and all the known terms to the second member, leaving the equation, or, after simplifying, (3) 48x+4x+3=60+8, 55 x 68. Finally, divide both members of the equation by the coefficient of x, that is to say, by 55, and find equation (4), which is equivalent to equation (1), Therefore equation (1) has but one root, which is 169. RULE. Therefore, it follows that, to solve an equation of the first degree in one unknown quantity x, it is necessary: (1), to remove the denominators; (2), to transfer all the terms involving x into one member and all the known terms into the other and collect the terms; (3), to divide the known term by the coefficient of x. The quotient thus obtained is the root of the given equation. 170. In practice, the steps in solving an equation of the first degree in one unknown quantity need not always be the same. Thus, consider the equation, Collecting all the terms in x in the first member and all the known terms in the second member; hence Remove the denominators and find 80x15x+8x280+8 171. REMARK.-It may happen in applying the rule in $169 that the coefficient of z in the first member and every known term in the second member are negative In this case, change the signs of all the terms in both members, as has been explained in §162. EXAMPLE. Solve the equation: 4 r Remove the denominators and obtain 15x-27-4x= 21x-57. Transfer all the terms which involve a to the first member and the known terms to the second; then 172. When the coefficient of x in the first member and the final known term in the second member have contrary signs, the root of the equation will be a negative number. Thus, consider the equation, 173. The Solution of More General Equations. EXAMPLE 1. Solve the equation, 43 for its root. After the denominators have been removed by multiplying all the terms of the equation by 60, it follows that 15(x 3) 10(2 c 5) = 41 +12(3 x 8) or by multiplying out, 4(5x + 6), 15x 45 20x50 41 + 36 x — 96 — 20 x — 24. or EXAMPLE 2. Solve for x: 3-x-2(x-1)(x + 2) = (x − 3)(5 — 2 x). Multiply out the parentheses: 3 x−2(x2 + x 2) 11 x 2x2-15 3-x-2x-2x+4=11x- 2x2 - 15. Transpose the terms involving the unknown quantities: -x-2x2-2x-11x +2x2 = - 3 -4-15 Combine first the terms in the first member of the equation, then Similarly, the terms in the second member combined give, FORMULAE FOR THE SOLUTION OF AN EQUATION OF THE FIRST 174. Every equation of the first degree in one unknown quantity can, as has been seen, be reduced by addition, subtraction, and multiplication to the form, ах = b, where a and b are known numbers. To arrive at this result, remove fractions, and render the equation integral throughout by multiplying both members by the least common denominator. Transpose all the terms in x to the first member, and all the known terms to the second. Then combine all the terms in a into one term, and, similarly, all the known terms into a single term. The equation having been reduced to the form, two principal cases can arise: either a, the coefficient of x, is different from zero or it is equal to zero. 175. When a is Different from Zero.-If a is different from zero, divide both members of the equation by a and form equation (2), equivalent to equation (1): Since a is different from zero, equation (1) has a determinate root, and this root is given by formula (2). 176. When a is Equal to Zero.-In case a is zero, it is no longer possible to divide both members of equation (1) by a. It is accordingly necessary to study this equation more minutely. Two cases can arise: at the same time that a is zero, b can be different from zero or equal to zero. 1. When a = 0, but b 0. In this case, no number substituted for x can satisfy equation (1), because the product of any number whatever by a, that is to say, by zero, is equal to zero, and consequently is different from b. The equation is therefore impossible. Suppose that instead of a's being zero, a is very small; then the equation, and accordingly will have a determinate root. If b remains fixed and the number a decreases indefinitely, and approaches zero, the root b a will increase indefinitely, and in case a is equal to zero, the equa tion is said to have an infinite* indeterminate root (873, 2). 2. When a = 0, and b = 0. Then any number put in place of x will satisfy the equation, because the product of any number whatever by 0 is equal to zero. The value of x, is then indeterminate (873, 1). 0 Consequently equation (1) has a determinate root 12.59259.... * An infinite number is one which is larger than any number one can choose. |