Equations of the First Degree. tiplied by B' and the second by B, the products become, by substitution, AB'x+BB'y+B' M=0, A·B x + BB' y+BM'=AB'x+BB'y+B'M=0; that is, the two given equations are equivalent to but one, and are, as in art. 144, indeterminate. so that both the terms of the value of y would also be Equations of the First Degree solved by Elimination by Substitution. 5. A says to B, "give me $100, and I shall have as much as you." "No," says B to A, give me rather $100, and then I shall have twice as much as you." How many dollars has each? Ans. A $500, and B $700. "Six 6. Said a lad to his father, "How old are we?" years ago," answered the latter, "I was one third more than three times as old as you; but three years hence, I shall be obliged to multiply your age by 24 in order to obtain my own." What is the age of each? Ans. The father 36, the son 15 years. 7. A cistern containing 210 buckets, may be filled by 2 pipes. By an experiment, in which the first was open 4, and the second 5 hours, 90 buckets of water were obtained. By another experiment, when the first was open 7, and the other 3 hours, 126 buckets were obtained. How many buckets does each pipe discharge in an hour? Ans. The first pipe discharges 15, and the second pipe discharges 6 buckets. 8. There is a fraction such, that if 1 be added to its nu merator its value becomes = denominator its value becomes ; and if 1 be added to its = 4. What fraction is it? Ans. 9. Required to find two numbers such, that if the first be increased by a, and the second by b, the product of these two sums exceeds the product of the two numbers themselves by c; if, on the other hand, the first be increased by Equations of the First Degree solved by Elimination by Substitution. a', and the second by b', the product of these sums exceeds the products of the two numbers themselves by c'. 10. A person had two barrels, and a certain quantity of wine in each. In order to obtain an equal quantity in each, he poured out as much of the first cask into the second, as the second already contained; then, again, he poured out as much of the second into the first as the first then contained, and lastly, he poured out again as much from the first into the second as the second still contained. At last he had 16 gallons of wine in each cask. How many gallons did they contain originally? Ans. The first 22, the second 10 gallons. 11. 21 lbs. of silver lose 2 lbs. in water, and 9 lbs. of copper lose 1lb. in water. Now, if a composition of silver and copper weighing 148 lbs. loses 143 lbs. in water, how many lbs. does it contain of each metal ? Ans. 112 lbs. of silver, and 36 lbs. of copper. 12. A given piece of metal, which weighs plbs., loses elbs. in water. This piece, however, is composed of two other metals A and B such, that plbs. of A lose a lbs. in water, and plbs. of B lose blbs. How much does this piece contain of each metal? 13. According to Vitruvius, the crown of Hiero, king of Syracuse, weighed 20 lbs., and lost 11 lbs. in water. Assuming that it consists of gold and silver only, and that 19,64 lbs. of gold lose 1 lb. in water, and 10,5 lbs. of silver lose 1lb. in Equations of the First Degree solved by Elimination by Substitution. water. How much gold, and how much silver, did this crown contain ? Ans. 14,77... lbs. of gold, and 5,22...lbs. of silver, 151. Problem. To solve any number of equations of the first degree with the same number of unknown quantities. Solution. Let there be three equations with three unknown quantities; these equations may, by art. 140, be reduced to the forms A" x+B" y +C" z+M"= 0. The value of x, given by the first of these equations, is which, being substituted in the other two equations, and the resulting equations being reduced, as in art. 140, gives = 0. (A B'—A' B) y +(A C'—A' C) z+A M'—A' M = 0, (AB"—A"B) y+(AC"—A"C)z+AM'—A"M These equations, being solved, as in art. 146, give y= %= (A'C"—A"C')M+(A"C—A C")M'+(A C—A'C)M" (A'B"—A"B')C+(A"B—AB")C'+(AB'—A'B)C in which the terms are arranged in groups in order to display the symmetry of the result; and these values, being substituted in the value of x, give x= (B"C—B'C'')M+(BC"—B"C)M'+(B'C—BC')M" (A'B"—A"B')C+(A′′B—AB")C'+(AB'—A'B)C" Examples to be solved by Elimination by Substitution. If this method of solution be applied to a greater number of equations, it will lead to similar results. 53xzy - 109, Ans. x= 64, y = 80, z = 100. 4. Solve the four equations x + y + x + u = 1, 16x8y+4x+2u=9, 81x+27y+ 9z+3u36, 256x+64 y + 16 z+4 u = 100. Ans. x, y=1, z=1, u=0. 5. The sums of three numbers, taken two and two, are a, b, c. What are they? Ans. (a+b—c), ≥ (a+c—b), ±(b+c— — a). 6. A, B, C compare their fortunes. A says to B, “ give me $700 of your money, and I shall have twice as much |