Ex. 6. Given (+7y=99, 14+7x=51, = to find the values of u and y. y Multiply each equation by 7, ...+49y=693, and y+49x=357; 21; .. by addition, 50x +- 50y=1050, 1050 and by division, x+y= 50 but since +49y=693, subtracting the upper equation from the lower, we have 48y=693-21=672, 672 y= =14, 48 whence =21-y=21-14=7. ac +2 + 8y 31, to find the 3 Ex. 7, Given values of a y+5+10x=192, J and y. 5 4 Cleariog the first equation of fractions, 3+2 +-24y=93; by transposition, a +-24y=91 : (1), Clearing the second equation of fractions, y+5+40x=768; .. by transposition, 40x+y=763 . . . (2.) Multiplying equation (1) by 40, and subtracting equation (2) from it, 40x +960y=3640; y= 763; .. 959y=2877, and by division, y=3; From equation (1), x=91 — 244, .. by substitution, x=91-24 X 3, or x=91-72, •, x=19. If from equation (2), multiplied by 24, equation (1) had been subtracted, an equation would have arisen involving only x, the value of which might be determined, and this being substituted in either of the equations, the value of y might also ข be found. Ex. 8. Given x+y=a;} to find the values d, &b, of x and y. a+b By addition, 2x=a +-6;.. 2 ab By subtraction, 2y=u-b,... y= 2 Ex. 9. Given {:+=13; } to find the va lues of x and y. 2 Molt. the 1st equation by , then +4y=24; 2nd 2, x - 4y= 8; By addition, 2x=32, 32 .., by division, r=Z=16. 2 By subtraction, By=16; .. by division, y= 2. Or, the values of x and y may be found thus ; From the first equation subtract the second and we have 4y=8,.. y=2. Add the first equation to the second, and ..=16. Ex. 10. Given 4x+3y=31, and 3x+2y=22 ; to find the values of x and y, Ans. 4, y=5. Es. 11. Given 5x—-4y=19, and 40+2y=36, to find the values of u and y. Ans. -7, y=4 Ex. 12. Given +9_-2y=2, and *--y 2x-44 +y= 23 -; to find the values of wand y. y. 3 5 5 24-9 +14=18, Ex. 13. Given to find the va 2y + x lues of x and and cf1619, y. Ans. c=5, and y=2. 2x + 3y +=S, 3 = y-10 Ex. 14. Given -S to find the va6 lues of u and 7y - 3 and -y=11, Jy: 2 Ans. =6, and y=8.. Ex. 15. Given 32+1=22, to find the va2x lues of x and y.. and 11y = 20, 5 Ans. x=5, and y=2, Ex. 16. Given x+1:y::5: 3, 25-y_41 2x-1 to find and 3 12 4 the values of x and y. Ans.w=4, and y=3, 2 10 Ex. 17. Given 5 3 4 to find 2y+4_2x+y _*+13 and 3 4 the values of u and y. Ans. t=7, and y=10. Ex. 18. Given x-+-15y=53, to find the values and yt 3x=27, of x and y. Ans. «= Es. 19. Given 4x + 9y=51,7 to find the values and 31-13y= 9, ) of x and y. Ans. x=6, and y=3, Ex. 20. Given 1=6, y . ? = 6 4 to find the values of 十三5号, Ans. x=12, and y=16. 8 =8, and y=3. c and y. RULE II. 248. Find the value of one of the unknown quantities, in terms of the other and known quantities, in the more simple of the two equations ; and substitute this value instead of the quantity itself in the other equation; thusan equation is obtained, in which there is only one unknown quantity; the value of which may be found as in the last Rule. Ex. 1. Given { 2 Ex. 1. Given § 3+2y=17, to find the va 3:--- y= 2, lues of x and y. From the first equation, x=17—24; Substituting therefore this value of x in the second equation, 3.(17-2y)-y=2, or 51-by-y=2; 7y=51--2=49, .. by division, y=7; whence, x=17--2y=17-14=3. Here a value of y might be determined from either equation, and substituted in the other; from which would arise an equation involving only x, the value of which might be found ; and therefore the value of y also might be obtained by substitution, thus ; - From the second equation, y=3x-2 ; substituting therefore this value of y in the first equation ; we have, |