Let us eliminate x. Substitute, in 4th and 5th, the value of x given in 1st, (6) 4y+4x+80=20x+5y; (7) 8y+8%+160+5y+20 z=480. Transpose and reduce in 6th and 7th, Substitute 16 for y, and 4 for z, in 1st, 3. The ages of 3 children are as follows. Four times A's age and 3 times B's make 27 years; 3 times A's added to C's make 15 years; and 4 times B's with twice C's make 32 years. Required the age of each. Let x, y, and z (years), represent the respective ages of A, B, and C. Then, (1) 4x+3y=27; (2) 3x+ z=15; (3) 4y+2x=32. In this example, each equation does not contain all the unknown quantities; nor is that necessary in order that the solution may be possible. Let us eliminate z by the first method. But since the 1st does not contain z, we have to eliminate that letter from the 2d and 3d only. Divide 3d by 2, (4) 2y+ z=16. Subtract 2d from 4th, (5) 2y-3x= 1. Bring down 1st, (1) 4x+3y=27. Let us now eliminate y from 5th and 1st. Multiply 5th by 3, and 1st by 2, (6) 6y-9x= 3; 1 3; } (7) 8x+6y=54. Subtract 6th from 7th, Substitute 3 for x in 1st, 12+3y=27,.. y=5 years, B's age. 9+z=15,..z6 years, B's age. Let the learner solve this question according to the second and third methods From what precedes, it is evident that the three modes of elimination, given in the last section, may be extended to any number of equations, provided the number of unknown quantities does not exceed the number of equations. The first method is applied to several equations by operating upon these equations taken two and two. In applying the second method to several equations, find, from each equation that contains it, the value of the unknown quantity to be eliminated, then put any two of these expressions for its value equal to each other. To extend the third method, we must, after having found, from one of the equations, the value of the unknown quantity to be eliminated, substitute this value in every other equation that contains this unknown quantity. If a question involves four unknown quantities, and gives rise to four independent equations, we first deduce from them three equations with only three unknown quantities; we then proceed with these three equations, as we have done with the preceding equations in this section. If either of the equations does not contain the unknown quantity to be eliminated, that equation may be put aside to be placed in the next set of equations, viz., those which contain one less unknown quantity. Either method of elimination may be used, but the first will generally be found the most convenient, because it does not give rise to fractions. The pupil is advised, however, to perform each problem in the three ways, in order to acquire skill and be able to judge which will be best in any particular case. It is not necessary that the same mode of elimination be pursued throughout the solution of a question, but either may be resorted to whenever it shall seem the most convenient. Let the learner find the values of x, y, and z, in the following sets of equations. ART. 76. Let the learner find the numerical value of the following expressions, when a=4, b=3, c=2, and d=5. 19. abc+d. 20. a2 b2+c3 d. 24. 2a+3c-2b. 26. 3a24b2-c-2 d. 27. (a+b) c. 28. ac (b+d). 29. (a+b) (c+d). 30. (a+b)(dc). 32. (b2+a2) (d—c). 33. (3 a2 — 2 b2) (4 d —3 c). Find the value of the a=5, b=3, m2, and n = 0. 41. a+b-m+n. 42. 3a-5n+4m. 43. ab+mn. 49. 50. 5 abn. 51. 52. 45. mn+3abn. 46. 4abcn-5 a b2. 3 mn 3 (a —b) (m — n) Substitute numbers in the following equivalent expressions, showing their identity, whatever numbers are put instead of the letters; provided, however, that the same value be given to the same letter in the two members of an equation. An identical equation is one in which the two members are exactly alike; they may differ in form, but both can be reduced to the same form. 53. (a+c) b=ab+bc. In this example, let a=1, b=2, and c=3. Then the first member becomes (1+3)2=4.2=8. The second member gives 1.2+2.3=2+6=8. The results, we perceive, are alike. 54. am-bmm (ab). 55. (a—m) (a+m) = a2 — m2. 56. (a+m)(a+m)=a2+2 am+m2. 64. (a+m)(a+n) = a2+a (m+n)+mn. |