CHAPTER VI. } Simple Equations. 57. When two algebraical expressions are connected by the sign (=), they are said to form an equation. When the equality is such that it is true for all values of the letters in the given expressions, it is called an identity. Thus, (x + a) (x + b) = x2 + (a + b)x + ab are identities. and (a + b)2 – (a - b) = 4 ab 58. When the condition of equality is such that some one or other of the letters must have particular values or a limited number of values, the statement of equality is termed an equation of condition, or, more briefly, an equation. Thus, it may be found on trial that the equality 4x + 2 3x + 5 is true only when x = 3. Such an expression is therefore an equation. 59. The letters of an equation to which particular or a limited number of values must be given are termed unknown quantities. Equations may contain one, two, three, or more unknown quantities. The determination of the particular value or values of the unknown quantities is called the solution of the equation, and each of the values which satisfies the equation is said to be a root of the equation. 60. The expressions on the left and right sides of the sign ( =) are termed the first and second sides respectively. It follows, therefore, that 1. If both sides of an equation be multiplied by the same quantity, the equation still subsists. 2. If both sides be divided by the same quantity, the equation still holds. 3. Any term may be transposed from one side to the other if the sign of the term be changed. Thus, if 3 x + a = b, we must have also = b a, 32 = CX — cx px + = mx + n. for this results from subtracting a from each side of the equation. 4. The equation holds if every term on both sides has its sign changed. Thus, if ax + b d, we may reason as follows: The quantity (ax + b) looked upon as a whole is given equal to the quantity (cu d) looked upon as a whole. If we change the qualities of these quantities, they will evidently be still equal. Hence, (ax + 1) (cx d) d. Now, this is the result of changing the sign of every term on both sides of the given equation. 5. The sides of an equation may be reversed without destroying the equality. Thus, if mx + n = 9, it must also follow that px + 9 6. The sides of an equation may be raised to the SAME POWER, or we may extract the same root of both sides, and the equation still subsists. 61. Simple equations are those in which the unknown quantities are not higher than the first degree, when the equations are reduced to a rational integral form. The following is the general method adopted in solving a simple equation involving only one unknown quantity 1. Clear of fractions if necessary. 2. Transpose all the terms involving the unknown quantity to the first side of the equation, and all the remaining terms to the second side. 3. Simplify both sides if necessary, and divide both sides by the coefficient of the unknown quantity. Ex. 1. Solve the equation 5 x + 6 = 3 x + 12. 12 - 6. Now, simplifying, we get 3 2 = 6; 2 3 = and dividing each side by the coefficient of the unknown quantity, viz., by 2, we have X = 6 ; 2 = 3. find x. VERIFICATION.—Putting the value 3 for æ in each side of the 2 given equation, the first side becomes 5 x 3 + 6 or 21; and the second side becomes 3 x 3 + 12 or 21. The value of a found therefore satisfies the given equation. 2 6 Ex. 2. Given 20 2 3 2 Clearing of fractions, by multiplying every term on each side by the L.C.M. of the denominators, viz., by 6, we get 3 (2 – 2) + 2 x 20 x 6 3 (ac 6). (Beginners often neglect to multiply integral terms such as 20 by the L.C.M.) or 3 x - 6 + 2 x = 120 - 3x + 18, or, transposing, C 3 x + 2x + 3x = 120 + 18 + 6, or, simplifying, 8 x = 144, 4 x or dividing each side by 8, the coefficient of x, we have x = 18, the value required. (It will be good practice for the student to verify this result as in the last example). 21 28 9- 7 x Ex. 3. +73+ = x + 34 - 8 It is sometimes convenient to first partially clear off fractions. Thus, multiplying each side by 72, we have 72 (4 x – 21) + 47 x 12 + 24 (7 x 28) 7 x) + 6; 9 (9 288 2 or 672, 72 x 3 + 564 + 168 x 7 72 C + 270 81 + 63 m + 6; or, transposing, 414 + 168 2 72 x – 63 x 270 81 + 6 + 216 - 564 + 672 ; or, simplifying, 744 x = 519; or, multiplying each side by 7, 519 x 7; ...3 7. 519 x + II 3. 2 x + a = 3 b. 4) + 2. + 12 3 6. 13 7 11 18 15 2 x + 3 1 7. 315. 9 12 8 (5a + 2) 1 17 2 - 2 53 + 80 x 8. 3 8 4 7 9. ax + bc ba + ac. a 3 b b b a 5 X 6 22C 10. C + CI + 19. + 12 20 с 1 1 1 + 13. 2x 6 x 20. (2c + a) (x + 6) (oc + c) (x + d). 21. (x – a) (x – 5) = (x a + 6) 1 1 1 b 22. } (a + b + c)2 – ( + + bæ асх abx) 6 1 1 23. 2 bc ab 7 2 2 ab + CX ax с a 2 ac + Ac CX + ac 62. When the unknown quantity is involved in both numerator and denominator, it is often convenient to reduce such fractions to mixed numbers. 7 12 x + 18 Ex. Solve the equation 3 5 7 19 By division we get X + 2 12 x + 18 38 and 4 + 3 5 3 Hence the given equation becomes 19 38 + 3.) 19 38 4 2; 5 19 38 or, transposing, 5 + 2 – 6 + 4; 19 38 3 or, dividing each side by - 19, we have- X of 2 5 |