Ex. 5. Let 3ax3— abx2 = ax3 + 2acx2: here, dividing the whole by ax2, we have 3x-6x+2c; therefore 2x = 2c + b, and x = 20 + b b = 21: then (by rule 3) 4x+3* abx=abcd, or bc + ac + ab × x — abcd; and confe = ax + xx; whence -xx + bx=- ab; which, by changing all the figns (in order that the highest power of x may be pofitive) gives xxbx ab. But the fame conclufion may be otherwife brought out, by first changing the fides of the equation ax + ab + bx = ax + xx; which thereby becoming ax + xx = ax + ab + bx, we thence get xxbx ab, the fame as before. and 5x = 15; whence, (by rule 4) 5x = 225, and -- = Ex. 13. Let V12 + x 2 + V: then (by rule 4) 12 + x = 4 + 4 + x; whence, by tranfpofition, 8 = 4√x; and, by divifion, 2 Vx; confequently 4 = x. √ a2 + x2 = a2 x2, and x × a2 + x2 = a4. 2a2x therefore 3a2x2='a2, and x2 = a4 a2 3a2 3 Ex. 16. Let x3-a3x-c: then, by cubing both fides, 3-a3x3-3cx2 + 3c2x-c2; whence 3cx2 — 3c2x = a3 — c3, and x2— cx = viding the whole by 3c. by di 3c 13 2 Ex. 17. Let Vaa + xx = †ba + x*; then, by raifing both fides to the fourth power, we have aa + xx = b+ + x2, that is, aa + 2a2x2 + x^ = 64 + x; and 64 confequently x = a2. a2 + x √ bb + xx — a. Here Ex. 18. Let x = x + a = √ a2 + x V bb + xx; which squared, gives x2+ 2ax+a2=a2 + x V bb + xx, or x2 + 2ax=x √bb + xx; divide by x, fo fhall x + 2a= √bb + xx; this fquared again gives x2 + 4ax + 4a2 = bb + xx; whence 4ax· bb-4a2, and therefore x = bb 4a -a. Of the Extermination of unknown quantities, or the reduction of two or more equations to a fingle one. It has been fhewn above, how to manage a fingle equation; but it often happens, that, in the folution of the fame problem, two, or three, or more equations are concerned, and as many unknown quantities, mixed promifcuously in each of them; which equations, before. any one of thofe quantities can be known, must be reduced into one, or fo ordered and connected, that, from thence, a new equation may at length arife, affected with only one unknown quantity. This, in moft cafes, may be performed various ways, but the following are the most general. 1o. Obferve which, of all your unknown quantities, is the leaft involved, and let the value of that quantity be found in each equation (by the methods already explained) looking upon all the rest as known; let the values thus found be put equal to each other (for they are equal, because they all exprefs the fame thing); whence new equations will arife, out of which that quantity will be totally excluded; with which new equations the operation may be repeated, and the unknown quantities exterminated, one by one, till, at last, you come to an equation containing only one unknown quantity. 2o. Or, let the value of the unknown quantity, which you would first exterminate, be found in that equation wherein it is the leaft involved, confidering all the other quantities as known; and let this value, and its powers, be fubftituted for that quantity and its respective powers in the other other equations; and with the new equations thus arifing repeat the operation, till you have only one unknown quantity, and one equation. 3. Or, laftly, let the given equations be multiplied or divided by fuch numbers or quantities, whether known or unknown, that the term which involves the highest power of the unknown quantity to be exterminated, may be the fame in each equation; and then, by adding, or fubtracting the equations, as occafion fhall require, that term will vanish, and a new equation emerge, wherein the number of dimenfions (if not the number of unknown quantities) will be diminifhed. But the use of the different methods here laid down will be more clearly understood by help of a few examples. EXAMPLE I. Let there be given the equations x + y = 12, and 5x + 3y=50; to find x and y. According to the first Method, by transposing y and 3ys we get = 12 ·3, and 5x = 50—3y; from the laft of which equations, x = 50-31: Now, by equating these two values of x, we have 12y = 50-31; 5 53503y: from which y is given = 5; and 2 (12-y=12-5)=7. According to the fecond Method; being, by the first, equation, 12-y, this value must therefore be fubftituted in the fecond, that is, 60. 5y must be wrote in the room of its equal 5; whence will be had 60 5y+3y= 50; and from thence y = ΙΟ 2 =5, as before. But according to the third Method, having multiplied the firft equation by 5, it will ftand thus, 5x+5y=60; from whence fubtracting the 2d equation, 5x+3y=50, there remains whence y 5, fill the fame as before: The The first of these three ways is much used by some Authors, but the last of them is, for the general part, the most eafy and expeditious in practice, and is, for that reason, chiefly regarded in the subsequent examples. EXAMPLE II. Let { 5x + 3y = 124 3x 2y= 20. Here the fecond equation being multiplied by 4 (in order that the coefficients of y in both equations may be the fame) we have 12x 8y80. Let this equation and the first be, now, added together; whence y will be exterminated, there coming out 17x=204; from which 204 12: therefore, 17 8 8 = 8. EXAMPLE III. Given { 5x-3y = 90 2x+5y=160. Here, multiplying the first equation by 2, and the fe cond by 5, in order that the coefficient of x may be the fame in both, there arises 10x- by=180 10x + 25y = 800. By fubtracting the former of which, from the latter, we have 313 620; hence y = 620 31 20; and fo, by (= 90 + 3y = 90+60) = 30. 5 5 But the value of x may be otherwife found, independent of the value of y; for, by multiplying the first equation by 5, and the fecond by 3, and then adding them together, y will be exterminated, and you will get 25 x + 6x = 450 + 480; whence x = the fame as before. |