both terms by the denominator of the fraction. In changing the first term of the equation, = 24, to 3x, it has been multiplied by 4, so that the second term must also be multiplied by 4. 853. In solving these examples by the algebraic method of "clearing of fractions," attention may be called to its similarity to the arithmetical method. To find the value of y in 2, the pupil multiplies 8 by 5 and divides the product by 2; as an example in arithmetic, he would divide 8 by , that is, he would multiply 8 by ; the only difference being that by the latter method he would the only difference betthat is, he would While 24 = 8 may be changed to 4 =4 by dividing both terms by 2, beginners are usually advised to begin by “clearing of fractions,” short methods being deferred to a later stage. 3x, 5x 854. 6 may be written O = 92. 8. 2 x should be reduced to an improper fraction, making the equation, 284 = 115. Make similar changes in 12, 14, 18, and 20. 855. 2. & + 5z = 100 Clearing of fractions, 7x + x = 3360, 8x = 3360, x=420, the greater number, in = 60, the less. Or, let x = less ; 7x = greater. x + 7x = 480, 8x = 480, x= 60, the less, 7x=420, the greater. The employment of the latter plan does away with fractions in the original equation. 11. 30x — x=522, or x - = 522. 13. Let x=number of plums; 4x = number of peaches. Then 2 x will be cost of plums, and 12x the cost of the peaches. 2x + 12x = 70. 2 – 3 = 80. 17. * — * --= 24. x + 12x +(143 X 3}) = 15. x+06+ 5x = 15. 15. 19. Let x=price per yard of the 48-yard piece ; 2x=price per yard of the 36-yard piece ; 48x will be the total cost of one, and 72x, of the other. 48x + 72x = 240. 20. 160 x + 120 x = 840. 856. The pupils should be permitted to give these answers without assistance. In Art. 857 is explained what is meant by “ transposing." 858. While these exercises are so simple that they can be worked without a pencil, they should be used to show the steps generally taken in more complicated equations. x + 37 = 56 In 1, for instance, the work should take the x = 56 – 37 form here indicated, only a single step being x=19 taken at a time. In 19, the first step is to clear the equation of fractions by multiplying by 6; the second step is to transpose the unknown 2x – 6= 16+ 0723 qu - quantities to the left side of the 12x - 36=96 + 3x – 2 x equation, and the known quantities 12x — 3x + 2x = 96 + 36 to the right; the third step is to com11x= 132 bine the unknown quantities into one, and to make a similar combinax = 12 tion of the known quantities; the last step is to find the value of x. After a little more familiarity with exercises of this kind, the pupil can take short cuts with less danger of mistakes; for the present, however, it will be safer to proceed in the slower way. 859. 5. x + (x + 75) + x + (x + 75) = 250. x + x + x + x = 250 – 75 — 75. NOTE. — The parentheses used here are unnecessary. They are employed merely to show that x + 75 is one side of the field. |