1 II. Make the value of r in the first equation equal to the value of x in the second, and a new equation will be formed, involving only y and z, with known quantities. III. Make the value of x in the first equation equal to the value of x in the third, and a second new equation will be formed, involving in like manner only y, z, and known quantities ». IV. Find the value of y and z in these two new equations, by either of the former methods; then, by substituting these values for y and z respectively, in either of the given equations, the value of x will be readily obtained. 23. Given x+y+z=9, x+3y-3 z=7, and x−4y+8z=8; required the values of x, y, and z? From the first equation x=9-y —z. From the third ..... x=8+4y-8z. New equations { 59-y-z=7-3y+3z. From the first new eq. 2y=4z-2, whence y=2x-1. From the second new eq. 5y=7z+1, whence y= 72+1 By making these two values of y equal, we have 2z-1= 72+1 5 whence 10z-5=7z+1, or 3 z=6, and z=2; also y= (2z-1=) 3, and x= (9-y-z, as above =) 9-3-2=4. 24. Given 2x+3y+4z=20, 3x-4y+5z=10, and 4x+5y —z=11, to find x, y, and z. h From an attentive consideration of the rule it will appear, that the object proposed is to exterminate one and the same unknown quantity from two of the given equations; when this is effected, we have two new equations only, involving two unknown quantities only: the subsequent part of the operation will therefore depend on the rules for two unknown quantities. From the first new eq. 60—9 y—12 z=20+8 y−10z, or y= 40-2 z 17 From the second, 80-12y-16z=22-10y+2z, or y= 29-9z. x+Y+z 3 25. Given +x=13, ++z=114, and x+y+z: x +y-z: 9:1, to find x, y, and z. These equations reduced as before, give y=6, z=8, and x=4. 26. Given x+y+z=15,x+y−z=3, and x-y+2=5, to find x, y, and z. Ans. x=4, y=5, z=6. 27. Given 2x-y-z-10, 3x-2y+z=23, and 5x-4y3 z=9, to find x, y, and z. Ans. x=12, y=9, z=5. 2 28. Given +23 +z=13, 2x−3y+2=4, and x+2y: 2y x y +z58, to find x, y, and z. Ans. x=4, y=3, z=10. 94. Second Method. RULE. Find the value of one of the unknown quantities in either of the equations, and substitute it for that quantity in both the remaining equations; these two equations will then contain only two unknown quantities, which may be found as before i. i The foregoing rule is similar to the first method for two unknown quantities, this is similar to the second. 29. Given x+y+z=15, x+y-z=11, and x-y-z-5, to find x, y, and z. From the first equation x=15-y-z; this value being substituted for x in the second and third equations, we shall have 15-y-z+y-z=11, whence 2 z=4, and 2=2; and 15-y-zy-z 5, whence 2y=10-2z; and y=-5-z= (5-2=) 3, and x=15-y-z (15-3-2=) 10. 30. Given 2x+3y+z=40, 3x+y-z=13, and 4x+5y=z +46, to find x, y, and z. From the third equation z=4x+5y-46; this value substituted for z in the first and second equations, gives 2x+3y+4x+ 5y-46=40, and 3x+y-4x-5y+46=13; whence by reduc86-8y 43-4 y and x=33-4y; whence 6 3 43-4 y 3 tion x= =33-4y, whence y=7; and x=33-4 y= (33-28=) 5, also z=4x+5y-46≈ (20+35—46—) 9. 31. Given x+y+z=90, x-2y+3 z=40, and x+4y-5x= 60, to find x, y, and z. Ans. x=40, y=30, z=20. 32. Given 2x+3y+4z=32, 4x+3y+z=23, and x-y+ 2z=11, to find x, y, and z. Ans. x=3, y=2, z=5. 95. PROMISCUOUS EXAMPLES FOR PRACTICE. It frequently happens, that the unknown quantities may be exterminated by methods more simple and easy than any of the foregoing; the application of these must be left to exercise the skill of the operator, as no general rules can be given that will apply to every case. 1. Given x+3y-4 z=10, 3x+5y+32=66, and 5x+2y+ 72=80, to find î, y, and z. From the sum of the first and third subtract the second, and there remains 3x=24, whence x=8; substitute this value for x in the first and second, and 8+3y-4z=10, also 24+5y+3z= 66; these two equations reduced, give y=6, and z=4. 