If the same quantity appear in each member of an equation affected with the same sign, it may be suppressed. (95.) III. The unknown quantity may be combined with known quantities by multiplication. Let the equation be 5 =25. If we divide each of the equal quantities 5x and 25 by the same quantity 5, the quotients will be equal, and we shall have x=5, the value required. ax=b, So, also, in the equation dividing each of these equals by a, the result is From this we conclude that When the unknown quantity is multiplied by a known quantity, the equation is solved by dividing both members by this known quantity. (96.) IV. The unknown quantity may be combined with known quantities by division. 4 by the same quantity 5, the products will be equal, and we shall have x=20, the value required. QUEST.-When may a quantity be suppressed? When the unknown quantity is multiplied by a known quantity, how is the equation solved? multiplying each of these equals by a, the result is x=ab, the value required. From this we conclude that When the unknown quantity is divided by a known quantity, the equation is solved by multiplying both members by this known quantity. (97.) V. Several terms of an equation may be frac tional. Multiplying each of these equals by 2, the result is 4 8 ཝ=3+་ Multiplying each of these last equals by 3, we ob tain 24 3x=4+5; and multiplying again by 5, we obtain 15x=20+24, an equation free from fractions. We might have obtained the same result by multiplying the original equation at once by the product of all the denominators. Thus, multiplying by 2×3×5, we have QUEST. When the unknown quantity is divided by a known quan. tity, how is the equation solved? When several terms of an equation are fractional, how should we proceed? multiplying successively by all the denominators, or by at, e at once, we obtain Canceling from each term the letter which is common to its numerator and denominator, we have cex=abe+acd, an equation clear of fractions. Hence it appears that An equation may be cleared of fractions by multiplying each member into all the denominators. We will now apply these principles to the solution of equations. Ex. 1. Given 5x+8=4x+10 to find the value of x. Transposing 4x to the first member of the equation, and 8 to the second member, taking care to change their signs (Art. 94), we have 5x-4x-10-8. Uniting similar terms, we find x=2. In order to verify this result, put 2 in the place of x wherever it occurs in the original equation, and we shall obtain That is, or, 5x2+8=4x2+10. 10+8=8+10, 18=18, which proves that we have found the correct value of x X X Ex. 2. Given x-7=3+, to find the value of x. Multiplying every term of the equation by 5, and also by 3, in order to clear it, of fractions (Art. 97), we obtain To verify this result, put 15 in the place of x in the original equation, and we have That is, 15 15 15-7= 5 +3. 15-7=3+5, 8=8, or, which shows that we have found the correct value of x. Ex. 3. Given 3ax-4ab=2ax-6ac, to find the value of x in terms of b and c. This result may be verified in the same manner as the preceding. (98.) Hence, in order to solve an equation of the first degree containing one unknown quantity, we have the following RULE. 1. Clear the equation of fractions, and perform in both members all the algebraic operations indicated. 2. Transpose all the terms containing the unknown quantity to one side, and all the remaining terms to the other side of the equation, and reduce each member to its most simple form. 3. Divide each member by the coefficient of the unknown quantity. Ex. 4. Given 5x+16-7-29 to find the value of x. Ans. x=4. Ex. 5. Given x+12=4x-6 to find the value of x. Ans. x=6. Ex. 6. Given 7-3x+12=25-4x-1 to find the value of x. х х Ans. x=5. Ex. 7. Given 2x+2+3=27 to find the value of x. х Ans. x=10. Ex. 8. Given 3x-3+4=6x-6 to find the value QUEST. Give the rule for solving an equation of the first degree. with one unknown quantity. |