Making these two values of x equal to each other, we shall have 17-3 y 6+2y 2 = 3 this equation cleared of fractions, (Art. 79.) becomes 51-9y=12+4y. Whence 13y=39, and y=(==) 3. 39 13 4. Given x+4y=18, and x-3y=4, to find x and y. Ans. x=10, y=2. 5. Given 4x+3y=25, and 5x-4y=8, to find x and y. Ans. x=4, y=3. -+3y=217, and √x: √y:: 5:4, to find x and y. Ans. x=100, y=64. 91. Second Method. RULE I. Find the value of either of the unknown quantities in one of the given equations, and substitute this value for that quantity in the other given equation; this equation will then contain only one unknown quantity, which may be found as before f. f This rule is evident, for it is plain that in any expression whatever, we II. Find the value of the other unknown quantity, as directed 1 the preceding rule. 9. Given x+y=20, and x-y=8. to find x and y. From eq. 2.x=8+y; this value being substituted for x in the rst equation, (x+y=20,) gives 8+y+y=20, or 2y+8=20, hence 2y=12 and y=5; this value of y being substituted for y the equation, x=8+y, gives x=8+6=14. 10. Given 2x+3y=7, and 3 x-2y=4, to find x and y. From eq. 1. x= 7-3y 21-9y whence 3x= ; this value sub 21-9y -2y=4, 2 ituted for 3x in the second equation, it becomes hich by multiplication and transposition, becomes 13 y=13, hence y=1; this value being substituted for y in the equation eing substituted in the given equation, we have, +2 5 5 9% =59; whence 59 z=59 x 15, or z=15; this value substituted 5 12. Given x+y=10, and x-y=2, to find x and y. Ans. =6, y=4. 13. Given 2x+3 z=38, and 6x+5 z=82, to find x and z. ns. x=7, z=8. 14. Given y-2 z=7, and 2y-z=20, to find y and z. Ans. =11, z=2. -n substitute their equals, instead of any of the quantities which compose it, Ethout altering the value of that expression. 15. Given 2x+3y=34, and x+y: x−y :: 7 : 1, to find z and y. Ans. x=8, y=6. 92. Third Method. RULE I. Multiply the first equation by the coefficient of one of the quantities in the second, and the second equation by the coefficient of the same quantity, in the first; the products will be two new equations, in both which the coefficients of that quantity will be the same. II. If the terms, with equal coefficients, have like signs, subtract one of the new equations from the other; but if they have unlike signs, add the new equations together: the result will be an equation with only one unknown quantity, which may be found as before *. 16. Given 2x+5y=23, and 7 x—3y=19, to find x and y. To exterminate x, multiply the first equation by 7, and the second by 2, and the products (or new equations) will be 14x+35y=161.7 And 14x-6y= 35.} New equations. Whence upper, whence y=( 41 y=123 by subtracting the lower from the 123 3. 41 Now to exterminate y, multiply the first given equation by 3, and the second by 5, and the new equations will be • We are at liberty to employ any process, where equals operate in a similar manner upon equals: under this restriction, we are authorised to make use of addition, subtraction, multiplication, division, involution, and evolution, according as it suits our purpose; in this rule equal multiplication is used, but sometimes equal division, when it can be used, makes the work shorter. This equation multiplied by (the least common multiple of its enominators, viz.) 15, gives (12 y+5y=) 17 y=272; whence y= 272 17 =) 16; this value substituted for y in the first given equation, Here multiplication is unnecessary; therefore by adding both quations together, we get 2x=(1 1 whence x=- and by 2 3 2 btracting the second from the first, 2y=(1—-—-—=) ——, whence 1 3 19. Given 2x+3y=13, and 5x-2y=4, to find x and y. "ns. x=2, y=3. 20. Given x+y=5, and x-y=1, to find x and y. Answer =3, y=2. 21. Given 7y+9z=169, and Sz−9y——2, to find y and z. ns. y=10, z=11. 22. Given 2x+9=5z+8, and 3 z-x=x+· +, to find x and x. 1 ns. x= Z= For three unknown quantities. 93. First Method. RULE I. Let x, y, and z, be the three unknown quantities, hose values are sought; first find the value of x in each of he three given equations. II. Make the value of x 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+Sz=8; required the values of x, y, and z? From the first equation x=9—y -Z. From the second ..... x=7-3y+3 z. From the third New equations {9_4_z=7—3y+3z. 19-y-2=8+4y-Sz. From the first new eq. 2y=4z-2, whence y=2 z—1. 72+1 From the second new eq. 5y=7z+1, whence y=· 5 By making these two values of y equal, we have 2z-1= (2 z−1=) 3, and x— (9—y—z, as above =) 9—3—2—4. 24. Given 2x+3y+4z=20, 3 x−4y+5z=10, and 4x+5y -=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. |