2. Given x+2y+2)2=144, x+y+2+=3, and 1 c+y−3z= From the square root of the first, subtract the square of the second, and there arises y=3; subtract twice the third from the square of the second, and 4z8, whence z=2; substitute these values of y and z in the square of the second, (viz. x+y+z =9,) and we shall have x=4. 3. Given x++=19, y+*+*=15, and z+ find x, y, and z. 2 3 x+y 4 =12, to From three times the second take twice the first, and 2 y―x= 7, or x=2y-7. From six times the second take twice the first, and 5y+z52, or z=52-5 y. In four times the third, viz. 4z+x+y=48, substitute the values of x and z as found above, and it becomes 208-20y+2y-7+y=48; whence y=9, x= (2y-7) 11, and z= (52-5 y=) 7. 4. Given xy=6, and x*+ya=97, to find x and y. Add twice the square of the first eq. to, and subtract it from the second, then find the square root of the sum and difference, which will be respectively x2+y2=13, and x2—y2=5; by adding these two together, we get 2x2=18, or x2=9, whence x=3; and by subtracting the latter of them from the former, 2 y2=8, or y2 4, whence y=2. 5. Given x2-y2=7, and xy=12, to find x and y. To the square of the first add four times the square of the second, the square root of the sum will be x2+y2=25; add the first equation to, and subtract it from this, and 2x2=32, x2=16, and x=4; also 2y2=18, y2=9, and y=3. Multiply the first by y+3, and the second by y-2, and we 50y+150 shall have xy=xy-50y+3x-150; whence x= 3 and xy=xy+200 y — 2x-400, whence x=100 y-200; therefore 50y+150 3 100 y-200, or 50y+150=300y-600; whence 250y=750, and y=3; also x=(100 y—200=) 100. 7. Given x 2 - x + y and yxx+y=48, to find x and y. 2 Multiply both equations together, and xy=(×48=) 32; but (y×x+y=) xy+y2=48; subtract the preceding equation from this, and y2=16, whence y=4; this value substituted for y in the equation xy=32, gives 4x=32, whence x=8. 8. Given x2+xy=15, and x2-y2=5, to find x and y. Subtract the second from the first, and xy+y2=10; add this to the first, and x2+2xy + y2=25; extract the square root of this, and x+y=5; substitute this value for x+y in the first equation (x2+xy=)x+yxx=15, and it becomes 5x=15, whence x=3; this value substituted for x in the equation x+y=5, gives y±2. = :60, and x+y× - 1=2, to find x and y. 9. Given x + y x· y 5 Multiply both equations together, and x+y=144, the square root of which x+y=12; substitute this for x+y in the first, and 12 x y 60, whence 12x=60y, and x=5y; this value substituted for x in the equation x+y=12, gives 6y=12, or y=2; whence (5y=) 10. = 10. Given xxx+y+z=36, yxx+y+z=27, and zxx+y+z -18, to find x, y, and z. Add the three equations together, and the sum is (x + y + z× x+y+z=)x+y+z81; extract the square root of this, and x+y+z=9; substitute this value in the three given equations, and 9x=36, or x=4; 9y=27, or y=3; and 9 z=18, or z=2. 11. Given x+yx3=540, and X- -y 5, to find x and y. Ans. x=100, y=80. 12. Given x+y: x-y:: 8:5, and x+y: 2y:: 8:3, to find x and y. Ans. x=65, y=15. 21, to find x, y, and z. Ans. x=12, y=9, z=3. 14. Given 2xx+y+z=18; 2×x+y+z=16, and 2x+y+z =13, to find x, y, and z. Ans. x=4, y=3, z=2. 15. Given 100-x-y=68; 68-y-z-48, and 48-x-2= 20, to find x, y, and z. Ans. x=20, y=12, z=8. |