at the 2-hour mark on XX and then rising 1 units for every unit it goes to the right; that is, it is represented by the line CB. The point B where these two lines meet corresponds graphically to the meeting of the automobiles, because at this point both have traveled the same distance as measured off on YY. The time that goes with this point, as measured off on XX, is seen to be 6 hours. Therefore, the second automobile will overtake the first one 6 hours after the first one starts, or, 4 hours after the second one starts. Ans. EXAMPLE 2. Two automobiles start from two towns that are 110 miles apart and travel towards each other, the one at 20 miles an hour and the other at 30 miles an hour. How long before they will be 10 miles apart? 120 Y SOLUTION. Take the same units as in Example 1. Then the two motions will be represented by the two lines in Fig. 71. Note that the line representing the machine traveling at 20 miles per hour starts at 0 hours and rises 1 unit for every unit it goes to the right; while the other line, which represents the machine traveling at 30 miles per hour, starts at the O hour also, but at the 110 mark, and it descends 1 units for every unit it goes to the right. X 100 +80 +60 +401 P 1201 0 X 0 1 2 3 4 Y Hours FIG. 71. Now, the question is, "When will the two machines be 10 miles apart?" Graphically, this means "What is the point on XX where the difference between the distance measured up to the lines will be 10 miles; that is, a unit?" is represented by PQ. The point on XX is seen to be 2. Hence, the answer to the example is 2 hours. * EXERCISES - MOTION This difference 1. A boy starts from home at noon and rides on a bicycle 10 miles an hour toward a certain town. At 3 o'clock in the afternoon a man starts out from the same place in an automobile and travels the same road at 30 miles an hour. Determine graphically how soon the boy will be overtaken, and also how far he will then be from home. 2. Answer the following question graphically: In a bicycle race A goes 22 yards per second, and B goes 12 yards per second. B has a start of 50 yards. Who will win the race? [HINT. Let each unit on XX represent 1 second, and each unit on YY represent 10 yards.] 3. Solve the following graphically: A man can row 18 miles per hour downstream, and 2 miles per hour returning. How far downstream can he go if he wishes to be back home in 10 hours from the time he starts out? CHAPTER XIII SIMULTANEOUS EQUATIONS SOLVED BY 128. Solution of Simultaneous Equations by Elimination. The solution of two simultaneous equations (such as those considered in § 125) may be readily found by a process called elimination. As there are several kinds of elimination, we shall first consider that which is called elimination by substitution. (1) (2) EXAMPLE 1. Solve the simultaneous equations x+y=6, 3x-2y=-2. SOLUTION. From equation (1) we have (by transposing the x) y=6-x. If we substitute (or place) this value of y in equation (2), we obtain the equation 3x-2(6-x)=-2, which is an equation containing only the one letter, x; in other words, we have obtained from (1) and (2) an equation in which the y has been removed or eliminated. This last equation (being like those considered in the earlier Chapters) may be at once solved. Thus, it reduces to which gives 3x-12+2x=-2, or 5 x=10, x=2. To get y, substitute the value x=2 in (1). The result is 2+y=6, or y=4. The required solution of (1) and (2) is therefore (x=2, y=4). Ans. CHECK. When x=2 and y=4, the first member of equation (1) becomes 2+4, which reduces (as this equation demands) to 6. Similarly, when x=2 and y=4 the first member of (2) becomes 3X2-2×4-6-8=-2, as required. This is an equation for the one letter x, and may therefore be solved. Thus, clearing it of fractions we obtain The required solution of (1) and (2) is therefore (x=-1, y=2). Ans. CHECK. When x=-1 and y=2, the first member of (1) becomes 2x(−1)+3×2, which reduces (as this equation demands) to 4. Similarly, when x=-1 and y=2, the first member of (2) becomes 5×(−1)−2×2, which reduces to −9, as required. NOTE. Observe that the way which we have just used for solving two simultaneous equations gives their solution without requiring us to draw the graph. For this reason it is usually shorter than the method studied in § 125, though the latter is very valuable in making clear what the solution means. From the two examples worked above we obtain the following rule. RULE FOR SOLVING TWO SIMULTANEOUS EQUATIONS BY SUBSTITUTION. Solve the first equation so as to get y in terms of x. Place the result in the second equation, which may then be solved to get the desired value of x. After x has been found, put its value in the first equation, which may then be solved to get the desired value of y. NOTE. This rule may be stated in a more general form as follows: Solve either equation for one of the letters in terms of the other one. Place the result in the other equation and solve. Having thus found the value of one of the letters, the value of the other may be found by substituting the known value into either of the given equations and solving the resulting equation. EXERCISES 1. Solve Example 1 of § 128 by the method of § 125 and compare the solution you thus obtain with that already worked out in § 128. They should agree. Solve each of the following pairs of simultaneous equations by substitution and check your answers. For further exercises on this topic, see Appendix, p. 308. |