From the first new eq. 60—9 y—12 z=20+8 y−10z, or y= 40-2z 17 From the second, SO-12y-16z=22-10y+2z, or y= 29-9z. Wherefore 40-2 z =29-9z, or 40-2z=493-153z, whence 20-3 y-4 z =) 1. 2 x + y + z 3 z=3; also y=(29—9 z=) 2, and x=( 25. Given +x=13, *++z=11;, 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+z=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. 28. Given Xx y +z=13, 2x−3y+2=4, and x+2y: 2y +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 2z=4, and z=2; and 15—y—2— y-z=5, whence 2y=10-2z; and y=-5-2= (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=: +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 43-4y and x=33-4y; whence 6 =) 3 3 tion x= =33—4y, whence y=7; and x=33—4y= (33—23=) 5, also z=4x+5y−46= (20+35—46=) 9. 31. Given x+y+z=90, x-2y+3 z=40, and x+4y-5= 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, 3 x+5y+3x=66, and 5x+2y+ 7 z=80, to find x, 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-4 z=10, also 24+5y+32= 66; these two equations reduced, give y=6, and z=4. 2. Given x+2y+ z)2=144, x+y+4=3, and to find x, y, and z. 1 -x+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 e square of the second, and 4 z=8, whence z=2; substitute ese values of y and z in the square of the second, (viz. x + y + z 9,) and we shall have x=4. From three times the second take twice the first, and 2y-x= - or x=2y-7. From six times the second take twice the first, id 5y+z=52, or z=52-5 y. In four times the third, viz. +x+y=48, substitute the values of x and z as found above, d it becomes 208—20 y+2 y −7+y=48; whence y=9, x= y-7=) 11, and z= (52-5 y=) 7. 4. Given xy=6, and x2+y+=97, to find x and y. Add twice the square of the first eq. to, and subtract it from e second, then find the square root of the sum and difference, hich will be respectively x2+ y2=13, and x2—y2=5; by adding ese two together, we get 2x2=18, or x2=9, whence x=3; and 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 econd, the square root of the sum will be x2+y2=25; add the rst equation to, and subtract it from this, and 2x2=32, x2=16, id x=4; also 2y2=18, y2=9, and y=3. =x- - 50, and =x+200, to find x and y. xy Multiply the first by y+3, and the second by y-2, and we all have xy=xy-50y+3x-150; whence x= 50y+150 and | = xy+200 y −2x-400, whence x=100-200; therefore y+150 3 =100 y-200, or 50y+150-300 y―600; whence 0y=750, and y=3; also x=(100 y—200=) 100. 7. Given х x + y 2 3 = and yxx+y=48, to find x and y. Multiply both equations together, and xy=(×48=) 32; 3 t (yxx+y=) xy+y2=48; subtract the preceding equation ›m this, and y2=16, whence y=4; this value substituted for y the equation xy=32, gives 4 x=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. 2 9. Given x+yx· —=60, and x + y×· =2. to find x and y. y y > 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 12x y =60, whence 12x=60y, and x=5y; this value substituted for r in the equation x+y=12, gives 6y=12, or y=2; whence x=(5y=) 10. 10. Given xxx+y+z=36,y×x+y+z=27, and z×x+y+z =18, to find x, y, and z. 2. Add the three equations together, and the sum is (x+y+zx x+y+z)x+y+2281; 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 9z=18, or z=2. 11. Given x+y×3=540, and "—", x=100, y=90. x-y -≈5, to find x and y. Ans. 12. Given x+y:x−y:: 8:5, and x+y: 2y:: 8:3, to find x and y. Ans. x=65, y=15. 13. Given 4x+2=54; 3x+ y+z =40, and 2 x = 21, to find x, y, and z. Ans. x=12, y=9, z=3. 14. Given 2xx+y+2=18; 2xx+y+z=16, and 2x+y+ 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—r—z= 20, to find x, y, and z. Ans. x=20, y=12, z=8. 3x+4y+5z 2 =56 and -65, to find x, y, and z. 8 Ans. x=40, y=30, 220. REDUCTION OF AFFECTED QUADRATIC EQUATIONS. k 96. An affected quadratic equation is that which contains both the first and second powers of the unknown quantity; every equation of this kind is comprehended under one of the three following forms, viz. First form, x2+ax= b. Where x is the unknown Second form, x2- -ax= b. quantity to be found, and a Third form, a2.—ax=—b.) and b known quantities. The value of the unknown quantity x, in each of these three formis, is found by one general method, as follows. 97. To find the roots of affected quadratic equations. RULE I. Range the terms of the given equation according to the dimensions of the unknown quantity; namely, let the term containing the square stand in the first place, (to the left,) and that containing the first power in the second, on that side of the equation in which the square will be affirmative. II. Transpose all the known quantities to the other side of the equation; and if the square of the unknown quantity have a multiplier or a divisor, it must be taken away by the methods employed in simple equations. III. Take half the coefficient of the unknown quantity in the second term, square it, and add this square to both sides of the equation, then will that side which contains the unknown quantity be a complete square'. k Quadratic is derived from the Latin quadratus, squared. The term adfected, or affected, (from affecto to pester or trouble,) was introduced by Vieta, the great improver of Algebra, about the year 1600: it is used to distinguish equations which involve, or are affected with different powers of the unknown quantity, from those which contain one power only, which are therefore called pure. Dr. Hutton sometimes calls the former compound equations: this term the venerable and learned Baron Maseres highly approves of, observing that it is less obscure, and therefore more proper, than that of affected or adfected equations. 1 Since the square of every simple quantity is a simple quantity, and the square of every binomial is a trinomial, it follows that no quantity in the form of a binomial can be a complete square; but that, in order to make it such, another term must be added to it, which term may in every case be found, from |