4. 7x19 5x + 7. = 5. 3(x-2)=2(x − 3). 6. 5(x+2)=3(x + 3) + 1. 8. 5(4-3x)=7(3-4 x). 9. 2(x-3)=5(x + 1) + 2 x − 1. 10. 4(1-x)+ 3(2 + x) = 13. 11. 2(x-2)+ 3 (x − 3) + 4 (x − 4) – 20 = 0. 13. 5+6(2+1)−7(+2)−8(2+3)=0. 29. (x2)(x-5)+(x − 3) (x-4)=2(x-4) (x5). 45. (α + x)+ } (2 a + x) + 1 (3 a + x) = 3 a. 50. (x + a + b)2 + (x + a − b)2 = 2 x2. 51. (x − a) (x − b) + (a + b)2 = (x + a) (x + b). (x-a-b+c) (x − a + b − c). 53. ax(x+a)+ bx(x + b) = (a + b) (x + a) (x + b). 54. (x − α)3 + (x − b)3 + (x − c)3 = 3 (x − a) (x − b) (x − c). - CHAPTER VII. PROBLEMS. 99. With some of the general methods of algebra now at our command, we return to the subject introduced in the first chapter; namely, the solution of problems. In order to solve a problem, the relations between the known and unknown quantities must be expressed by means of algebraical symbols: we thus obtain equations, the roots of which are the required values of the unknown quantities. 100. In the present chapter we shall only consider problems in which there is one unknown quantity, and in which the relation between the known and the unknown quantities is expressed algebraically by means of a simple equation. Of such problems the following are examples: Ex. 1. A has $20, and B has $3.75. How much must A give to B in order that he may have just four times as much as B? Let x be the number of dollars that A gives to B. Then A will have 20 x dollars, and B will have 3.75 +x dollars. But A now has four times as much as B. Hence we have the equation NOTE. - It should be remembered that x must always stand for a number. It is also to be noticed that in any problem all concrete quantities of the same kind must be expressed in terms of the same unit; for example, in the above all sums of money were expressed as dollars. How Ex. 2. A man has 12 coins, some of which are half-dollars and the rest dimes, and the coins are worth 4 dollars altogether. many are there of each kind? Let x be the number of half-dollars; then 12 x will be the number of dimes. The half-dollars are worth x dollars, and the dimes are worth (12 – x) dollars. Hence, since the coins are worth 4 dollars altogether, we have the equation Therefore that is Hence there are 7 half-dollars and 5 dimes. Ex. 3. A father is six times as old as his son, and in four years he will be four times as old. Let the son be x years old. How old is each? Then the father must be 6x years old. After four years the son will be x + 4, and the father will be 6x+4 years old. Hence by the question Hence the son is 6 years old, and the father is 36 years old. Ex. 4. A can do a piece of work in 12 hours which B can do in 4 hours. A begins the work, but after a time B takes his place, and the whole work is finished in 6 hours from the beginning. How long did A work? Let x the number of hours that A worked. Then 6 x = the number of hours that B worked. Since A can do the whole work in 12 hours, the part done by A in 1 hour is 12. Therefore the part done by A altogether is х 12 Since B can do the whole work in 4 hours, the part done by B in 1 hour is }. Therefore the part done by B altogether is (6 − x). But A and B together do the whole of the work. Hence we have the equation Ex. 5. Find the time between 3 and 4 at which the hands of a clock are together. Suppose that the hands are together at x minutes after 3 o'clock. At 3 o'clock the hour-hand is 15 minute-spaces in front of the minute-hand, and after 2 minutes they are together. Hence while the minute-hand moves through x minute-spaces the hourhand will move through x 15 such spaces. But the minutehand moves twelve times as fast as the hour-hand, and therefore in any time the minute-hand passes over twelve times as many minute-spaces as the hour-hand. |