the least common multiple of the denominators is a3b2; hence clearing the fractions, we obtain a+bx-2a2bc2x+4a2b2=4b3c2x—5a6+2a2b2c2 — 3a3b3. Second Transformation. 69. When the two members of an equation are entire polynomials, to transpose certain terms from one member to the other. 1. Take for exam, ie the equation 5x-6-8+2x. If, in the first place we subtract 2x from both members, the equality will not be destroyed, and we have -5x-6-2x=8. Whence we see that the term 2x, which was additive in the second member becomes subtractive in the first. What do you QUEST.-69. What is the second transformation? understand by transposing a term? Give the rule for transposing from one member to the other. In the second place, if we add 6 to both members, the equality will still exist, and we have 5x-6-2x+6=8+6. Or, since 6 and +6 destroy each other, we have 5x-2x=8+6. Hence the term which was subtractive in the first member, passes into the second member with the sign of addition. 2. Again, take the equation ax+b=d-cx. If we add cx to both members and subtract b from them, the equation becomes or reducing ax+b+cxb=d-cx+cx-b. ax+cx=d-b.* When a term is taken from one member of an equation and placed in the other, it is said to be transposed. Therefore, for the transposition of the terms, we have the following RULE. Any term of an equation may be transposed from one member to the other by changing its sign. 70. We will now apply the preceding principles to the resolution of equations. 1. Take the equation 4x-3=2x+5. By transposing the terms -3 and 2x, it becomes If now, 4 be substituted in the place of x in the given equation it becomes or, 4x-3=2x+5, 4x4-3=2×4+5. 13=13. Hence, the value of x is verified by substituting it for the unknown quantity in the given equation. 2. For a second example, take the equation By making the derminators disappear, we have 10x-32x-312-21-52x, a result which may be verified by substituting it for x in the given 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 two polynomials, and thus be able to disengage the unknown quantity x, from the known quantities. Having done that, the equation becomes, 3a2-ax-3ab+bx+2ax=4bx+4ab, or, by transposing —ax+bx+2ax-4bx =4ab+3ab—3a2, Hence, in order to resolve an equation of the first degree, we have the following general RULE. I. If there are any denominators, cause them to disappear, and perform, in both members, all the algebraic operations indicated; we thus obtain an equation the two members of which are entire polynomials. II. Then transpose all the terms affected with the unknown quantity into the first member, and all the known terms into the second member. III. Reduce to a single term all the terms involving x: this term will be composed of two factors, one of which will be x, and the other all the multipliers of x, connected with their respective signs. IV. Divide both members by the number or polynomial by which the unknown quantity is multiplied. QUEST.-70. What is the first step in resolving an equation of the first degree? What the second? What the third? What the fourth? |
