3. Divide a--b by varb, vada Wat do * Here a, the first term of the dividend, being divided by Va, the first term of the divisar, gives va for the first term of the quotient. For asa', and vā at, and the difference of the exponents is 1 -, or ; therefore a' divided by a gives art=a= vā, as above. Or it may be considered thus : ask what quantity being multiplied by vā will give a, and the answer is vā; then the divisor being multiplied by va, the prodụct is avab ; but there being no term in the dividend, that Here it is obvious, that the division cannot terminate without a remainder ; therefore we write the divisor under the remainder with a line between them, and add the fraction to a me, the other two terms, to complete the quotient. But when the dividend does not precisely contain the divisor, then we generally express the whole quotient as a fraction, having reduced it to its lowest terms, or rejected the letters and factors, that are found in every term of the dividend and divisor. 5. Thus, that corresponds to wab, the second term of this product, we subtract a-nab from amb, the dividend, and the sign of the quantity -vab being changed, the remainder is tvab b. Now +vab, the first term of this remainder, divided by vă, the first term of the divisor, gives -vā for the second term of the quotient, by which we multiply the divisor, and the product, viz. +vab -b, being subtracted from the aforesaid res mainder, pothing remains ; and the quotient is wat vb. or 5. Thus, a'bxfacx' tax divided by adx tu an% ao bx+Fac%*Fax 3 abte**** gives adxanx dt. Here the quotient a'bx-track? tax is reduced to adx-tang abfcxt** , by dividing every term of its numerator and d-ton denominator by ax. 6. And atab+d divided by co-acta'c gives a tab to da a--acta c** Here the quotient cannot be reduced to lower terms, be cause the factor a is not to be found in the term d”. But it is to be observed, that though a fraction cannot be reduced to lower terms by a simple divisor, yet it may sometimes be so reduced by a compound one ; as will appear in the reduction of fractions. 7. Divide a3 + x3 by a+x. Ans. a'max+x4. 8. Divide a---m3a+y+zaya—y3 by amy. Ans. a'-_2ay+y*. 9. Divide 6x4--96 by 3x-6. Ans. 2x34-4** +8x+-16. 10. Divide a---5a*x+1023x2—101* x3 +5ax4-45 by a---2ax-*-*. Ans. a-30*x+zax*-** FRACTIONS, ALGEBRAIC FRACTIONS have the same names and rules of operation, as fractions in Arithmetic. PROBLEM PROBLEM 1. to find the greatest common measure of the terms of a fraction. RULE. 1. Range the quantities according to the dimensions of some letter, as is shewn in division. 2. Divide the greater term by the less, and the last division by the last remainder, and so on till nothing remain 3 then the divisor last used will be the common measure required. Note. All the letters or figures, which are common to each term of any divisor, must be rejected before such divisor is used in the operation. EXAMPLES. 《 ex+x 1. To find the greatest common measure of ca? ta** cx*xca* ta** +x)ca* ta*x{a* cao ta** or Therefore the greatest common measure is ct*. 2. To find the greatest common measure of 03-62x **+25x75 x* to2bx+b)x3-b* X(X *3 +26**-+6** x + bx or Therefore x tb is the greatest common measure. 3. To 3. To find the greatest common measure of xy +y Ans. xti * -64 4. To find the greatest common measure of * + Ans. ** tab PROBLEM 11. To reduce a fraction to its lowest terms. BULE: 1. Find the greatest common measure, as in the last problem. 2. Divide both the terms of the fraction by the como mon measure thus found, and it will be reduced to its lowest terms. EXAMPLES. 1. Reduce cx**? to its lowest terms. cx+*]ca* tax ca* ta' Therefore cefox is the greatest common measure ; cx + 2 and c+x) is the fraction required. tax 2. Having |