Ex. 2. Given fx +y=7, and žu + y=8, to find the values of x and y. Multiplying both equations by 6, and we shall bave 3x +2y=42, and 2x + 3y=47, 42-2y From the first of these equations, x= 3 48 - 3y and from the second, w= 2 42—2ý _48—34, -3y 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 haye 42-24 18 6. 3 3 . Ex. 3. Given 8x+18y=94, and 82—13y=1, to find tbe values of x and y. 47-9y From the first equation, u= 4 1+13y and from the second, x= 8 47-9y_1+13y ; 4 8 And multiplying both sides of this equation, by 8, 94-18y=1+13y; =1+134 ... 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 · ye 31 1 +13y 1+39 40 8 8 8 whence a 5. Ex. 4. Given to y=a, ? to find the values bx+cy=de, ) of æ and y. From the first equation, x=a--y; and from the second, x= b de-cy de-cy ...a-Y= ; b ab-by=de--cy; ' by collecting the coefficients, (c—b) y=de-ab; de- ab :.. by division, y= -b de-ab whence w= =Q-y=a -b ca--ab-detab -de that is, a cob C C ca C 250. If in the above equations, there existed, between the coefficients, these relations, =b, and ca> or <de; then, de- ab y= 0 0 And therefore, (Art. 233), the two proposed equations would be contradictory. ca a 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 and 26 0 0 Where the values of x and y are both infinites and therefore, under these relations, there can be no finite values of u 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, x+y =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 v 0 0 and g=%; 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. 0 Ex. 5. Given 3x+7y=79, and 2y— =9, to find the values of x and y. Ans. x=10, and y=7. , Ans. =11, and y=45 C +y Ex. 6. Given **4+1=6, and 7+3=4, to to 2x3 67 Ex. 7. Given +y=7, and 5x-13y= 2 2' find the values of u and y. 1 Ans. x=8, and y= 2 3x—74_2x+y+1, and 8 - y Ex. 8. Given 3 5 =6, to find the values of x 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 y=3. Ex. 10. Given 3x~-5y=13, and 2x +-7y=81, find the values of x and y. Ans. x=16, and y=7. = +10x= 3 4 192, to find the values of x and y. y Ans. x=19, and y= Ex. 12. Given 24,-4+14=18 , and 2y+ x =3. 2 3 to find the values of « and y. Ans. x=5, and y=3. Ex. 13. Given 8 to find the va 6 3 lues of x and 7y-3x and =11+y, yo 2 Ans. x=6, and y=8. Ex. 11. Given **2 +8y=31, and y+5 2x + 3y = = 251. Examples in which the preceding Rules are applied, in the Solution of Simple Equations, Involv. ing two unlonown Quantities. 8-Y 2+3 3x-2y Ex. 1. Given 2y- =7+ 4 5 to find 2x+1 and 40 = 241 3 2 the values of x and y. Multiplying the first equation by 20, 40-5x-15=140+120---By ; ... by transposition, 487–170=155. Multiplying the second equation by 6, 24x --16+2y=147-6x-3; .. by transposition, 2y+30x=160.,. · (A). Multiplying this by 24, we have 48y+720x=3840; but 48 - 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=0; • y=5. = 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. 2 4十1 十 2y 8x-2 Ex. 2. Given 1. X --y 2 18 36 3 6 3- 8 6 4+y + |