Clearing these equations of fractions, and simplifying the results, we get 7x−y+2=35, or 7x - y = 33. 12y x - 10 9, or 12 y (1), From the first equation, y = 7 x — 33, and this value substituted for y in the second, gives and y. Multiply the first of these equations by 3, and the second by 2; then Again, multiply the first equation by 7 and the second by 3, then 2 4. Given ax + by = c, and a, x + b1y = c1, to find the values of x and y. Multiply the first equation by a, and b, separately, and the second equation by a and b, separately; then we have ab1x+bb1y = b1c This may be considered as the general solution of simple equations containing two unknown quantities. Clearing these equations of fractions, by multiplying the terms of the first by 30, and those of the second by 132, we get 9x+12y+9-4x-14+2y=150+6y-48, or 5x + 107, - 80. = To four times the first of these, viz., 20x + 32 y 51x y = 31 x · 33 y = = 80, = and we get Substitute this value for y in the equation, 5x+8y = 107, and we get 5x+8 (51x — 348) = 107, or 5x+408x-2784 = Hence also .. 413x=2784+ 1072891, and x= y = 51 x 348 357 348 = 9. 6. Let a (x + y2) − b (x 2 — y3) = 2a, = 4 ab, to find the values of x and y. 107; 7. and (a3 — b3) (x2 —y3) = 2 a From the first we have (a - b) x2 + (a + b) y2 and from the second (a - b) x2 — (a2 — b2) y2 = 4 ab Multiply the terms of the former by a+b, then (a2 — b3) x2 + (a2 + 2 a b + b2) y3 = 2 a2 + 2 a b ; (1), (2). but, Again, multiply the terms of (1) by a ― b, then (a2 — 2 a b + b2) x2 + (a2 − b2) y2 = 2 a2 - 2 ab; but, (2 a 2 ab) x = 2 a2+2ab, or 2 a (a - b) x2 = 2a (a+b); (x+7) (y-2)+3=2xy-(x+1) (y-1).) 14. xy3 37, 3x4y = 0. Ans. x=4, y=3. = 19.√(x+2y-1)−1 = x, √√ (y2+3x)-1=y. Ans. x=3, y=4. SIMULTANEOUS EQUATIONS OF THE FIRST DEGREE, CONTAINING THREE OR MORE UNKNOWN QUANTITIES. 98. The methods to be employed in the solution of equations containing several unknown quantities are precisely similar to those which have been employed in the solution of equations involving two unknown quantities. It may readily be shown that if the values of three unknown quantities are to be determined, it is necessary that there should be given three independent and consistent equations, and in general there must be given as many independent and consistent equations as there are unknown quantities to be determined. By eliminating one of the unknown quantities from the three equations, there will arise two equations involving the two other unknown quantities, and these may be treated by the methods already detailed. Proceed in a similar manner with equations containing several unknowns. = EXAMPLES. 1. Given x + y + 2 z = 19, 2x + 3y - 5z7, and 5x but = = 38, . 31. (1). Again, multiply the first equation by 5, then will 5x+5y+10 z 95, 3 z = 47 = 66z= - 217, 264, and z = 4. 31 = 36 · 31 y= = 9z x = 19 y 2 z = 19 = = 5, 8 = 6. 17, y + 2 z = 11, and 3x 42 = 9, to find the Subtracting this last equation from the third equation, gives 7; 5, or x-4 z = — 5. 12, hence z = = 5. 3, EXAMPLES FOR PRACTICE. 1. x+y=30, x+z=21, y+z=27. 2. x-y=6,x−z=4, y + z = 34. 3. 5x+3y=84, 4x+6z=108, 5. 3x+4y+5 z=38, 4x+3y-4x=1, 6x-2y+7= = 34. Ans. x 2, y = 3, z= 4. 6. x+y+z=15, x+y− z=3, x − y + z = 5. Ans. x 4, y = 5, z = 6. 7.3x+2y-z 20, 2 x + 3y+6z70, xy+6z= 41. QUESTIONS PRODUCING SIMPLE EQUATIONS. 99. When a question is proposed to be solved by the principles of algebra, its meaning ought to be perfectly understood, and its condition or conditions exhibited in the clearest manner. If there is only one unknown quantity to be determined, one condition will be sufficient, and if several quantities are to be determined, we must have as many independent conditions as there are unknown quantities to be found. In reducing problems to algebraical equations no general rule can be given, but the following hint may be useful. Suppose that the values of the quantities to be determined are actually found, and consider by what operations the truth of the solution may be verified. Then, instead of known numbers, make use of unknown symbols for the quantities to be determined, and let the same operations be performed with these letters as were performed with the numbers, and we shall obtain certain equations expressing the conditions of the question. The solution of these equations will give the values of the unknown quantities. Thus if the question were to find the number which exceeds its fourth part by 27, then supposing the number were found, viz. 36, the process of verification would be as follows: Now if x denote the number required, we should, by following the same steps, arrive at the equation and the solution of this equation would give the number required; for, multiplying by 4, we get 4 x - x = 108, or 3 x = 108; hence x = 36. 100. When several quantities are to be determined, there must be given an equal number of independent conditions, and for the purpose of expressing these conditions we may find it convenient to satisfy one or more of them by a judicious selection of an unknown quantity which has some known relation to the required quantity. Thus, if the sum of two quantities be given equal to 2s, we may denote their difference by 2x; then s+x and s- x would denote the greater and less numbers. In a similar manner, if the difference of two numbers be given equal to 2 d, then x+d, and x d might be employed to denote the two numbers, for the condition of their having a given difference would be fulfilled. Also, if the ratio of two numbers be given, as m to n, then mx and nx might be put for the numbers, and one condition would be fulfilled; and if the product of two numbers be given equal to p, then P might represent the two numbers, for then xx given product, and so on. x and P = p = the x |