{ X 11. Given 1209}=60x+777 to find u and y. 24x-35y=-152 Ans. x=73; y=94. 13+, 33–87 + 7 =x+y-53 3 12. Given to find a and y. 8-(y-) Ans. x=1; y=2. by_4y=19_*+ 20–2y -=t 6 3 3 13. Given to find a and y. х+5y 2y+21 +5= 6 3 Ans. x=5; y=7. 13 3 3+2y+3 4х — бу +6 14. Given to find x and y. 19 Ans. x=7; y=8. y. a-72 26 15. Given ** ==8 Ans. x= Equations of the First Degree containing more than Two Unknown Quantities. 152. If we have three simultaneous equations containing three unknown quantities, we may, by the preceding methods, reduce two of the equations to one containing only two of the unknown quantities; then reduce the third equation and either of the former two to one containing the same two unknown quantities; and from the two equations thus obtained, the unknown quantities which they involve may be found. The third quantity may then be found by substituting these values in either of the proposed equations. Take the system of equations 2x+3y+4z=16, (1.) 3x+2y-52= 8, (2.) 5a-6+3x= 6. (3.) Multiplying (1) by 3, and (2) by 2, we have 6x+9y+122=48, (4.) 6x+4y-10z=16. (5.) Subtracting (6) from (4), 5y+22z=32. 5), + (6.) Multiplying (1) by 5, and (3) by 2, we have 10x+15+202=80, (7.) 10x— 12y+6z=12. (8.) Subtracting (8) from (7), 27y +14z=68. (9.) (10.) (11.) 2=1. Substituting this value of x in (6), 5y+22=32; whence Y=2. Substituting the values of y and z in (1), 2x+6+4=16; wlience 2=3. z 153. Hence, to solve three equations containing three un. known quantities, we have the following RULE. From the three equations deduce two containing only two unknown quantities; then from these two deduce one containing only one unknown quantity. 154. If we had four simultaneous equations containing four unknown quantities, we might, by the methods already explained, eliminate one of the unknown quantities. We should thus obtain three equations between three unknown quantities, which might be solved according to Art. 152. So, also, if we had five equations containing five unknown quantities, we might, by the same process, reduce them to four equations containing four unknown quantities, then to three, and so on. By following the same method, we might resolve a system of any number of equations of the first degree. Hence, if we have m equations containing m unknown quantities, we proceed by the following RULE. 1st. Combine successively any one of the equations with each of the others, so as to eliminate the same unknown quantity; there will result m-1 new equations, containing m-1 unknown quantities. 2d. Combine any one of these new equations with the others, so as to eliminate a second unknown quantity; there will result m-2 equations, containing m—2 unknown quantities. 3d. Continue this series of operations until there results a single equation containing but one unknown quantity, from which the value of this unknown quantity is easily deduced. 4th. Substitute this value for its equal in one of the equations containing two unknown quantities, and thus find the value of a second unknown quantity; substitute these values in an equation containing three unknown quantities, and find the value of a third; and so on, till the values of all are determined. Either of the unknown quantities may be selected as the one to be first eliminated. It is, however, generally best to begin with that which has the smallest coefficients; and if each of the unknown quantities is not contained in all the proposed equations, it is generally best to begin with that which is found in the least number of equations. Sometimes a solution may . be very much abridged by the use of peculiar artifices, for which no general rules can be given. EXAMPLES. Solve the following groups of simultaneous equations: 2x+4y—3z=22 x=3. 1. } 42—2y+52=18 Ans. {y=7. 6x+7-2=63 2=4. x+y=a 2. x+2=0 y+z=C Note. Take the sum of the three preceding equations. 2+y+z=29 x= 8. 3. x+2y+3z=62 Ans. 18+}y+z=10 2=12. 3+ }y+z=32 4. 3x+4y+}z=15 (x+}y+z=12 x+y-2=1320 x=987. 5. X-Y+z= 654 Ans. y=654. . -+y+z=-12) z=321. = Y= 9. |