SIMPLE EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. (ART. 61.) We have thus far considered such equations only as contained but one unknown quantity; but we now suppose the pupil sufficiently advanced to comprehend equations containing two or more unknown quantities. There are many simple problems which one may meet with in Algebra, which cannot be solved by the use of a single unknown quantity, and there are also some which may be solved by a single letter, that may become much more simple by using two or more unknown quantities. When two unknown quantities are used, two independent equations must exist, in which the value of the unknown letters must be the same in each. When three unknown quantities are used, there must exist three independent equations, in which the value of any one of the unknown letters is the same in each. In short, there must be as many independent equations as unknown quantities used in the question. An independent equation may be called a primitive or prime equation-one that is not derived from any other equation. Thus, x+3y=a, and 2x+6y=2a, are not independent equations, because one can be derived from the other; but x+3y =a, and 4x+5y=b, are independent equations, because neither one can be reduced to the other by any arithmetical operation. The reason that two equations are required to determine two unknown quantities, will be made clear by considering the following equation: x+y=20 =1, and y=19, This equation will be verified if we make x= or x=2, and y=18, or x=1, and y=191, &c., &c., without limit. But if we combine another equation with this, as x—y=4, then we have to verify two equations with the same values to x and y, and only one value for 2 and one value for y will answer both conditions. That is, we have found a value for x and another to y (12 and 8), so that their sum shall be 20, and their difference 4; and no other possible numbers will answer. A merchant sends me a bill of 16 dollars for 3 pairs of shoes and 2 pairs of boots; afterward he sends another bill of 23 dollars for 4 pairs of shoes and 3 pairs of boots, charging at the same rate. What was his price for a pair of shoes, and what for a pair of boots? This can be resolved by one unknown quantity, but it is far more simple to use two. Let x= the price of a pair of shoes, Then by the question 3x+2y=16 These two equations are independent; that is, one cannot be converted into the other by multiplication or division, notwithstanding the value of x and of y are the same in both equations. Equations are independent when they express different conditions, and dependent when they express the same conditions under different forms. To reduce equations involving two unknown quantities, it is necessary to perform some arithmetical operation upon them, which will cause one of the unknown quantities to disappear. These operations are called elimination. There are three principal methods of elimination. 1. By comparison. 2. By substitution. 3. By addition or subtraction. All the operations rest on the axioms. FIRST METHOD. (ART. 62.) Transpose the terms containing y to the right hand sides of the equations, and divide by the coefficients of x, and Put the two expressions for x equal to each other (Ax 7), An equation which readily gives y=5, which, taken as the value of y in either equation (C) or (D), will give x=2. SECOND METHOD. (ART. 63.) To explain the second method of elimination, resume the equations 3x+2y=16 4x+3y=23 (4) (B) The value of x from equation (A) is x=3(16—2y). Substitute this value for a in equation (B), and we have 4×}(16—2y)+3y=23, an equation containing only y. Reducing it, we find y=5, the same as before. Observe, that this method consists in finding the value of one of the unknown quantities from one equation, and substituting that value in the other. Hence, it is properly called the method by substitution. THIRD METHOD OF ELIMINATION. (ART. 64.) Resume again 3x+2y=16 4x+3y=23 (4) (B) When the coefficients of either x or y are the same in both equations, and the signs alike, that term will disappear by subtraction. When the signs are unlike, and the coefficients equal, the term will disappear by addition. To make the coefficients of x equal, multiply each equation by the coefficient of x in the other. To make the coefficients of y equal, multiply each equation by the coefficient of y in the other. Multiply equation (A) by 4 and 12x+8y=64 Multiply equation (B) by 3 and 12x+9y=69 Difference y=5, as before. To continue this investigation, let us take the equations Multiply equation (A) by 2, and equation (B) by 3, and Equations in which the coefficients of y are equal, and the signs unlike. In this case add, and the y's will destroy each other, giving 19x=76 Or . x=4 Of these three methods of elimination, sometimes one is preferable and sometimes another, according to the relation of the coefficients and the positions in which they stand. No one should be prejudiced against either method; and in practice we use either one, or modifications of them, as the case may require. The forms may be disregarded when the principles are kept in view. Here, it would be very inexpedient to take the first method of elimination. Observe that the coefficients of y are alike in number, but opposite in signs. A skillful operator takes great advantage of circumstances, and very rarely goes through all the operations of set rules; but this skill can only be acquired by observation and practice. Add the two equations. Why? |