243. It is important to remark that the same formule would still serve, when the proposed equations should not have all their terms affected with the sign + If we had for example, 3x-9y+3x-41; -5x+4y+22=-20; The comparison of these with the general equations (Art. 207), by having regard to the signs, will give a=3, b-9, c=+8, d=+41, a"+11, b"-7, c"-6, d" +37. In substituting these values in the formulæ (c), (d), and (e); we must determine the sign which every term ought to have, according to the signs of the factors of which it is composed: it is thus that we would find, for instance, that the first term of the common denominator, which is ab'c", becoming +3x+4x-6, changes the sign, and the product is -72. By observing the same with regard to the other terms, in the numerators, as well as in the common denominator, collecting into one sum those that are positive, and into another, those that are negative, we shall find 244. Having fully explained what concerns the elimination of unknown quantities in simple equations, and also illustrated the characters by which it may be known, if the proposed equations be determinate, indeterminate, or impossible; we may now proceed to the resolution of examples in determinate equations of the first degree: the practical rules that are necessary for this purpose, shall be pointed out in the two following sections. § II. RESOLUTION OF SIMPLE EQUATIONS, Involving two unknown Quantities. 245. When there are two independent simple equations, involving two unknown quantities, the value of each of them may be found by any of the following practical rules, which are easily deduced from the Articles in the preceding Section. RULE I. 246. Multiply the first equation by the coefficient of one of the unknown quantities in the second equation, and the second equation by the coefficient of the same unknown quantity in the first. If the signs of the term involving the unknown quantity be alike in both, subtract one equation from the other; if unlike, add them together, and an equation arises in which only one unknown quantity is found. Having obtained the value of the unknown quantity from this equation, the other may be determined by substituting in either equation the value of the quantity found, and thus reducing the equation to one which contains only the other unknown quantity. Or. Multiply or divide the given equations by such numbers, or quantities, as will make the term that contains one of the unknown quantities the same in each equation, and then proceed as before. Ex. 1. Given of x and y. (2x+3y=23, to find the values Multiply the 1st equ" by 5, then 10x+15y=115; 2nd by 2, then 10x ... by subtraction, 19y=95, 95 4y= 20; by division, y=19' ...y=5. Now, from the first of the preceding equations we shall have x= 8 4. 4 23-3y 23-15 =(since y=5) 2 2 The values of x and y might be found in a similar manner, thus: Mult. the 1st equation by 2, then 4x+6y=46; 2nd. by 3, then 15x—6y=30; Now, from the first of the preceding equations, Ex. 2. Given values of x and y. S4x+9y=35, to find the Mult. the 1st equation by 6, then 24x+54y=210; 2nd Now, from the first of the preceding equations, we shall have x 35-9y 4 4 The values of x and y may be found thus ; Mult. the 1st equation by 3, then 12x+27y=105; 2, 12x+24y 96; 2nd 4 The numbers 3 and 2, by which we multiplied the given equations, are found thus ; The product of two numbers or quantities, divided by their greatest common measure, will give their least common multiple, (Art. 146). 6X4 12 the least common multiple. 2 12 Then - 4 3, the number by which the first equa 12 tion is multiplied; and -2, the number by which 6 the second equation is multiplied. By proceeding in a similar manner with other equations, the final equation will be always reduced to its lowest terms. Ex. 3. Given lues of x and y. S5x+4y=58, 3x+7y=67, } to find the va Mult. the 2nd equation by 5, then 15x+35y=335; 1st. 39 15x+12y=174; 30 whence, 5x=58-4y=58-28=30, and .. x= -6. 5 If the second equation had been multiplied by 4, and subtracted from the first when multiplied by 7, an equation would have arisen involving only x, the value of which might be determined, and thence, by substitution, the value of y. Ex. 4. Given values of x and y. 56x-2y= 14, to find the 5x-6y=-10, Mult. the 1st equation by 3, 18x-6y= 42; but 5x-6y=-10; ... by subtraction, 13x=52, and x=4. |