a xb—arr)2 tar xa-br2=cx a-bri +b2 cr2-abcr4 a3r4+aber4+ab2y3———a2cr3 —— 262 cr2 2a2 br2 — 2abr2+a3r+3abcr = ca ab2. Left the Learner fhould not be oble to make out the 16, or 17, Steps, I thought fit to fet it down here as an Explanation, viz. ab2—2a2br2+a3r4 ber-aber? +ar = ct a-br Which, if this laft reduc'd out of Fractions according to the common Rules of Algebra, will give the Equation as the 18th Step exhibits. Equat: And And Then the above Equations will stand, viz. 1 xxxy2x2+x2=b=y2x3ż+x3% 4 *3 = b b d 7 x 4=5=6.7 zyyyzz+yyy+zz 8 byx2+by czy2+cz by2+ bz2y = dzy2 + dz dy then 7 cydy For z2 put its Value in the above Equations |13| xxy2 + 1 = ly. Now this 13th dyc 1-2 and a3+a2e+ae2+e3=c } Quere a, and e. Subftitutexate, and z=ae, then 2xx = 2a1e + 2ae2 3 x3-——2xx = a3+a2e+ac2+e3 = 6; and by fubftituting for its Value z, in the firft 3+x= b. Now finding the Value of x, and then fubftituting its Value in the fecond Step, we fhall have the Value of x. Substitute ya, and for e. in the two given = Equations, then you will have x6+2b for the first, where you will by our Method of Converging Series hereafter laid down find the Value of x. In the other Equation you have (first multiplying by y3, and dividing by x3) y6+y++y2 + 1 =——33. Confequently you will have the Value of y. x3 EXAMPLE 18. 3% Suppose 2x2+33113 + 6x13 = a. Then z= a √2+ 3+ 40 =Answer. Q. E. J. Before I conclude this Part, I fhall give a few Examples how to manage Surds, let them be never so much complicated, the young Reader may by what is here done be able to folve any Equation, especially if he will obferve the following, as Prefatory RULE S. The Characters used by Algebraifts are in moft Authors shall not infert them here. Where Note, that the Square, Cube, Biquadratic, &c. of x+y is x2+2x+y2. x3 + 3x2+3xy2+y3. x4+4x3y+6x2y2 +4xy3+y4, &c. but there is a new Way for Notation in Operations as are used by the most acute Mathematicians; as *+ Square of x+y=x2+2xy+3y. x+y)* = Biquadrate of x+y=x++ 4x3y + 6x2y2+ 4x3+4. Fifth Power of x+, &c. &c. Confequently the Square Root of y3+aay is √y3+aay), or by the new Way=y3+a2y{} Cube Root of y3+a2y= √y3+a2y,)or y2+a2)}} Cube Root of the Square of y3+a2y=√6+2a2y++a+y2) =y3+a3‚}'}} &c. &c. UNIVERSALLY. 3⁄41⁄4√ a2b—c3+d3‚) or a2bc3+d3), is the Cube Root of the Difference between a2b+d3 and c3. a3±√ab+8+nx), or a3+ab+8+nx), is the Cube Root of a3 added or fubtracted to or from the Square Root of ab+g+nx. And */ m++Wa3p5+b3c3—dd √aa‡ce, or = m4)2 +a3p5+b3ç3—ddxaa+cc\}); is the Biquadrate Root from m+ added to the Cube Root a3p5+ b3c3dd, multiplied into the Square Root of aa+cc. And from these the young Algebraift will be able to determine the rest. But Surds may be reduced to facilitate Operations, which to do please to observe this RULE. Divide the Surd by the greatest Square, Cube, Biquadrate, &c. or any other higher Power, by which you can discover, is contained in it, and it will measure it without any Remainder, and then prefix the Root of that Power before the Quotient, or Surd fo divided, and this will produce a new Surd of the fame Value with the former, but in more fimple Terms; as L 2 49abb 49bb 49abb ; the Root of which Quotient (4962) is 7b, which by the Rule must be prefix'd thus, 76 √ a, or 76 xa. Allo xy3zxy3z will, by the fame Way of Reasoning bey xxy xxx. And this Reduction is of great Ufe; and befides it looks neater and workmanlike to exprefs Quantities in their moft fimple Terms. Suppofe you would Square, Cube, &c. any Surd Root, it is no more than to Square, Cube, &c. the Power retaining the fame Note of Radicality; as for Inftance. Suppofe you would Cube Vaataab, it will be aab =aab. Again, fuppofe mx3y=mx3y were to be Cub'd, it will be/mx3y=mx3y. So the Biquadrate Root of xy=xy is x2y2, as being the Square of the Square of, and the Cube of x x = √xx xx xx will be = **), or x. I 2 I 2 But it is better, where it can be done, to take one half, one third Part, &c. of the Exponent of the Root; on the contrary, if you would extract the Square, Cube, &c. Root of any Surd, you must double, triple, &c. the Exponents of the Radicality, thus the Square Root of √5mpx 5mpx is = 5mpx=5mpx, and the Square Root of Vaxy-axy is Vaxy = axy). I It will not be amiss to shew how may be mul ==m But to note by the Way this Ex |