If a b be divided by +a, the quotient must beb, be b gives If + a b be divided by ab. a, the quotient must be — b, be b gives + ab. If a b be divided by -a, the quotient must be + b, beax+b gives a b. cause The rule for signs therefore is the same as in multiplication. When the signs are alike, that is, both or both, the sign of the product must be +; but when the signs are unlike, that is, one + and the other —, the sign of the quotient must be By the reasoning above we derive the following rule for division of compound numbers. Arrange the dividend and divisor according to the powers of some letter. Divide the first term of the dividend by the first term of the divisor, and write the result in the quotient. Multiply all the terms of the divisor by the term of the quotient thus found, and subtract the product from the dividend. The remainder will be a new dividend, and in order to find the next term of the quotient, proceed exactly as before; and so on until there is no remainder. Sometimes, however, there will be a remainder, such that the first term of the divisor, will not divide either term of it; in which case the division can be continued no farther, and the remainder must be written over the divisor in the form of a fraction, and annexed to the quotient as in arithmetic. Divide 2 a3- 11 a1 b + 11 a3 b2 + 13 a b3 by 2 a-b. In this example, the division may be continued until the remainder is 4 65, which cannot be divided by a, therefore it must be written over the divisor 2 a b as a fraction and added to the quotient. The above rules are sufficient to solve all equations of the first degree. First, clear it of fractions by multiplying by the denomina tors. * Let the learner prove his results by multiplication. Expressing the multiplication, we have (a b2 x-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). = - Performing the multiplication it becomes 9 a b2 x 18 a c-3 a b3 x + 6 b c — 30 a2 c = 45a3 b x 15 a b2 x 15 a b2 x +5 a b3 x. - a2 Transposing all the terms which contain x into the first member, and those which do not contain it into the second member, it becomes 9a2 b2x-3 a b3 x — 45 a3 b x + 15 a2b2 x + 15 a2 b2 x— 5 a b3x · --- Uniting the terms which are alike 39 a2 b2 x-8 a b3 x-45 a3 b x 18 a c-6 b c +30 a2c. = Separating the first member into factors (39 a2 b2 -8 a b3-45 a3 b) x 18 ac-6 b c +30 a* 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 r 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 sum 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. Let x = the greater and y the less. 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? |