By making the denominators disappear, we have 10x32x 312= 21 52x by transposing a result which, being substituted for x, will verify the giver. equation. 3. For a third example let us take the equation (3a − x) (a − b) + 2ax = 4b (x + a). It is first necessary to perform the multiplications indicated, in order to reduce the two members to polynomials, and thus be able to disengage the unknown quantity x from the known quantities. Having performed the multiplications, the equation be 93. Hence, in order to resolve any equation of the first degree, we have the following general RULE. I. If the equation contains fractional terms, reduce it to one in which all the terms shall be entire, and then transpose all the terms affected with the unknown quantity into the first member, and all the known terms into the second. II. Reduce to a single term all the terms involving the unknown quantity: this term will be composed of two factors, one of which will be the unknown quantity, and the other all its co-efficients connected by their respective signs. III. Then divide both members of the equation by the multiplier of the unknown quantity. Questions producing Equations of the First Degree, involving but one Unknown Quantity. 94. It has already been observed (Art. 81), that the solution of a problem by Algebra, consists of two distinct parts. 1st. The statement; and 2d. The solution of the equation. We have already explained the methods of solving the equation; and it only remains to point out the best manner of making the statement. This part cannot, like the second, be subjected to any welldefined rule. Sometimes the enunciation of the problem furnishes the equation immediately; and sometimes it is necessary to discover, from the enunciation, new conditions from which an equation may be formed. The conditions enunciated are called explicit conditions, and those which are deduced from them, implicit conditions. In almost all cases, however, we are enabled to discover the equation by applying the following RULE. Represent the unknown quantity by one of the final letters of the alphabet, and then indicate, by means of the algebraic signs, the same operations on the known and unknown quantities, as would verify the value of the unknown quantity, were such value known. QUESTIONS. 1. Find a number such, that the sum of one half, one third, and one fourth of it, augmented by 45, shall be equal to 448. Let the required number be denoted by Then, one half of it will be denoted by · x. તું ૦૨ મા જ છે | 4 By making the terms of the equation entire, we obtain Let us see if this value will verify the equation of the problem. We have 372 372 372 2 + + +45 186 + 124 +93 + 45 = 448. 2. What number is that whose third part exceeds its fourth, by 16. Let the required number be represented by x. Then 3. Out of a cask of wine which had leaked away a third part, 21 gallons were afterward drawn, and the cask being then gauged, appeared to be half full: how much did it hold? Suppose the cask to have held a gallons. by changing the signs of both members, which does not destroy 4. A fish was caught whose tail weighed 97b.; his head weighed as much as his tail and half his body, and his body weighed as much as his head and tail together: what was the weight of And since the body weighed as much as both head and tail |