Expressing the multiplication, we have (a b2x-2 c) (3 a—b) (3) — (2 a c) (5 a) (3) = (a b x) (5 a) (3 a — b) (3) — (b2 x) (5 a) (3 a —b). = Transposing all the terms which contain x into the first member, and those which do not contain it into the second member, it becomes 3 a b3 x - 45 a3 b x + 15 a2 b2 x + 15 a2 b2 x — 5 a b3 x 18 a c Uniting the terms which are alike 6 b c +30 a2 c. 6 b c +30 a2c. 39 a2 b2 x 8 a b3 x 45 a b x 18 a c Separating the first member into factors (39 a2 b2 — 8 a b3 — 45 a3 b) x = 18 ac-6 b c +30 a2 c. 2. Find the value of x in the following equation; 3. What is the value of x in the following equation? 4. What is the value of x in the following equation? 5. What is the value of x in the following equation? XXII. Miscellaneous Examples producing Simple Equations. 1. A merchant sent a venture to sea and lost one fourth of it by shipwreck; he then added $2250 to what remained, and sent again. This time he lost one third of what he sent. He then added $1000 to what remained, and sent a third time, and gained a sum equal to twice the third venture; his whole return was equal to three times his first venture. What was the value of the first venture? 2. A man let out a certain sun of money at 6 per cent., simple interest, which interest in 10 years wanted but £ 12 to be equal to the principal. What was the principal? 3. A man let out £ 98 in two different parcels, one at 5, and the other at 6 per cent, simple interest; and the interest of the whole, in 15 years, amounted to £81. What were the two parcels? 4. A shepherd driving a flock of sheep in time of war, met a company of soldiers, who plundered him of one half the sheep he had and half a sheep over; the same treatment he received from a second, a third, and a fourth company, each succeeding company plundering him of one half the sheep he had left and one half a sheep over. At last he had only 7 sheep left. How many had he at first? 5. A man being asked how many teeth he had remaining, answered, three times as many as he had lost; and being asked how many he had lost, answered, as many as, being multiplied into part of the number he had left, would give the number he had at first. How many had he remaining, and how many had he lost? After this question is put into equation every term may be divided by x. 6. There is a rectangular field whose length is to its breadth as 3 to 2, and the number of square rods in the field is equal to 6 times the number of rods round it. Required the length and breadth of the field. 7. What two numbers are those, whose difference, sum, and product, are to each other, as the numbers 2, 3, and 5 respectively? 8. Generalize the above by putting a, b, and c instead of 2, 3, and 5 respectively. α 7. clearing of fractions beac⇒ a b x — a2 x + b c — a e 8. by transposition abx-a3x=2 a c Solve the 7th Ex. by these formulas; also try other numbers. 9. When a company at a tavern came to pay their reckoning, they found that if there had been three persons more, they would have had a shilling apiece less to pay; and if there had been two less, they would have had to pay a shilling apiece more. How many persons were there, and how much had each to pay? 10. A sum of money is to be divided equally among a certain number of persons. Now if there were three claimants less, each would receive 150 dollars more; and if there were 6 more, each would receive 120 dollars less. How many persons are there, and how much is each to receive? 11. What fraction is that, to the numerator of which if 1 be added, its value will be, but if 1 be added to its denominator its value will be 4. 12. What fraction is that, to the numerator of which if a be added, its value will be but if a be added to its denominator m Solve the 11th example by these formulas. 13. What fraction is that, from the numerator of which if a be subtracted, its value will be its denominator, its value will be m but if a be subtracted from N. B. The answers to the 12th and 13th differ only in the signs of the denominators. The learner will find by endeavoring to solve particular examples from these formulas, that he will not always succeed. If in making examples for the 12th, he selects his numbers, so that n p is greater than m q, the formula will fail; but if he takes the same numbers, and applies them according to the conditions of the 13th, they will answer those conditions. When mq is greater than n p the numbers will not suit the conditions of the 13th, but they will answer to those of the 12th. The numbers in example 11th will not form an example according to the 13th. The following numbers will form an example for the 13th but not for the 12th. 14. What fraction is that, from the numerator of which if 3 be subtracted, its value will be §, but if 3 be subtracted from its denominator, its value will be ? The reason why numbers chosen indiscriminately will not satisfy the conditions of the above formulas will be explained hereafter. Equations with several Unknown Quantities. XXIII. Questions involving more than two unknown Quantities. Sometimes it is necessary to employ, in the solution of a question, more than two unknown quantities. In this case, the question must furnish conditions enough to form as many distinct equations as there are unknown quantities. 1. A market woman sold to one man, 7 apples, 10 pears, and 12 peaches, for 63 cents; and to another, 13 apples, 6 pears, and 2 peaches, for 31 cents; and to a third, 11 apples, 14 pears, and 8 peaches for 63 cents. She sold them each time at the same rate. What was the price of each? Let y = z= Then we shall have 1. 2. 3. a pear, 7x+10y+12 z = 63 13x+6y+ 2z=31 11x+14y+8x=63. The second being multiplied by 6, the z will have the same coefficient as in the first. If the second be multiplied by 4, the z will have the same coefficient as the 3d. |