5+/25-22 10. Given 5-√/25-x2 =9, to find the values of x. The value of a fraction not being altered by multiplying the numerator and denominator by the same number; therefore, multiply the numerator and denominator of the fractional part by the numerator, and 2 5+√/25—x2 :9; the equation becomes 5+√/25—x2 extracting the square root, =±3; x :. 5+√/25—x2=±3x ; by transposition, 25-x2+3x-5; EXAMPLES FOR PRACTICE In Pure Quadratics, &c. of one and two unknown Quantities. 1. Given 4x2-35-3x2-10, to find the values of x. 4. Given x+y: x-y::7:1, to find the values and x-y=63, of x and y. Sx=+12. Ans. 圭 5. Given x2—y2 ; x2+y2:: 5; 13, to find the va and xy=12, lues of x and y. Ans. Sx=6. 6. Given x2+2xy+y2: (x—y)2:: 16:1, to find and 3x2-2y2=57, S the vaAns. Jx=±5. lues of x and y. 7. Given x2-y2: x+y y=±3. 2:1, to find the values Sof x and y. x=5 or -3. Ans. y=3 or −5. to find the va and x-y 2 lues of x and y. y=2. 9. Given x3-y3: x2y—xy2:: 7:2, to find the va and x+y=6, 14. Given x+6xy. (x1+y1)+4x3y3+yo=46656— 12x2y2. (x2+xy+y2), and x-y=2, to find the values of x and y. Sx=4, or Ans. y=2, or 2. 4. 19. Given y3152, to find the values of x and y. and xy 2,5 20. Given x+xy3+x3y+y®-8301256384, and a to xy3+y=2159536, find the values of x and y. Ans. x=20. y=42. EXAMPLES FOR PRACTICE In the application of Pure Quadratics, &c. of one and two unknown quantities. NOTE.-The application of Pure Quadratics to the solution of problems, is precisely similar to the application of Simple Equations to the solution of problems already given. 1. Bought a piece of cloth for £26. 9s., at as many shillings per yard as the piece contained yards. Required its length. Ans. 23 yards. 2. Find two numbers, the sum of whose squares is to their difference as 5 to 3, and whose product is 18. Ans. 6 and 3. 3. A farmer has two square fields, the length of a side of the greater is 5 chains more than a side of the less, and their areas are in the proportion of 9 to 16. Required the area of each. Ans. (22a. 2r. 40a. Or. 4. Bought a piece of cloth and a piece of silk for £9, and paid as many shillings per yard for the cloth as there were yards of silk, and for the silk as many shillings per yard as there were yards of cloth, and the square of the number of yards of silk is to the square of the number of yards of cloth as 25 to 4. Required the number of yards of each. Ans. 15 yards of silk, and 6 yards of cloth. 5. Upon a rectangular piece of building ground, whose length is to its breadth as 5 to 3, there is to be a house erected upon the end of it the whole width of the ground, to take up one fifth of the whole, and to leave 300 square yards for a garden. Required the dimensions of the house and garden. Ans. The house 15 yards by 5, and the garden 20 yards by 15. 6. Divide the number 75 into two such parts, that the quotient of the less divided by the greater, may be to the quotient of the greater divided by the less, asz to 4. Ans. 35 and 40. 7. A company of merchants built a ship, which cost £5000, and divided their interest therein into a certain number of shares, portioning to each as many shares as there were merchants in company; the square 3 of the number of merchants, is to of the number 5 of shares +5, as 5 to 4. Required the number of merchants, and the sum contributed by each. Ans. 5 merchants, and £1000 the sum contributed by each. 8. Bought a number of yards of silk for £3. 15s., and a quantity of cloth, which was in proportion to the silk as 4 to 5: I found that 5 yards of cloth cost |