converting proportion into equation in the 1st, 2x=3y 4. Dividing by 2, x=. 3y 2 5. Making 2d and 4th equal,+y= xy 6 6. Clearing of fractions, 18y+12y=2xy 7. Dividing by y, 18+12=2x 8. Transposing and uniting, 2x=30, x=15 and y=10. VII. Let x=the number in the greater army, and y=the number in the less 1. By the first condition, x+y=21110 2. By the second condition, 2x+3y=52219 VIII. 1. By the first supposition, 2x+y=16 2. By the second 66 3. Multiplying the first by 3, 6x+3y=48 IX. 1. By the first condition, x+y=14 2. By the second x-y=2 3. Subtracting the second from the first, 2y=12, and y=6. X. Let x=the lower part, and y=the upper part |