86. Take the equation, 2x . 10 vo Clearing of fractions (Art. 83), and performing the operations indicated, we have Transposing all the unknown terms to the first member, and the known terms to the second member (Art. 84), we have, 20x 15x = 240 15 210. Reducing the terms in the two members, we have 5x = 15. Dividing both members by the coefficient have, 2 = 3. In the same way, all equations of the first degree, containing but one unknown quantity, may be solved. Hence, the following RULE. I. Clear the equation of fractions, and perform ale. the indicated operations. 1. Transpose all of the unknown terms to the first member, and all the known terms to the second menber. III. Reduce all the terms in the first member to a sin. gle term, one factor of which is the unknown quantity; the other factor will be the algebraic sum of its coefficicnts. IV Divide both members by the coefficient of the un known quantity: the second member will be the value of the unknown quantity. 2 a Clearing of fractions, 110x — 12x – 6 = 772 + 165 – 60; transposing and reduring, 21% = 105 ; dividing by 21, = 5. 2. Solve the equation, 6 bx Clearing of fractions, 62 a’bx = abla transposing and reducing, - (ab+ ab)x = - (a? + 32); a? + 62 1 dividing by – (a3b + ab), C ab + ab3 ax a> ; ah 4. 1 + Ans. 2 = 12. 87. A PROBLEM is a question proposed, requiring a solution. 86. The SOLUTION of a problem is the operation of finding a quantity, or quantities, that will satisfy the given conditions. The solution of a problem consists of two parts; the statement, and the solution of the equation or equations of the problem. The STATEMENT consists in translating the conditions of the problem into algebraic language, the resulting equations being called the equations of the problem. The solution of the equations is made by the general rules for solving equations. The statement is made by representing the unknown quantities of the problem by some of the final letters of the alphabet, and then operating upon these so as to comply with the conditions of the problem. The me:hod of stating a problem will be best learned from practical examples. 1. What number is that to which if its fifth part be added, the sum will be equal to 24 ? Let X denote the number. Then will 5 X 5 Clearing of fractions, 5.0 + x = 120, reducing, 120, dividing by 6, 2 = 20, the number required 6x = 2. The sum of two numbers is 30, and their differ ence 6. What are the numbers ? Let denote one number. Then will 30 X, denote the other. From the conditions, (30 — ) transposing and reducing, – 2x = -- 24; dividing by – 2, ac = 12) the two minbers. .'. 30 - x = 181 - 2 = 6; |