24. House and lot, or 3 lots +1 lot, or 4 lots = $8100; 1 lot=$810044-$1800; house = $1800 × 31.

25.

$18 × (20 × 12) × (15 × 12) × (6 × 12)

1000 × 8 × 4x-2

27. 36 yd. 8 in. =1304 in.; 13 yd. 1 ft. 9 in.

=

1304

quantity left 1304 in. - 489 in. = 815 in.; fraction left = 815 =; decimal left = .625; per cent left = 621⁄2. Ans.

28. Assessed value = 80% of $30000 $24000. Taxes on 24 thousand dollars = $21.60 × 24.

846. 2. See Arithmetic, Art. 1251; angles E, F, G, and H; and M, N, O, and P.

3. Art. 1251; angles A and B, C and D.

4. Angles I and J, K and L. The scholars should understand that two lines can be perpendicular without one being a horizontal line and the other a vertical line.

5. The size of an angle does not depend upon the length of the lines that form the angle. Two short lines may meet at a very obtuse angle, and two long lines may form a very acute angle.

13. If the pupils have in their drawing lessons constructed triangles by means of compasses, these may be used; otherwise, let them manage as best they can, no great accuracy being required.

15. Children are accustomed to seeing an isosceles triangle in only one position: they should learn that if a triangle has two equal sides, it is isosceles, no matter whether the unequal side is vertical, horizontal, or oblique.

16-22. Accustom the scholars to the occasional employment of an oblique line as a base in constructing squares, rectangles, etc. See Arithmetic, Art. 1265. A card may be used to make a square corner.

24. See Arithmetic, Art. 929, No. 8, for a rectangle, a rhombus, and a rhomboid, having equal bases and equal altitudes.

shows three rhomboids of equal bases and equal altitudes, but differing in shape.

25. See Art. 929, No. 8.

847. 1. (of 15 x 20) sq. in. The length of the third side. does not enter into the computation.

6. Let the scholars find the area of the rectangle, 66 ft. by 63 ft., and the two triangles, 31 ft. each by 63 ft., and find the sum of the areas. Then lead them to see that bringing the righthand triangle to the left of the rhombus would make a rectangle 97 ft. by 63 ft., whose area is the sum above found.

7. Find the area in square meters, saying nothing more about the meter than that it is largely used on the continent of Europe, and is a little longer than a yard,

8-10. Give no rules yet for calculating the areas of trapezoids and trapeziums. Let the pupils ascertain the areas of the figures from the data supplied.

XIII

NOTES ON CHAPTER TEN

The formal study of algebra belongs to the high-school; but some so-called arithmetical problems are so much simplified by the use of the equation that it is a mistake for a teacher not to avail herself of this means of lightening her pupils' burdens.

In beginning this part of her mathematical instruction, the teacher should not bewilder her scholars with definitions. The necessary terms should be employed as occasion requires, and without any explanation beyond that which is absolutely necessary.

849. Very young pupils can give answers to most of these questions; so that there will be no need, for the present, at least, of introducing a number of axioms to enable the scholar to obtain a result that he can reach without them.

850. Pupils will learn how to work these problems by working a number of them. They may need to be told that x stands for lx; and that, as a rule, only abstract numbers are used in the equations, the denomination-dollars, marbles, etc. being supplied afterwards.

While the scholars should be required to furnish rather full solutions of the earlier problems, they should be permitted to shorten the work by degrees, writing only whatever may be

necessary.

4. x+2x=54. 5. +5 =78.

6. 7x+5x=156.

7. 9x 3 x

=

66.

8. x+2x+6x=27000. 9. 2+5=72. 10. x+2x+3x=54. 11. x+6x=42.

20. Let x= each boy's share; 2x= each girl's share.

2x+4x=240.

21. x = number of days son worked; 2x= number father worked. 3x son's earnings; 8x= father's earnings.

=

3x+8x165.

22. x= number of dimes; 2x = number of nickels; 6x= number of cents.

Susan's number; 2x = Mary's; 3x= Jane's.

852. Pupils already know that means 34, so that they can understand that 3x

4

means 3x÷4, or 4 of 3x. When of

something (3x) is 24, the whole thing (3x) must be 4 times 24, or 96; that is, when

3 x

=24, 3x=96.

4

When

42=20, 4 z = 20 × 5, or 100.

5

From these examples can be formulated the rule for disposing of a fraction in one term of an equation, which is, to multiply

3 x 4

=

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 cancel.

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

115. Make similar changes in 12, 14, 18, and 20.

9. Let 5x

numerator; 7x= denominator. 7x-5x = 24; 12. The numerator, 5x, will be 5 times 12, or 60; the denominator will be 84; and the fraction, . Ans.

2x

=

24; x

x+