3 EXAMPLES. 1 1. Given x+y=3x and x+y=x, to find the values of x and y. Put x=P; then x=P3 and 2=P3 And y=Q; then y=Q3 and y=Q 2 Now the primitive equations become P3+Q2=3P2, and P+Q=P2 From the 1st, Q2=(3-P)P2 From the 2d, Q =(P-1)P : Put the two values of Q2 equal to each other, rejecting or dividing by the common factor P2, and we have 36, to 2. Given x+y+2x+2y=23, and find the values of x and y. Ans. x=27 or 8, y=8 or 27, (Art. 112.) No additional principles, to those already given, are requisite for the solution of problems containing three or more unknown quantities in quadratics. As in simple equations, we must have as many independent equations as unknown quantities. As auxiliary to the solution of certain problems, particularly in geometrical progression, we give the following problem: Given x+y=s, and xy=p, to find the expressions for x2+y2, x3+y2, x2+y2, and xs+ys, in terms of s and p. Squaring the first, x2+2xy+y2=s2 Subtract twice the second, 2xy =2p x2+y2 =s-2p (x+y)(x2+y2)=x2+x2y + xy2+y3=s3-2ps Subtract = ps (A) x+y=s3-3ps (B) xy(x+y) 2d result, Square (A), and 3d result, x2+2x2y2+y=s-1s2p+1p2 Multiply (A) by (B) and Subtract 4th result, x+x3y2+x2y3+y=s5-5s3p+6sp2 Questions producing Quadratic Equations. (Art. 113.) The method of proceeding to reduce the question into equations, is the same as in simple equations, and in fact many problems which result in a quadratic may be brought out by simple equations, by foresight and skill in notation. Others again are so essentially quadratic, that no expedient can change their form. EXAMPLES. 1. A person bought a number of sheep for $240. If there had been 8 more, they would have cost him $1 a piece less. What was the cost of a sheep, and how many did he purchase? Let x= the number of sheep; then 240 = cost of one. = cost of one. Clearing of fractions, 240x+1920-240x+x2+8x Resolving gives 240 or -48; but a minus number will not appply to sheep, the other value only will apply to the problem as enunciated. This question can be brought into a simple equation thus: Let x-4= the number of sheep, then 8 more would be expressed by x+4, and the equation would be Clearing of fractions, ax+4a=ax-4а+х2-16. Transposing, x2=8a+16=8(a+2)=16×121 Extracting square root, x=4×11=44. Hence, x-4=40, the number of sheep. Divide 240 by 40, and we have $6 for the price of one sheep. (Art. 114.) In resolving problems, if the second member is negative after completing the square, it indicates some impossibility in the conditions from which the equation is derived, or an error in forming the equation, and in such cases the values of the unknown quantity are both imaginary. 2. For example, let it be required to divide 20 into two such parts that their products shall be 140. Let x= one part, then 20-x= the other. Completing the square, x2-2x+10040 Or, x =10 ±2-10 This result shows an impossibility; there are no such parts of 20 as here expressed. It is impossible to divide 20 into two such parts that their product shall be over 100, the product of 10 by 10, and so with any other number. The product of two parts is the greatest possible, when the parts are equal. 3. Find two numbers, such that the sum of their squares being subtracted from three times their product, 11 will remain; and the difference of their squares being subtracted from twice their product, the remainder will be 14. Let x=the greater number, and y=the less. By the conditions, 3xy-x2-y2=11 These are homogeneous equations; therefore, put x=vy; Conceive (A) divided by (B) and the fraction reduced, we have 30-02-1 11 20-02+1 14 Clearing of fractions and reducing, we find 2502-2003. See Equation 14, (Art. 101,) from which we infer v Put this value in equation (A) and we have Multiply by 25, and 45y2-9y2-25y2=11×25 Or lly2=11×25 y2=25 or y=5. Hence, x=3. 4. A company dining at a house of entertainment, had to pay $3.50; but before the bill was presented two of them went away; in consequence of which, those who remained had to pay each 20 cents more than if all had been present. How many persons dined? Ans. 7. 5. There is a certain number, which being subtracted from 22, and the remainder multiplied by the number, the product will be 117. What is the number? Ans. 13 or 9. 6. In a certain number of hours a man travelled 36 miles, but if he had travelled one mile more per hour, he would have taken 3 hours less than he did to perform his journey. How many miles did he travel per hour? Ans. 3 miles. 7. A person dies, leaving children and a fortune of $46,800, which, by the will, is to be divided equally among them; but it happens that immediately after the death of the father, two of the children also die; and if, in consequence of this, each remaining child receive $1950 more than he or she was entitled to by the will, how many children were there? Ans. 8 children. 8. A gentleman bought a number of pieces of cloth for 675 dollars, which he sold again at 48 dollars by the piece, and gained by the bargain as much as one piece cost him. Ans. 15. What was the number of pieces? This problem produces one of the equations in (Art. 107.) 9. A merchant sends for a piece of goods and pays a certain sum for it, besides 4 per cent. for carriage; he sells it for $390, and thus gains as much per cent. on the cost |