Three quantities being here given as equal to one another, we can form two equations among them by pairing them in either of two ways. Thus we may have as one pair of equations: or else 3x+5y-70=x+y+8 3x+5y-70= X 8y 5 3 If we adopt the former arrangement, the equations be Three quantities being given as equal to one another, we have the option of forming a pair of equations in two different ways. If we take the pair 227. If three distinct and compatible equations are presented between three unknown quantities, one of them has usually to be eliminated, so as to produce two equations between the remaining two, and these two quantities being determined by the methods already given, the one which was first eliminated can then also be found, and the solution thus completed. Ex. x+2y-33=-4, (1) x+2% y = 9, (2) 2y+2% x = 15. (3) It will be most convenient to eliminate x. (1) + (3) gives 4y— z = 11, (5) (2)+(3) 4× (5)+(6) y+4% = 24, (6) 228. In the following example it happens that the solution can be effected without that successive elimination of unknown quantities, which has been made in the last examples. CHAPTER VI. PROBLEMS PRODUCING SIMULTANEOUS EQUATIONS. 230. Simultaneous simple equations will now be employed in solving problems. We gain, it may be anticipated, increased power in expressing, algebraically, the facts of a problem, when we can introduce more than one unknown quantity to assist us in representing the elements to be determined. 231. It has been seen (210) that we can find values for two unknown quantities when we have two distinct and compatible equations between them. So, also, we can find values for three unknown quantities, if we have three distinct and compatible equations among them. Hence, generally, as many symbols as are assumed to designate unknown quantities in a problem, so many distinct and compatible equations must we obtain from the conditions of the problem, if the solution is to be effected. 232. 1. A purse contains shillings and sovereigns. Add a shilling, and then there are twice as many shillings as sovereigns. Add a sovereign to the original contents of the purse, and then there would be more shillings than sovereigns by 2. Find the number of shillings and sovereigns in the purse originally. Suppose that there were originally in the purse x sovereigns and y shillings. When a shilling is added there are x sovereigns and y+1 shillings, and the first given condition is When a sovereign is added there are +1 sovereigns and y shillings, and the second given condition is or there were originally 4 sovereigns and 7 shillings in the purse. 233. 2. A certain number, consisting of two digits, is equal to six times the sum of the digits, and if 117 be subtracted from three times the number, the digits are reversed. Find the number. Let x be the digit in the place of tens, and 10y+x is the number when the digits are reversed. .. 10x+y=6(x+y). 3(10x+y)-117 = 10y+x. Hence 4x 5y, and the number is 54. 29x-7y=117. .. x = 5, y = 4, (1) (2) 234. 3. A fraction becomes equal to 2 when 7 is added to its numerator, and equal to I when I is subtracted from its denominator. Find the fraction. |