RULE I. Multiply each of the equations by such multipliers as will make the resulting coefficient of one of the unknown quantities the same in all the equations, then by adding or subtracting these equations, a new set of equations, one less in number, and containing one unknown quantity less than before, will be obtained; with which proceed as before. 77. RULE II. Find a value of one of the unknown quantities, from each of the equations, in terms of the other unknowns; then equate these values, and a new set of equations will be obtained, containing one unknown quantity less than formerly; with which proceed as before. 78. RULE III. Find a value of one of the unknown quantities from one equation, and substitute this value, instead of it, in the others, which will evidently give one equation less than formerly, containing all the unknown quantities, except that whose value was found, with which equations proceed as before. 79. RULE IV. Multiply all the equations except one by some conditional multipliers, as m, n, p, &c., then add all the multiplied equations and the unmultiplied one together, and a new equation will be obtained, in which if we make all the coefficients of the unknown quantities except one equal to 0, a set of equations among the conditional multipliers will be obtained, which will determine their values, and these values substituted in the resulting equation will give the value of the other unknown quantity whose coefficient was not considered to be 0. The above rules, as well as those given in Case I. are all evidently true from the axioms. (4x+y+2=30 2x+7y-82-24 to find the values of x, y, and z. Multiplying the first equation by 4, the second by 6, and the third by 3, they become by Rule 1, (16)÷12 The solution by Rules 2d and 3d are left as exercises for the pupils. The solution by Rule 4th is as follows: multiplying the first of the given equations by m, the second by n, and adding these products to the 3d of the given equations, we obtain the following: (3m+2n+4)x —(2m—7n—1)y+(6m—8n+1)=22m+24n+30. Let now m and n be such as to make the coefficient of y and z each equal to 0, and the following equations are obtained, in which m and n are the unknown quantities. 2m-7n-1-0. 6m-8n+1-0. 1 6m-21n-3-0. (2)-(3) 13n+4=0. Transp. (5)+13 In the same manner, or by substitution, we obtain m= 15 and since the coefficients of y and z are each=0, we 26 have (3m+2n+4)x=22m+24n+30, in which, by substituting the values of m and n, there results, hence 137 98. 43x=258, by mult. by 26. In the same manner, if the coefficients of x and z be equated to 0, we can find the value of y; or if the coefficients of x and y be equated to 0, we can find the value of z. to find the values of x, y, and z. 1. Given x+y-z=10 x-y+z=6 y+z―x=2 Ans. x=8, y=6, z=4. x+ y + z=35 to find the values of x, x+2y+32=66 x+y+12=13) Ans. x=12, y=15, z=8. 2. Given y, and z. to find the values of x, y, and z. Ans. x=16, y=12, z=10. 2x-20-3y-40 to find the values of x, y, and z. x+20=2%+5 Ans. x=35,y=30,z=25. PROBLEMS PRODUCING EQUATIONS WITH TWO OR MORE UNKNOWN QUANTITIES. 1. Find two numbers such that the first with twice the second shall be 21, and twice the first, with half the second, shall be equal to 14. Ans. 5 and 8. 2. Find two numbers such that one half the first, with one third the second, may be 7, and one third of the first, with one fourth of the second, may be 5. 3. Find two numbers such, that if 5 be first, the sum will be twice the second, and if four times the second be increased by 3, the sum will be three times the first. Ans. 13 and 9. Ans. 6 and 12. added to the 4. Find a fraction such, that if its numerator be increased by 1, and its denominator diminished by 2, its value will be, and if its numerator be increased by 3, and its denominator diminished by 2, the value will still be 3. Ans. 22. 5. Find a number, consisting of two digits, the sum of whose digits is equal to of the number, and the product of whose digits is equal to of the number. Ans. 36. 6. There is a certain number consisting of two figures, which is equal to 6 times the figure in the unit's place, and if 27 be added to the number, the digits will be inverted: what is that number? Ans. 36. 7. There is a certain number consisting of three figures, the sum of the digits is 7, twice the sum of the extreme digits is equal to 5 times the mean, and if 297 be subtracted from the number, the digits will be inverted: what is the number? Ans. 421. 