100. Let us now consider the case of three equations involving three unknown quantities. Take the equations, 5x 15. 6y+4z 2x + y + 6z = 46. To eliminate z from the first two equations, multiply the first equation by 3 and the second by 4; and since the co-efficients of z have contrary signs, add the two results together: this gives a new equation Multiplying the second equation by 2, a factor of the co-efficient of z in the third equation, and adding them together, we have 43x-2y = 121 16x + 9y = 84 The question is then reduced to finding the values of x and y, which will satisfy these new equations. Now, if the first be multiplied by 9, the second by 2, and the results be added together, we find 419x= 1257, whence x = By means of the two equations involving x and y, we may determine y as we have determined x; but the value of y may be determined more simply, by observing, that by substituting for x its value found above, the last of the two equations becomes, 48 +9y=84, whence y 84- 48 = 4. In the same manner, by substituting the values of x and y, the first of the three proposed equations becomes, 15 -24 + 4z = 15, whence 2= 24 = 6. 101. Hence, if there are m equations involving a like number of unknown quantities, the unknown quantities may be eliminated by the following RULE. I. To eliminate one of the unknown quantities, combine any one of the equations with each of the m1 others; there will thus be obtained m - 1 new equations containing m -1 unknown quantities. II. Eliminate another unknown quantity by combining one of these new equations with the m 2 others; this will give m tions containing m-2 unknown quantities. -- 2 equa III. Continue this series of operations until a single equation is obtained containing but one unknown quantity, the value of which can then be found. Then by going back through the series of equations the values of the other unknown quantities may be successively determined. 102. It often happens that some of the proposed equations do not contain all the unknown quantities. In this case, with a little address, the elimination is very quickly performed. Take the four equations involving four unknown quantities, By examining these equations, we see that the elimination of z in equations (1) and (3), will give an equation involving x and y; and if we eliminate u in the equations (2) and (4), we shall obtain a second equation, involving x and y. In the first place, the elimination of z, in (1) and (3) gives 7y-2x= 1 that of u, in (2) and (4), gives 20y + 6x = 38 Multiplying the first of these equations by 3, and and adding, we have whence Substituting this value in 7y-2x 1, we find Substituting for x its value in equation (2), it becomes, 4u630, whence And substituting for y its value in equation (3), there results Of indeterminate Problems. 103. In all the preceding reasoning, we have supposed the number of equations equal to the number of unknown quantities. This must be the case in every problem, in order that it may be determinate; that is, in order that it may admit of a finite number of solutions. Let it be required, for example, to find two quantities such, that five times one of them, diminished by three times the other, shall be equal to 12. If we denote the quantities sought by x and y, we shall have and any two corresponding values of x, y, being substituted in the given equation, will satisfy it equally well: hence, there are an infinite number of values for x and y which will satisfy the equation, and consequently, the problem is indeterminate; that is, it admits of an infinite number of solutions. If, however, we impose a second condition, as for example, that the sum of the two quantities shall be equal to 4, we shall have a second equation, x + y = 4; and this, combined with the equation already considered, will give determinate values for x and y. If we have two equations, involving three unknown quantities, we can eliminate one of the unknown quantities, and thus obtain an equation containing two unknown quantities. This equation, like the preceding, would be satisfied by an infinite number of values, attributed in succession, to the unknown quantities. Since each equation expresses one condition of a problem, therefore, in order that a problem may be determinate, its enunciation must contain at least as many different conditions as there are unknown quantities, and these conditions must be such, that each of them may be expressed by an independent equation; that is, an equation not produced by any combination of the others of the system. If, on the contrary, the number of independent equations exceeds the number of unknown quantities involved in them, the conditions which they express cannot be fulfilled. For example, let it be required to find two numbers such that their sum shall be 100, their difference 80, and their product 700. The equations expressing these conditions are, Now, the first two equations determine the values of x and y, viz., x = 90 and y = 10. The product of the two numbers is therefore known, and equal to 900. Hence, the third condition cannot be fulfilled. Had the product been placed equal to 900, all the conditions would have been satisfied, in which case, however, the third would not have been an independent equation, since the condition expressed by it, is implied in the other two. EXAMPLES. 1. Given 2x+3y=16, and 3x-2y=11 to find the values of x and y. Ans. x = 5, y = 2. +7y=99, and 2+7x= 51 to find the values 2 12= 1+8, and to find the values of x and y. 1. What fraction is that, to the numerator of which, if 1 be added, its value will be one third, but if one be added to its denominator, its value will be one fourth. 2. A market woman bought a certain number of eggs at 2 for a penny, and as many more, at 3 for a penny, and having sold them again altogether, at the rate of 5 for 2d., found that she had lost 4d.: how many eggs had she? |