68. An equation which involves the unknown quantity in the first degree is called a simple equation. SOLUTION OF SIMPLE EQUATIONS. 69. The process of solving a simple equation depends only upon the following axioms: 1. 2. If to equals we add equals the sums are equal. If from equals we take equals the remainders are equal. 3. If equals are multiplied by equals the products are equal. If equals are divided by equals the quotients are equal. 4. 70. Consider the equation 7x=14. It is required to find what numerical value x must have consistent with this statement. Again, in the equation 7.x-2.x-x=23+15-10, by collecting terms, we have 4x=28. ... x = 7. TRANSPOSITION OF TERMS. 71. To solve 3x-8=x+12. Here the unknown quantity occurs on both sides of the equation. We can, however, transpose any term from one side to the other by simply changing its sign. This we proceed to show. Subtract x from both sides of the equation, and we get 3x-x-8=12......... Adding 8 to both sides, we have 3x-x=12+8 .....(Axiom 2). ...(Axiom 1). Thus we see that +x has been removed from one side, and appears as -x on the other; and -8 has been removed from one side and appears as +8 on the other. It is evident that similar steps may be employed in all cases. Hence we may enunciate the following rule: RULE. Any term may be transposed from one side of the equation to the other by changing its sign. It appears from this that we may change the sign of every term in an equation; for this is equivalent to transposing all the terms, and then making the right and left hand members change places. Example. Take the equation -3x-12= x-24. or Transposing, - x+24=3x+12, 3x+12=x+24, which is the original equation with the sign of every term changed. 72. We can now give a general rule for solving a simple equation with one unknown quantity. RULE. Transpose all the terms containing the unknown quantity to one side of the equation, and the known quantities to the other. Collect the terms on each side; divide both sides by the coefficient of the unknown quantity, and the value required is obtained. Example 1. Solve 5(x−3) −7(6−x)+3=24−3(8−x). transposing, thus 5x+7x-3x=24−24+15+42-3; ... x=6. Example 2. Solve 5x-(4x-7) (3x-5)=6—3(4x−9)(x−1). Simplifying, we have 5x − (12x2-41x+35)=6−3(4x2−13x+9). And by removing brackets 5x-12x2+41x-35-6-12x2+39x-27. Erase the term -12x2 on each side and transpose; 5x+41x-39x=6-27+35; NOTE. Since the sign before a bracket affects every term within it, in the first line of work of Ex. 2, we do not remove the brackets until we have formed the products. Example 3. Solve 7x-5[x-{7-6(x−3)}]=3x+1. 73. It is extremely useful for the beginner to acquire the habit of occasionally verifying, that is, proving the truth of his results. Proofs of this kind are interesting and convincing; and the habit of applying such tests tends to make the student self-reliant and confident in his own accuracy. In the case of simple equations we have only to show that when we substitute the value of x in the two sides of the equation we obtain the same result. Example. To show that x=2 satisfies the equation 5x − (4x —7) (3x-5)=6−3(4x−9) (x−1)..............Ex. 2. Art. 72. When x=2, the left side 5x-(4x-7) (3x-5)=10-(8-7)(6-5) =10-1 The right side 6-3(4x-9)(x-1)=6-3(8-9) (2−1) =6-3(-1) =9. Thus, since these two results are the same, x=2 satisfies the equation. 7. 5 -6(c−5)=2(+5)+5( −4). 10. 3(169-x)-(78+x)=29.x. 11. 12. 13. 5x-17+3x-5=6x-7-8x+115. 118-65x-123-15x+35-120x. 14. 157-21(x+3)=163-15 (2x-5). 18. 5x-(3x-7) - {4-2x-(6x-3)}=10. 22. (x+1)2 - (x2 − 1) = x(2x+1)−2(x+2)(x+1)+20. 23. 2(x+1)(x+3)+8=(2x+1)(x+5). 24. 6(2-3x+2)-2(x2-1)=4(x+1)(x+2)- 24. 28. 20(2-x)+3(x−7) −2[x+9−3{9−4(2-x)}]=22. 32. 3(x-1)2-3(x2-1)=x-15. 33. (3x+1)(2x-7)=6(x-3)2+7. 34. x2-8x+25=x(x-4)-25(x-5) - 16. 35. x(x+1)+(x+1)(x+2)=(x+2)(x+3)+x(x+4)−9. 36. 2(x+2)(x− 4) = x (2x+1)− 21. 37. (x+1)2+2(x+3)2=3x(x+2)+35. 38. 4(x+5)2- (2x+1)2=3(x-5)+180. 39. 84+(x+4)(x − 3)(x+5)=(x+1)(x+2)(x+3). 40. (x+1)(x+2)(x+6)=x3+9x2+4(7x-1). CHAPTER IX. SYMBOLICAL EXPRESSION. 74. IN solving algebraical problems the chief difficulty of the beginner is to express the conditions of the question by means of symbols. A question proposed in algebraical symbols will frequently be found puzzling, when a similar arithmetical question would present no difficulty. Thus, the answer to the question "find a number greater than x by a" may not be selfevident to the beginner, who would of course readily answer an analogous arithmetical question, "find a number greater than 50 by 6." The process of addition which gives the answer in the second case supplies the necessary hint; and, just as the number which is greater than 50 by 6 is 50+6, so the number which is greater than x by a is x+a. 75. The following examples will perhaps be the best introduction to the subject of this chapter. After the first we leave to the student the choice of arithmetical instances, should he find them necessary. Example 1. By how much does x exceed 17? Take a numerical instance; "by how much does 27 exceed 17?" The answer obviously is 10, which is equal to 27 – 17. Hence the excess of x over 17 is x-17. Similarly the defect of x from 17 is 17-x. Example 2. If x is one part of 45 the other part is 45-x. Example 3. If x is one factor of 45 the other factor is Example 4. How far can a man walk in a hours at the rate of 4 miles an hour? In 1 hour he walks 4 miles, In a hours he walks a times as far, that is, 4a miles. Example 5. If $20 is divided equally among y persons, the share of each is the total sum divided by the number of persons, or $ 20 |