first term of the equation, 3 x 4 both terms by the denominator of the fraction. In changing the 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 cancel. = 8 may be changed to = 4 by dividing both terms У by 2, beginners are usually advised to begin by "clearing of fractions," short methods being deferred to a later stage. 8. 27x should be reduced to an improper fraction, making the equation, 23x 8 115. Make similar changes in 12, 14, 18, and 20. 9. Let 5x== numerator; 7x= = denominator. 7x-5x=24; 2x=24; x= 12. The numerator, 5x, will be 5 times 12, or 60; the denominator will be 84; and the fraction, §4. Ans. x + The employment of the latter plan does away with fractions. in the original equation. Then 2x will be cost of plums, and 12x the cost of the peaches. 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. 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. In 1, for instance, the work should take the form here indicated, only a single step being taken at a time. In 19, the first step is to x +37=56 x= 56-37 x= 19 clear the equation of fractions by multiplying by 6; the second 2x = 6 16+ х 2 3 step is to transpose the unknown quantities to the left side of the 12x3696+3x-2x equation, and the known quantities 12x-3x+2x=96+36 to the right; the third step is to combine the unknown quantities into one, and to make a similar combination 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. Other pupils may solve the problems in this way : 7x=635-320 315, x= 45, price, in dollars, of a cow; x+80=125, price, in dollars, of a horse. 14. x= number of dimes; x + 11 = number of five-cent pieces; 10x= value of dimes (in cents); 5x+55= value of five-cent pieces. 15. x= 10x+5x+55= 100. = x+x+ 2400 + x + 2400 + 2400 18000. 18. Let xless; x+33 = greater. x+33-3x=11. Bringing known quantities to the left side of the equation, and the unknown quantities to the right, |