the equation is satisfied. No matter how we may obtain the value of an unknown letter, even if it be by mere guessing or by inspection, if it stands this test of substitution, it is a solution. 5. The process of solving a conditional equation consists of obtaining and substituting for a given equation another equation which has all of the solutions of the first, and, if possible, no more solutions, and which is of such form that the relations expressed between the letter whose value is to be found and the remaining quantities in the equation is less complicated. This process is continued until, if possible, an equation is finally obtained which may be solved by inspection. 6. Suggestions Concerning the Solution of Simple Equations Containing One Unknown Number (i) Remove fractional coefficients, if there be any, by multiplying both members of the equation by the least number which contains the denominators of the different fractions exactly as divisors. (ii) Perform all such indicated operations as are necessary to separate the terms of the equation into two distinct groups, group consisting of all of the terms containing the unknown number (and no other terms), and a second group consisting of all terms which do not contain the unknown number. one The terms which contain the unknown number are commonly transposed to the first member of the derived equation, and all terms which are free from the unknown are transposed to the second member of the equation. (iii.) All numerical or all monomial or polynomial factors not containing the unknown numbers, which are common to all of the terms of both members of the equation, should be removed by division and rejected as soon as discovered. (iv.) Combine into one term all of the terms containing the unknown number, and into another term the remaining terms which are free from the unknown. (v.) Divide both members of the equation by the coefficient of the unknown number. (vi.) The expression found for the unknown number should be reduced to simplest form. EXERCISE XI. 1 (MENTAL AND WRITTEN EXAMPLES) Solve the following equations, verifying all results by substitution. The first sixty-four examples may be solved mentally: 4. 7 x 9. 13x+4=11x + 10. 10. 6 x 11 2x + 9. 11. 13x18x + 6. - 12. 8x · 13 = 3 x 53. 13. 22+ 15 = 19 x 12. 14. 15+ 37 = 3x + 13. 15. 12x+1 = 28 + 3 x. 16. 1917 x = 59 3x. 17. 15x 1329 + 8x. 18. 21+ 22x = 8 x 35. 19. 7x+ 2 = 4 x + 7. 20. 17 18x=87 25 x. 21. 1511x 795x. 22. 2x + 23 = 5 x + 2. 23. 4 x 23 = 1 − 4 x. 33-7x= 3 x — 1. 33. 195 + 2 x 39. 47. 199 · 12x+6= 2 x + 35 + 3 x. 49. 8 4x 50. 7x+9 2+9x=7+2x-19 51. 3+13 + 5 x − 7 = x + 7 + 2 x + 2. 52. 141 + 3 x + 5 = 7 x + 2 - - 4x-8. 53. 5x (2x+3)= 12. 64. 8(x7) 6 (x − 5) = 5 (x-4) 4 (x3). 70. (x+3)(x+5)= x2 + 31. 79. 8}x. x = 80. 1-fx. 3x 81.x = − = 1)(4x+3). (10 x − 3)(x + 1) + 8. 89. x + x = §. x-5=3x-4. + f x = f. 90. 81. 8+ 2x. 91. (1) 20. 85. x + x = 2. 86. xx 4. 87. 4x-4x= 12. 88. xx 4. (x+7)= 1% (x+6). 98. (51) 8} (4x-2). 4 99. (1) 1% (2x) = 11⁄2 (3 + x). 100. 101. 102. 103. (128) 4 (14x+7)=84. 104.1+x= } (6 x 9) — 3 x. Equations in which Decimal Fractions appear among the Coefficients Ex. 105. .5 x = .015. By multiplying both members of the equation by 1000 we shall obtain an equivalent equation in which the decimal coefficients are replaced by integral coefficients. 7. A problem is a question proposed for solution. 8. A problem is said to be determinate if it has a limited or finite number of solutions. In the contrary case, it is said to be indeterminate. 9. To solve a given problem is to find the values of certain unknown quantities whose relations with one another and with certain known quantities are given. The relations between the known and unknown quantities are called the conditions of the problem. In solving a problem which admits of algebraic solution, the first step to be taken is to discover the relations between the unknown and the known quantities, as given in the statement of the problem. 10. The beginner will find it helpful, whenever relations are given between general numbers, to consider an analogous arithmetic problem, choosing definite numerical values in place of the given general numbers. E. g. By how much does a exceed 15? Consider a similar example in numbers. By how much does 21 exceed 15? The excess of 21 over 15 is the difference between 21 and 15, that is, 21-15. By the same reasoning it appears that a must exceed 15 by the difference between a and 15, that is, by a - . 15. ALGEBRAIC EXPRESSION MENTAL EXERCISE XI. 2 1. By how much does x exceed y? 5. By how much does a exceed 1? 6. What number must be added to x to obtain a ? 7. By what number must x be diminished to equal z? 8. What number is less than 10 by a? 9. What number is greater than 15 by b? 10. If a represents an integer how may the next greater integer be represented? The next less? 11. If b represents an odd integer how may the next greater odd integer be represented? The next less? 12. If 2 c represents an even integer how may the next greater even integer be represented? 13. Find an expression for three consecutive integers of which a is the least. 14. Find an expression for three consecutive integers of which b is the greatest. 15. Find an expression for three consecutive integers of which c is the one between the other two. 16. If a number represented by x is separated into two parts, one of which is 5, what is the other part? 17. If a number represented by a is separated into two parts one of which is b, what is the other part? 18. Find an expression for the greater of two numbers if the less is I and the difference is d. 19. A man sold a horse for $h and gained $g on the cost. pression for the cost of the horse. 20. A boy is 15 years old now. Find an ex Find an expression for his age x years ago. How old will he be in y years? 21. If a boy is y years old now how old will he be 5 How old was he three years ago? |