By inspecting these equations, we see that the elimination of z in the two equations, (1) and (3), will give an equation involving x and y; and if we eliminate u in the equations (2) and (4), we will obtain a second equation, involving x and y. These two last unknown quantities may therefore be easily determined. In the first place, the elimination of z in (1) and (3) gives That of u in (2) and (4), gives Multiplying the first of these equations by 3, 7y-2x=1 20y+6x=38 41y=41 y= 1 and adding Whence it becomes 4u-6-30, whence And substituting for y its value in equation (3), there results EXAMPLES. u= 9 2=5 1. Given 2x+3y=16, and 3x-2y-11 to find the values of х 3. Given -+7y=99, and 2+7x=51, to find the values of 7 Ans. x=2, y=4, z=3, u=3, t=1. 103. In all the preceding reasoning, we have supposed the number of equations equal to the number of symbols employed to denote the unknown quantities. This must be the case in every problem involving two or more unknown quantities, in order that it may be determinate; that is, in order that it may not admit of an infi. nite number of solutions. Suppose, for example, that a problem involving two unknown quantities, x and y, leads to the single equation, 5x-3y=12; we (y=1, x=3); (y=2, x=18); (y=3, x=21); &c. 5 substituted for x and y in the equation, will satisfy it equally well. If we had two equations involving three unknown quantities, we could in the first place eliminate one of the unknown quantities by means of the proposed equations, and thus obtain an equation, which, containing two unknown quantities, would be satisfied by an infinite number of systems of values taken for these unknown quantities. Therefore, in order that a problem may be determined, 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. QUESTIONS. 1. What fraction is that, to the numerator of which, if 1 be add Whence 3x+3=y, and 4x=y+1. Therefore, by subtracting, x-3=1 or x= 4. y=15. 2. A market woman bought a certain number of eggs at 2 for a penny, and as many others, 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? 3. A person possessed a capital of 30,000 dollars for which he drew a certain interest; but he owed the sum of 20,000 dollars, for which he paid a certain interest. The interest that he received exceeded that which he paid by 800 dollars. Another person pos sessed 35,000 dollars, for which he received interest at the second of the above rates, but he owed 24,000 dollars, for which he paid interest at the first of the above rates. The interest that he received exceeded that which he paid by 310 dollars. Required, the two rates of interest. Let x and y denote the two rates of interest: that is, the interest of $100 for the given time. To obtain the interest of $30,000 at the first rate denoted by x, we form the proportion And for the interest $20,000, the rate being y. But from the enunciation, the difference between these two interests is equal to 800 dollars. We have, then, for the first equation of the problem, 300x-200y=800. By writing algebraically the second condition of the problem, we obtain the other equation, 350y-240x310. Both members of the first equation being divisible by 100, and those of the second by 10, we may put the following, in place of them : To eliminate x, multiply the first equation by 8, and then add it to the second; there results 19y=95, whence y=5. Substituting for y, in the first equation, its value, this equation becomes 3x-10=8, whence x=6. Therefore, the first rate is 6 per cent., and the second 5. |