x+2.(3x-2)=17, .. by transposition, 7x=17+4=21, 21 by division, x=8; andy3x-2=3X3-2=9-2=7. Ex. 2. Given values of x and y. § 3x+y=60, From the first equation, y=60—3x; to find the Let the value of y be substituted in the second equation, and it becomes, 5x+10=78+(60-3x) Then, by transposition, 8x=78+60—10; 128 16. = and by division, x= 8 2nd by 3, then x-y=186-6x (1); (2); From equation (1), we have x=198-7y, (2), ... · 7x-y=186; By substituting the above value of x, in the last equation, it becomes 7(198—7y)—y=186, or, 1386-49y-y=186; by transposition, -50y-186-1336-1200, by changing the signs, 50y=1200, 1200 ... by division, y= =24. 50 Whence, x=198-7y=198-7X24-198-168, x=30. Ex. 4. Given {+2y=80;} to find the values of x and y. 2x+y=60,$ From the second equation, x=60→y : By substituting this value of x in the 1st equation, we have, 60-y+2y=80, by transposition, y=80—60, ...y=20. And x 60-y-(by substitution) 60-20, . Ex. 5. Given fues of x and y. 40. x+2y=17,2 to find the va3x- y= 2, S From the 1st equation, x=17-2y. And this value substituted in the second, 3(17-2y)-y=2, or 51-6y-y=2, by transposition, &c, 7y=49, .. by division, y=7, whence, x=17-2y-17-2X7=17-14, of x and y. From the first equation, x=5-y, squaring both sides, x2=(5—y)2. And by substituting this value for 2 in the second equation, it becomes, x+64y=1552, .. by transposition, 1552-64y. And substituting this value for x, in the second equation, it becomes, *+8(1552–64y)=131, by reduction, y+99328-4096y=1048, by transposition, 4095y=98280, 98280 by division, y= 4095 ..y=24. Whence x=1552-64y=1552-64 × 24, or x=1552-1536; ..x16. The value of y might be found from the second equation, in terms of x and the known quantities; which value of y substituted for it in the first, an equation would arise involving only x, the value of which might be found; and therefore the value of y also may be obtained by substitution. Ex. 9. Given 15y+45x=300, and x+15y=36, to find the values of x and y. Ans. x=6, and y=2. Ex. 10. Given 3x+y=60, and 5x+10=78+y, to find the values of x and y. Ans. x=16, and y=12. to find the values of x and y. Ex. 11. Given 10x-3y=-38, and 3x-y=11, Ans. 5, and y=4. Ex. 12. Given x+y=198-6y, and x-y=186 -6x, to find the values of x and y. Ans. x=30, and y=24. Ex. 13. Given+y=26, and 2+8x=131, 8 Ans. x 6, and y=12. Ans. x 8, and y=2. Ex. 15. Given 4x+y=34, and 4y+x=16, to find the values of x and y. Ex. 16. Given 3x+2y=54, and xy:: 4:3, to find the values of x and y. Ans. 12, and y=9. +6y=21, and y+6 +5x = 3 23, to find the values of x and y. Ans. x4, and y=3, RULE III. 249. Find the value of the same unknown quantity in terms of the other and known quantities, in each of the equations; then, let the two values, thus found, be put equal to each other; an equation arises involving only one unknown quantity; the value of which may be found, and therefore, that of the other unknown quantity, as in the preceding rules. This rule depends upon the well known axiom, (Art 47); and the two preceding methods are founded on principles which are equally simple and obvious. Ex. 1. Given lues of x and y. S x+3y=100, to find the va From the first equation, x=100-3y, and from the second, x=. 100-y 2 Multiplying by 2, 100-y=200—6y, by transposition, 6y-y=200-100, whence, or, 5y=100; .. by division, y=20. 100-3y=100-3X20; •*. x=40. Here, two values of y might have been found, which would have given an equation involving only ; and from the solution of this new equation, a value of , and therefore of y, might be found, |
