Ex. 2. Given +y=7, and + y=8, to find the values of u and y. Multiplying both equations by 6, and we shall have 3x +2y=42, and 2x+3y=47, 42-2y From the first of these equations, x= 3 and from the second, x= 48—34, 2 42—24 _48—34, 3 2 • 84-4y=144 -9y; or 5y=60;..y=12. And, by substituting this value of y, in one of the values of x, the first, for instance, we shall have 42-24 18 6. 3 3 Ex. 3. Given 8x+18y=94, and 8. - 13y=1, to find tbe values of x and y. 47--9y From the first equation, 2= 4 1 and from the second, x= 8 47—9y_1+134 ; 8 And multiplying both sides of this equation, by 8, 94–18y=1+134 ; 4 .. by transposition, -18y-13y=-94+1; Changing the signs, or what amounts to the same thing, multiplying both sides by -1, and we shall have 18y +- 13y=94-1, or 3ly=93; 93 3; 1+134 1+39 40 whence a 5. 8 8 ... y=31 8 Ex. 4. Given a + y=a, ? to find the values bx+cy=de, ) of æ and y. From the Girst equation, š=Q-Y; de and from the second, i= b -cy de-cy Oly i b and multiplying by b; we shall have ab-by=de-cy; by transposition, cy-by-de--ab; by collecting the coefficients, (c—b) y=de-ab; de-ab ,, by division, y= -b de- ab whence w=Q-Ya 6 -ab- de tab -de that is, c-b ca ca 250. If in the above equations, there existed, between the coefficients, these relations, cab, and ca> or <de; then, de- ab =, and y= 0 And therefore, (Art. 233), the two proposed equations would be contradictory. ca-de 3C y= = and x= In order to give a numerical example, let c=h= 4, a=3, and de=10; then, by substituting these values, we shall have 10-12 2 12-10 2 0 0 09 Where the values of x and y are both infinite, and therefore, under these relations, there can be no finite values of x and y, which would fulfil both equations at once; this is what will still appear more evident, if we substitute these values in the proposed equations; for then, we shall have, æty =3, and 4x+4y=10; which are evidently contradictory; since, if we multiply the first by 4, and subtract the second from the result, we should have 0=2. Again, if c=b=4; a=3, and de=12; then <= 0 0 0 and y=;; therefore, under these relations, the two proposed equations would be indeterminate ; and, in fact, this appears evident by inspection only; for the second furnishes' no condition, but what is contained in the first, since the two proposed equations, in this case, would become x+y=3, and 4x+4y=12. Ex. 3. Given 30+7y=79, and 2y =9, to find the values of x and y. Ans. x=10, and y=7. c Ex. 6. Given and 3 7 find the valaes of x and y. Ans. <=11, and y=45 to 9 y=3. 2x-3 67 Ex. 7. Given ty=7, and 5x-13y= 2 2 find the values of x and y. 1 Ans. x=8, and y 2 3r—7y_2x+y +1, and 3Ex. 8. Given + X-Y 3 5 5 =6, to find the values of u and y. Ans. x=13, and y=3, Ex. 9. Given x+y=10, and 2x ---- 3y=5, to find the values of x and y. Ans. x=7, and Ex. 10. Given 3.0-5y=13, and 2x + 7y=81, to find the values of x and y. Ans. x=16, and y=7. +2 Ex. 11. Given = and +10x= 3 y. and y= Ex. 12. Given =, 2 3 to find the values of x and y. Ans. x=5, and y=3. Ex. 13. Given to find the va. 6 3' lues of x and 7y - 3.1 and y 2 Ans. x=6, and y=8. y+5 2x-Y+14=18, and 2y+* = 3; 2x+3y=8 =11+y, 251. Examples in which the preceding Rules are applied, in the Solution of Simple Equations, Involv. ing two unlonown Quantities. 8-Y=24 2+3 3x-2y Ex. 1. Given 2y- =7+ 4 5 to find 2x+1 2 40y-5x-15=140+12x ---8y; .. by transposition, 487–17x=155. Multiplying the second equation by 6, 24x --16+2y=147-6.0-3; by transposition, 2y+30x=160 . , (A). Multiplying this by 24, we have 48y +720x=3840; but 487 – 17= 155; .. by subtraction, 737x=3685, and by division, x=5. From equation (A), 2y=160-30x ; :. by substitution, 2y=160—150, 10 by division, y = jy=5. 2 The values of x and y might be found by any of the methods given in the preceding part of this Section ; but in solving this example, it appears, that Rule I, is) the most expeditious method which we could apply. 4+ 뼇 2y 8.0 -2 Ex. 2. Given 1+y,x-o :1 X-ry? + 3 6 4+y*-Y 1 y_4x — 1 |