10. Three towns, A, B, and C, are connected by straight roads, which form a triangle. From A to B by way of C is 25 miles; from B to C by way of A is 30 miles; from C to A by way of B is 35 miles. How far apart are the towns?

11. A, B, and C have certain sums of money. B would have the same as A if A gave him $100; C would have twice as much as A if A gave him $100; and C would have four times as much as B if B gave him $100. How much has each?

12. I have 35 coins, all nickels, dimes, and quarters, whose combined value is $3.75, and there are half as many dimes as nickels. How many are there of each denomination?

13. A and B can do a piece of work in 10 days; A and C can do it in 8 days; and B and C can do it in 12 days. How long will it take each to do it alone?

[HINT. Let x, y, z= the number of days in which A, B, C, respectively, can do the work. Then find the part that each can do in one day. Compare Ex. 17, p. 227, and Ex. 21, p. 171.]

14. I have $90 on deposit in the First National Bank, $51 in the State Savings Bank, and $75 in the Postal Savings Bank. If I have $144 more to deposit, how shall I distribute it among the three banks so as to make all three deposits equal?

15. If I have $a in bank A, $b in bank B, and $c in bank C, how should I distribute $d more among them so as to equalize the accounts? Check your answer by seeing if it gives (in particular) the answer you have found for Ex. 14.

EXERCISES - REVIEW OF CHAPTERS XII-XIII

1. What is meant by the solution of a pair of simultaneous equations?

2. Draw the graphs of the following simultaneous equations, using the same axes for both. State any conclusions you thereby

reach regarding their solution.

x-y=-2, 2x+y=-16.

3. If we have three simultaneous equations between the two letters x and y, and their three graphs all pass through one and the same point (point of intersection), then there will exist a certain value of x and a certain value of y that will satisfy all three equations at the same time. Why? Try this for the three equations

x-2y=-3,
3x-y=1,
x-y=-1.

4. In Ex. 3 what can be said if the three graphs do not all have the same point of intersection? Try this for the equations

x − y = −1, x-2y=-3,

2x-3y=-5.

Solve each of the following pairs of equations by any method.

*15. What is meant by the solution of three simultaneous equations between three unknown letters, x, y, and z? Describe how, in general, it is obtained.

* 16. Solve for x, y, and z in the following

x+y+2z=2(a+b), |x+z+2y=2(a+c),

y+z+2x=2(b+c).

*17. In a race of 500 yards, A can beat B by 20 yards, and C by 30 yards. By how much can B beat C in a good 500 yard race? For further exercises on this chapter, see Appendix, pp. 307-308.

CHAPTER XIV

SQUARE ROOT

134. Definitions. The square root of a number is one of its two equal factors.

Thus, 2 is the square root of 4 because 2x2=4. Likewise, 3 is the square root of 9, etc.

The square root of a number is denoted by the radical sign V placed over it.

Thus, √4=2, √9=3, √16=4, etc.

To find the square root of a number is called extracting its square root.

135. Extracting Square Roots in Arithmetic. Many times we can pick out the square root of a number by inspection. Thus, √144 is seen to be 12, because 144=12×12. Similarly, 19614. But in finding such a root as √74529, we cannot get the answer by mere inspection. Here we may follow the process studied in arithmetic, which is indicated below and which should now be reviewed by the pupil. PROCESS. 7'45'29 273 Ans. 4

EXPLANATION. We first separate the number into periods of two figures each, beginning at the right. That is, we write it in the form

7'45'29. Find the greatest square in the left-hand period and write its root for the first figure of the required root. This gives the 2 appearing in the answer.

Square this root (giving 4), subtract the result from the left-hand period and to the remainder annex the next period for new dividend. This gives the 345 appearing in the process.

Double the root already found, with a cipher annexed (giving 40) for a trial divisor and divide the last dividend (345) by it. The quotient (or, in some cases, the quotient diminished) gives the second figure, 7, of the root.

Add to the trial divisor the figure last found (7), giving the complete divisor, 47. Multiply this complete divisor by the figure of the root last found (7), giving the 329 appearing in the process.

Subtract this from the dividend, and to the remainder annex the next period for the next dividend. This gives the 1629 of the process.

Proceed as before, and continue until a final new dividend equal to O is obtained. In the example above, this happens at once, giving 273 as the required root.

This process is usually studied in elementary arithmetic, and is stated here as a review of the arithmetic method. We shall see in § 136 that a similar process may be used in extracting square roots of algebraic expressions. A thorough mastery of the process for numbers will assist materially in understanding the algebraic case.

In the example just worked the root comes out exactly because 74529 is a perfect square; that is, it is like one of the numbers 1, 4, 9, 16, 25, 36, etc. If, on the other hand, we had started with a number which was not a perfect square, the process would not be different except that we should not finally reach a new dividend which was equal to 0. For such a number, in fact, the process goes on indefinitely, but if we stop it at any point we have the desired root correct (decimally) up to that point. For example, in getting the square root of 550 correct to two places of decimals, the process is as follows: