also (by ax. 5), 11−6x+3y_ 9x+6y—4 multiplying this new equation by 8, by transposition, 44+4=24x+9x+6y-12y, 9x+6 33x-48 .. (by ax. 5th), 6 multiplying this equation by 24, by transposition, 132x-27x=192+18, Substituting this value of x, in the equation E, Also, substituting the values of x and y, in the equa tion A, 2= whence, x=2, y=3, and z=4, NOTE.-The value of z may be found by substituting the values of x and y, in any of the equations, A, B, or C. Also, the value of y may be found by substituting the value of x, in the equations D or E. RULE II. Multiply or divide the given equations by any number that may be convenient, to make one unknown quantity in each equation the same; then, if these quantities have like signs, the difference of the equations must be taken, in order to exterminate one of the unknown quantities. But if the equal quantities have unlike signs, then their sum must be taken, in order to exterminate one of the unknown quantities. 1. Given 3x+2y=18, to find the values of x and 2xy= 5, S and y. Here, multiplying the second equation by 2, 4x-2y=10; but 3x+2y=18, (the first equation,) by addition, 7x =28*; .. x=4. Substituting this value of x, in any of the above equations, y will be found = 3. Or thus: Multiplying the first equation by 2, and the second by 3, 6x+4y=36; (A) and 6x-3y=15; by subtraction, 7y=21+; .. y=3. Substituting this value of y, in the equation (A), NOTE.-The-2y and 2y are like quantities, and have unlike signs; therefore, the sum of the equations is taken in order to exterminate the y. + Here the unknown quantity a is exterminated. Substituting this value of x, in one of the above equations, y=3. 2. Given 3x+2y+4z=28, to find the values of x, 6x-3y+22=11, and 9x+6y-8z= 4, y, and z. Here, multiplying the second equation by 2, 12x-6y+4z=22; but 3x+2y+4z=28; (the 1st equation) by subtraction, -9x+8y-6. (A) NOTE.-By this operation, z is exterminated. Also, multiplying the first equation by 2, and adding the product to the third equation, 6x+4y+8x=56, (twice the 1st equation,) but 9x+6y-8z= 4, (the third equation,) by addition, 15x+10y=60. (B) By this operation the z is also exterminated, therefore, here are two equations, viz. A and B, each containing two unknown quantities. Multiplying the equation A by 10, and the equation B by 8, -90x+80y=60 (10 times the equation a), 210x =420, by subtraction; .. x=2. Substituting this value of x, in the equation в, 15×2+10y=60, or 30+10y=60; Also, substituting the values of x and y in the first equation, viz. 3x+2y+4z=28, it becomes Find the value of one of the unknown quantities, in terms of the rest of the equation, and substitute its value, thus found, in the other equation. 1. Given 3x+2y=18,2 to and 2xy= 5, S find the values of x and y. 18-2y (A) 3 From the first equation, by Rule I, x= Substituting this value of x in the second equation, 2. (18–24)—y=5, or 36–44—y=5; multiplying this equation by 3, 36-4y-3y=15; by transposition, 36-15-4y+3y, Substituting this value of y, in the equation (A), From the second equation, y=2x-5, (A) substituting this value of y, in the first equation, 3x+2.(2x-5)=18, or 3x+4x-10=18; by transposition, 3x+4x=18+10, or 7x=28; .. x=4. Substituting this value of x, in the equation (A), y = 3. NOTE. This method of solution is applicable to the second example, or to any question containing four or more equations, with the same number of unknown quantities. EXAMPLES, With their solutions, by the 2nd. Rule. clearing the second equation of fractions, dividing this equation by 2, 3x+3y=84; but 4x+3y=96; by subtraction, x=12; whence, by transposing the equation (A), 3y=84-3x; substituting this value of x, in this equation, 3y=84-3648; .. y=16. 2. Given 5x-20=3y+ 2,1 to find the values of x and 3x+6=7y-12,5 and y. From the first equation, by transposition, 5x-3y=22; (4) from the second equation, by transposition, H |