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30. One cask contains 18 gallons of wine and 12 gallons of water; another, 4 gallons of wine and 12 of water. How many gallons must be taken from each cask so that when mixed there may be 21 gallons, half wine and half water?

IV. PROBLEMS ON MOTION 31. A and B run a race of 500 yards. In the first trial, A gives B a start of 80 yards and wins by 5 seconds. In the second trial, A gives B a start of 140 yards and B wins by 10 seconds. Find the rates of A and B in yards per second.

(HINT. Let x=A's rate, and y=B's rate. Then, the time it takes A to run the whole 500 yards is 500/x seconds (see § 107). In the first trial B runs but 420 yards and the time it takes him to do this is 420/y. In the second trial B runs but 360 yards and the time it takes him to do this is 360/y. Now form two equations which express the facts about the times as given in the problem.]

32. Two cities are 40 miles apart. To travel the distance between them by automobile takes 3 hours less time than by bicycle, but if the bicycle has a start of 24 miles, each takes the same time. What is the rate of the automobile, and what the rate of the bicycle?

33. A boy can row 10 miles downstream in 2 hours, and the same distance upstream in 31 hours. What is his rate of rowing in still water, and what is the rate of the stream ?

[Hint. Let x=his rate of rowing in still water, and y=the rate of the stream. Then, his rate of rowing downstream is x+y, and his rate of rowing upstream is x-y.]

34. A boy rows 18 miles down a stream and back in 12 hours. He finds that he can row 3 miles downstream while he rows 1 mile upstream. What is his rate, and what is the rate of the stream?

35. A man rows for 4 hours down a stream which runs at the rate of 3. miles an hour. In returning it takes him 141 hours to reach a point 3 miles below. his place of starting. Find the distance he rowed downstream and his rate in still water.

132. Literal Simultaneous Equations. Literal simultaneous equations are those in which letters are used to represent one or more of the known numbers. The letters used for this purpose are usually the first ones in the alphabet, a, b, c, etc. Such equations may be solved like those already considered in this Chapter, but it must be remembered that the solution must always be in terms of the known letters. (See § 106.)

EXAMPLE. Solve the equations (1)

ax+by=m, (2) :

cx+dy=n. SOLUTION.

(1) Xd gives (3)

adx+bdy=dm. (2)Xb gives (4)

bcx+bdy = bn (3)–(4) gives (adbc)X = dm-bn Hence

x=

ad- bc (1)Xc gives (5)

acx+bcy=cm. (2)Xa gives

acx+ady = an (5)-(6) gives (bc-ad) y = cm-an Hence

dm-bn. Ans.

(6)

cm- an. Y=bc-ad

Ans.

NOTE. In solving literal simultaneous equations it is usually best to perform all eliminations by addition or subtraction, thus avoiding substitution.

EXERCISES

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3.

ax=by,

Solve each of the following pairs of equations for x and y. Sax by=m,

2 bx+ay=a+b, 1 cx dy=n.

1 ab(x - y) = a? 52. [See if your answer is correct when a=2, b=1, c=3, d=4, m=2,

1x + =1, n=1, and thus check your work.]

a'b 2. Jax+by=m, 1 bx - ay=n.

11+1-1, 1x+y=ab.

J x'y a' Check when a=2, b=1.

11_1-1 4 Sax+3y=10 a,

1 x y b
14 x — 2 y=6 a.

[Hint. See § 131.]
Ja(x,y)=5,
1 bx cy=n.

1 X Y

10. sx-y=2 n,

b_ a = -1. | mx ny=ma+na.

1 x y 11. The sum of two numbers is a, and their difference is b. Find the numbers.

12. One half the sum of two numbers is g, while one fourth their difference is h. What are the numbers ?

13. I have c coins, all silver dollars and quarters, whose value is $d. How many of each denomination are there? (Compare Exs. 25, 26, 27 p. 228.)

14. A grocer has two kinds of sugar, one worth a cents and the other b cents per pound. How many pounds of each must he take to make a mixture of c pounds worth d cents per pound? (Compare Exs. 28, 29, 30, p. 228.)

la_b=-1,

15. A and B run a race of a yards. In the first trial A gives B a start of b yards and wins by s seconds. In the second trial A gives B a start of c yards and B wins by t seconds. Find the rates of A and B in yards per second. (Compare Exs. 31, 32, 33, 34, p. 229.)

16. If A gives $d to B they have equal sums. If B gives $e to A, then A has 3 times as much as B. How much has each?

* 133. Simultaneous Equations in Three Unknown Letters. Sometimes we meet with three linear equations containing three unknown letters, instead of two equations in two letters. Such equations may be solved by the process of elimination, as will appear from a study of the following:

EXAMPLE. Solve the three simultaneous equations (1)

x+y+z=6, 2 x-y+3 2=9,

x+2 y-z=2. Solution. Eliminate one unknown, say y, from (1) and (2).

(2) (3)

Thus,

(6)

(1)+(2) gives (4)

3 x+4 z=15. Eliminate the same unknown, y, from (2) and (3), thus:

(2)X2 gives (5)

4 x–2 y+6 z=18 From (3)

x+2y-z=2 (5)+(6) gives

5 x+5 z=20, or (7)

x+2=4. Equations (4) and (7) contain only x and 2 and hence may be solved for these letters, as in § 129. Thus,

from (4) (8)

3 x+4 z=15 (7)X3 gives (9)

3 x+3 2=12 (8)–(9) gives

2=3.

Substituting z=3 in (7) gives

x+3=4. Therefore

x=1. Substituting z=3 and x=1 in (1) gives

1+y+3=6. Therefore,

y=2. The desired solution is therefore x=1, y=2, 2=3. Ans. Сивск.

1+2+3=6, as required by (1). 2X1-2+3x3=2–2+9=9, as required by (2).

1+2x2–3=1+4-3=2, as required by (3). An examination of the preceding solution gives the following rule.

RULE FOR SOLVING THREE SIMULTANEOUS EQUATIONS. First eliminate one of the unknown letters from two of the given equations; then eliminate the same unknown from another two of the equations. This gives two new equations which contain only two of the unknown letters, and hence these two letters can then be found as in the solution of equations with two unknowns. The third letter can now be found by substitution of the two letters already obtained into one of the three original equations.

EXERCISES

Solve for x, y, and z in the following and check your answers in the first four.

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