8. Find three numbers, so that the first, with half the other two, the second with one-third of the other two, and the third with one-fourth of the. other two, may each be equal to 34. Ans. 10, 22, 26. 9. Find a number consisting of three figures, whose digits are in arithmetical progression, such that if this number be divided by the sum of its digits, the quotient will be 48; and if from the number 198 be subtracted, the digits will be inverted. Ans. 432. 10. If A and B together can perform a piece of work in 8 days, A and C together in 9 days, and B and C together in 10 days, in what time will each of them perform it alone? Ans. A in 143, B in 17, and C in 233 days. QUADRATIC EQUATIONS. 80. QUADRATIC EQUATIONS may be divided into two kinds, namely, such as contain only the square of the unknown quantity, and those which contain both the square and first power of the unknown quantity; the first are called Pure Quadratics, and their solution may be effected by Rule V. of Simple Equations, with one unknown quantity; the second are called Adfected Quadratics, and they are of one of the four following forms: ax2+bx=+c, All these are included in the general formula, ax2±bx c; and we proceed now to solve this equation. If the first term were multiplied by 4a, it would become the square of 2ax; let both sides be multiplied by this quantity, viz. 4a, and the general equation becomes 4a2x2+4abx =4ac; the first side of this equation evidently wants something of being the square of a binomial, of which the first term is 2ax; let this quantity be p2, then adding this quantity to both sides, the equation becomes 4a2x2-4abx +p-p4ac: now if the first side be a complete square, its square root can be no other than 2axp, (Art. 29 ;) hence squaring this, its square must be identical with the first side of the last equation, but its square is 4a2x2± 4apx+p2.. to 4a2x2+4abx+p2; hence taking away the common terms from both sides, 4apx=4abx, and dividing both sides by 4ax, we have pb, and therefore the quantity which must be added to both sides of 4a2x2 4abx4ac, so as to make it a complete square, is b2, that is, the square of the coefficient of x. If now the root of both sides be extracted, we obtain 2ax+b=√b2±4ac, bb4ac, which is a general 2a formula in which, if we insert the proper values of a, b, and c, with their proper signs, (the sign of a being always +), we will have two values of x, both of which fulfil the conditions of the algebraic equation. From the above investigation the following rule is derived: RULE. 81. 1. Transpose all the terms containing the unknown quantity to one side of the equation, and so that the term containing the square of the unknown quantity may be positive, and the known terms to the other. 2. Multiply both sides of this equation by four times the coefficient of x2. 3. Add the square of the coefficient of x in the first equation to both sides, then will the first side be a complete square. 4. Extract the root of both sides, and the result will be a simple equation, which may be solved by the previous rules. 82. SCHOLIUM. When the equation, after being transposed as in the Rule, can be divided by the coefficient of x2, without introducing fractions, it may be solved conveniently as follows:-Perform the division, then add the square of half the new coefficient of x to both sides, which will make the first side a complete square; then extract the square root of both sides, and the equation will be reduced to a simple one, which may be solved as before. EXAMPLE 1. Required the values of x in the equation 3x2-4x-160. 36x2-48x=1920, by multiplying by 12=(4×3), 36x2-48x+16=1936, by adding (4)2 to both sides, 6x-4-44, 6x=4 44, x=8, or -6, by extracting the square root, by transposition, by dividing by 6. Either of these values substituted in the given equation will make the sides equal, and they are therefore both roots of the equation. EXAMPLE 2. Find two numbers whose sum is 100, and whose product is 2059. Let x the one; then since their sum is 100, the other may be represented by 100-x; and hence their product will be x(100-x), which by the question is equal to 2059. Hence 100x x2-2059. x2-100x=-2059, by changing the signs. .'. x=50±21=71, or 29, which are the two parts required, and therefore x has come out either the greater or the less part. The solution by the rule is left for the exercise of the pupil. |