CHECK. When x=6 the first member of the given equation becomes 5×6+6x3=30+18=48, while the second member becomes 12+36=48. Since both have the same value, namely 48, the root (or x=6) is correct.

WRITTEN EXERCISES

Solve and check your answer in each of the following equations.

1. 3(x+4)=18.

2. 2(x+5)+3(x-6)=12.

3. 5(x+4)+2(x-3)=4(x-1).

4. (x+2)(x+3)=x(x+3)+10.

5. (x-4)(x+5)=x(x-2).

6. (r-3)(r-5)=(r+4) (r−2)-7.

7. (y+1)(y-2) = (y-3) (y−4).

8. (s+3)(s+8)—s(s+6)=0.

9. (x2-x-1) (2 x+4)= 2 x2(x+1)−7(x−1).

10. x2−(2x+3)(2 x−3)+(2 x − 3)2 = (x+9) (x − 2) — 2. 11. 3(4−x)2-2(x+3) = (2 x −3)2 — (x+2) (x −2)+1. 12. 3x2- {5 x-[4— (x − 1) (2 x −3) −7 x]+(x−3)2 } = 0. 13. 5x+1−2{2x−3[x−(x+1)(x+3)]−3(x+2)2} =0. For further exercises on this topic, see Appendix, p. 299.

These problems resemble those on pp. 76, 77, yet they are different because they cannot be worked without a knowledge of the principles of multiplication studied in this Chapter.

1. Divide 29 into two parts so that three times the one part added to five times the other part shall equal 105.

SOLUTION. Let x=one part. Then the other part =29-x.
The problem then says that

3x+5(29-x)=105.

Multiplying (see § 51), this becomes

3x+145-5x=105.

Transposing gives

-2x=-40.

Dividing both sides by −2 gives

x=20.

The one part is therefore 20, and the other is 29-20, or 9. Ans.

2. The difference between the ages of a father and his son is 45 years and in a year's time the father will be six times as old as the son. How old is each?

=

[HINT. Let x = the number of years in the son's age. Then x+45 the number of years in the father's age. A year from now their ages will be x+1 and x+46.]

3. A father is four times as old as his son; five years ago he was five times as old. How old is each?

4. A has $37 and B has $11. How much must A give B in order that A may then have twice as much as B?

5. A box of 28 chocolates was divided between 6 boys and 4 girls, each girl receiving 2 more than a boy. How many did each girl receive?

6. In buying material for a dress a lady bought 8 yards of silk and 4 yards of trimming, paying 90 cents more per

yard for the trimming than for the silk. The bill was $18.00. How much did each cost per yard?

[HINT. Work in cents.]

7. A bag contains 52 silver coins, some of them quarterdollars and the rest dimes. The value of the whole is $10.00. How many coins of each kind are there?

8. A man has $100 in one bank and $25 in another. If he has $125 more to deposit, how should he divide it between the two banks so that the first account may become four times as great as the second?

[HINT. Compare Ex. 44, p. 26.]

9. A father made this offer to his son, "For every day on which your standing in algebra is as high as 80% I will give you a nickel, but on every day that your standing falls below 80% you are to give me a penny." At the end of 12 days the son had 18 cents. For how many days was his standing above 80%?

D

10. In the figure the shaded part, or border, represents the area between an inner square (lettered abcd) and an outer square (lettered ABCD). If this border contains 32 square feet and is everywhere 1 foot wide, how long must one of the sides of the inner square be?

[HINT. Let x= the length of one of the sides of the inner square. Then x2-the area of the inner square, and x2+32=the area of the inner square and border.

a

ft

A

FIG. 23.

But the area of the inner square and border is the same as that of the outer square, and this is (x+2)2, since each side of the outer square

=

x+1+1, or x+2.]

FIG. 24.

11. It is desired to lay out a square flower garden with a gravel walk around it, the gravel walk to be 2 feet wide and 6 inches deep. How long can each side of the garden be if only 5 loads of gravel are to be used, it being given that a load contains 1 cubic yard?

[HINT. First find how much area the gravel will cover when laid 6 inches deep.

12. Suppose that the garden mentioned in Ex. 11 is rectangular, with one of the long sides measuring 2 feet more than a short side, and suppose as before that the gravel border is to be 6 inches deep and 2 feet wide, ten loads of gravel to be used. Show that in this case the equation for the shorter side, x, becomes x(x+2)

FIG. 25.

+540=(x+4)(x+6). Whence, find the dimensions (length and breadth) of the rectangle.

Pond

Area
Overflowed

FIG. 26.

13. During a spring freshet a certain circular pond overflowed its banks so that an acre of ground all around was inundated (under water). It was found afterwards that the overflow had carried the water back from the bank about 20 feet on all sides. What was the diameter of the pond in ordinary weather?

T=

[HINT. See Ex. 25 page 22. Take T-34. Remember the diameter =twice the radius.]

PART II. SPECIAL CASES IN MULTIPLICATION AND FACTORING

57. Product of Two Binomials. We have to multiply binomials together so often in algebra that it is desirable to know how to get the answer quickly by mere inspection. This is possible in many cases.

the answer could easily be obtained in the following way: First, square the x, giving x2, then add (mentally) the 3 and the 4, giving 7, and multiply this by x, giving the 7 x of the answer. Finally, multiply (mentally) the 3 by the 4, giving the 12 which appears at the end. By adding the x2, the 7x and the 12 obtained in this way, we have the answer.

Let us take another illustration. Suppose we wish to multiply x+2 by x+3. Reasoning as before, the answer will be x2+(2+3)x+2×3, or x2+5x+6. Check this yourself by multiplying x+2 by x+3 in the long way.

Similarly, we can shorten the work of multiplying even when negative numbers occur. For example, in multiplying x-2 by x+1 the answer will be x2+(-2+1)x+(-2) X1, or x2-x-2.

Other illustrations follow and should be carefully examined. 1. (x+7)(x-2)= x2+(7-2)x+7(−2) = x2+5x-14. Ans. 2. (x−8)(x+3)= x2+(−8+3)x+(−8)3 = x2 − 5 x − 24. Ans.

3. (x-6)(x-4)= x2+(−6−4)x+(−6)(−4)

= x2-10x+24. Ans.

H