The same observation applies to this method of solution, as did to the last. In some particular equations, two unknown quantities may be eliminated at once. Ex. 2. Given x+y+z=31 x+y—z=25 to find the values of x, y, Adding the first and third equations, 2x40; Subtracting the second from the first, 2z=6; and subtracting the third from the second, &t z. ...y=20. ..2=3; 2y=16;.. y=8. x-y=2, Ex. 3. Given x-2=3, to find x, y, and z. Here subtracting the first equation from the second, we have y-z=1; which is identically the third. Therefore, the third equation furnishes no new condition; but what is already contained in the other two; and, consequently, the proposed equations are indeterminate; or, what is the same, we may obtain an infinite number of values which will satisfy the conditions proposed. 204. It is proper to remark, that in particular cases, Analysts make use of various other methods besides those pointed out in the practical rules; in the resolution of equations, which greatly facilitate the calculation, and by means of which, some equations of a degree superior to the first, may be easily resolved, after the same manner as simple equations. We shall illustrate a few of those artifices by the following examples. By adding the three equations, we shall have 22 2 1:1 1 121 ++ -= tāt: x y Or, dividing by 2, 8 9 10 360° From this subtracting each of the three first equations, and By adding a to each member of the first equation, y to the second, and z to the third, we shall get which values being substituted in the first equation, we have but, by the fourth equation, u=x-14; 3x whence x=40: consequently y= 30, z=24, and u=r 14=26, 4 Ex. 6. Given 4x-4y—4x=24, to find the values of x, y, . and z. 6y-2x-2x-24, and 7z-y-x=24,. By putting x+y+z=S, the proposed equations become 8x-4S 24, 8y-2S-24, 8z-S=24; .. x=3+1S, y=3+1S, z=3+7S. By adding these three equations, we have x+y+x=9+78; whence S72. Substituting this value for S, in x, y, and z, we shall find Ex. 7. Given x=39, y=21, and z=12. zty+z=90, to find the values of x, y, 2x+40=3y+20, and z. and 2x-4x+40=10, J Ans. x=35, y=30, and z= =25. Ex. 8. Given +α= y+z, to find the values of x, y, y+a=2x+2z, and z+a=3x+3y, and z. Ex. 9. It is required to find the values of x, y, and z, in the following equations; x+y=13, x+z=14, and y+2=15. Ans. x= =6, y=7, and z=8. Ex. 10. In the following it is required to find the values of Ex. 11. Given x+y+z=26,) to find the values of x, y, x-y = 4, and x = 6, and z. Ans. 12, y=8, and z=6. Ex. 12. Given x+y+ z= 9, to find the values of x, y, x+2y+32=16, and z. and xy-2x= 3,. Ans. x=4, y=3, and z=2. Ex. 13. Given x+y+ z=12 and 2. x+2y+3x=20, and +y+ z= 6, to find the values of x, y, Ans. x 6, y=4, and z=2.. Ex. 14. Given x+y—z=8, x+z-y=9, and y+z—x= 10; to find the values of x, y, and z. Ans. x=81, y=9, and z=91. Ex. 15. Given x+2y=100, y+32=100, and z+4x=100; to find the values of x, y, and z. Ans. x=64, y=72, and z=84. Ex. 16. Given x+4y=357, y+1z=476, z+4u=595, and u+3x=714; to find the values of x, y, z, and u. Ans. x=190, y=334, z=426, and u=676. § IV. SOLUTION OF PROBLEMS PRODUCING SIMPLE EQUATIONS, Involving more than one unknown Quantity. 205. The usual method of solving determinate problems of the first degree, is, to assume as many unknown letters, namely, x, y, z, &c., as there are unknown numbers to be found; then, having properly examined the meaning and conditions of the problem, translate the several conditions into as many distinct algebraic equations; and, finally, by the resolution of these equations according to the rules laid down in Chapter IV, the quantities sought will be determined. It is.proper to observe that, in certain cases, other methods of proceeding may be used, which practice and observation alone can suggest. PROBLEM I. There are two numbers, such, that three times the greater added to one-third the lesser is equal 36; and if twice the greater be subtracted from 6 times the lesser, and, the remainder divided by 8, the quotient will be 4. What are the numbers ? Let a designate the greater number, and y the lesser number. Multiplying equation (A) by 6, 6y+54x=648; but 6y-2x= 32; .. by subtraction, 56x=616, and by division, x=11. From equation (A), y=108-9x; .. by substitution, y=108-99, or y=9. PROB. 2. After A had won four shillings of B, he had only half as many shillings as B had left. But had B won six shillings of A, then he would have three times as many as A would have had left. How many had each ? Let x designate the number of shillings A had, and y= the number B had; then y-4=2x+8, and y+6=3x-18; ... by subtraction, 10=x-26, and by transposition, 36=x, or x=36; by substitution, y+6=3x36-18; and by 'ransposition, y=84; A had 36, and B 84. PROB. 3. What fraction is that, to the numerator of which if 4 be added, the value is one-half, but if 7 be added to the denominator, its value is one-fifth? Let z= its numerator, then the fraction y= denominator, Add 4 to the numerator, then y x+4 =,.. 2x+8=y; :=}, ... 5x=y+7; 3+7 by subtraction, 3x-8=7; by transposition, 3x=15; .. x=5; and y=2x+8;.. by substitution, y=10+8=18, PROB. 4. A and B have certain sums of money, says A to B, give me 15/ of your money, and I shall have 5 times as much as you have left: says B to A, give me 5l of your money, and I shall have exactly as much as you will have left. What sum of money had each ? Let x A's money, then +15 what A would have, y= B's, after receiving 15/ from B. y-15 what B would have left. Again, y+5=what B would have after receiving 57 from A. x-5 what A would have left. Hence, by the problem, x+15=5×(y—15)=5y—75, and y+5=x-5. |