EQUATIONS WITH FRACTIONS

95. When an equation involves fractions we usually begin its solution by clearing it of fractions. This operation rests on the principle that when equals are multiplied by equals, the results are equal.

Observe that this equation is meaningless and absurd when m = n, for in that case it involves a division by zero. For the same reason we cannot have m =— - n, for then m + n = 0. Let us assume that neither denominator is zero.

Then the 1. c. d. = · (m + n) (m − n). Multiply both sides by

Observe that if x = 1, or x2 = 1, this equation represents an absurdity, since it involves division by zero.

On the supposition that x2 #1, the 1. c. d. = 4(x2 — 1).
Multiply both sides by 4(x2-1), 4 x(x+1)+4(x − 3)=5(x2−1). (2)
Simplify,
4x2+4x+4x-12= 5 x2 - 5.

Transpose,

Factor,

x2-8x+7=0.

(x-7)(x-1)=0.

Hence,

x = 7, x = 1.

We

Substitute in the original equation (1); we find x = 7 satisfies it, but x = 1 does not satisfy it, since it gives rise to division by zero. must reject x = 1; x = 7 is the only correct solution of (1).

It is of interest to notice that, while x=1 does not satisfy (1), it does satisfy (2). In other words, equation (2) is satisfied by two values of x, while equation (1) is satisfied by only one value of x.

It is clear that the new root was introduced in the act of multiplying both sides of (1) by 4(x2-1). Such new roots

which do not satisfy the original equation, but do satisfy an equation that is derived from it, are called extraneous roots.

When we multiply both sides of an equation by an expression involving x, we obtain a new equation some of whose roots may not satisfy the original equation and are not correct solutions of it; they are extraneous roots. This may happen even though all operations are performed free of error.

Hence, to make sure that our values of x are correct, we must substitute the values found for x in the original equation, to see whether it is satisfied or not.

Solve and tell, by substitution in the original equation, which of the roots, if any, are extraneous.

MISCELLANEOUS PRACTICAL PROBLEMS

97. Applications of algebra, such as occur in the problems which follow, are frequently made by builders and manufacturers. Certain formulas which have been established by engineers and mathematicians are used in computing the desired results. We shall assume the formulas as true without giving the mode of deriving them. Care must be taken to use in every problem the proper units of measure.

1. Find correctly, to two decimal places, the side of a square whose area is 35.06 sq. in.

2. Compute the altitude of an equilateral triangle whose sides are a inches.

3. If one side of a rectangle is a inches long, and one of the adjacent sides is three times as long, find the length of the diagonal.

4. A ladder of the length 9 a is placed against a wall, with its foot at a distance 4 a from the wall. How high above the ground is the top of the ladder?

5. In placing blackboards in schoolrooms, architects determine the height of the chalk rail above the floor by the formula

=

h=25+ 3(g−4),

where h height of the chalk rail in inches, g the number of the grade to which the pupils in the room belong. Find the height of the chalk rail for pupils in the 5th grade; also for pupils in the 8th grade and in the 3d grade.

6. The Baldwin Locomotive Works in Philadelphia have derived a formula, R=3+, for finding the approximate relation between the resistance, R, offered by a railway train, and the velocity, the train traveling on a straight and level track. In this formula R is the resistance in pounds per ton weight of the train, v is the velocity of the train in miles per hour.

What is the resistance per ton of a train running at the

rate of 40 mi. an hour?

if it weighs 100,000 T.?

What is the resistance of this train,

7. The load that may be safely applied to an iron chain is given by the formula, = 7.11 d2, where is the load in tons l 7 (2000 lb.), and d is the diameter of the iron chain in inches. Find the largest safe load that may be applied to a chain for which d is in. Find d when l = 13 T.

8. The distance which one can see from an elevation at the sea has been found to be d= 1.23√h, where d is the number of miles one can see, and h is the elevation of the observer in feet. How far is the horizon from a man standing at the seashore, whose eye is 6 ft. from the ground?

9. In Ex. 8, how far can one see from a cliff 27 ft. high? 67 ft. high? How high must a cliff be to afford a view of 20 mi.?

10. From a cliff I can just see the lights of a seaport 12 mi. across the sea. If these lights are 20 ft. above the sea, what is the height of my eye?

11. According to a rule sometimes used by architects, the "rise" (r inches) in the steps of a staircase (Fig. 8) is connected with the "tread " (t inches), by the formula = } (24 — t), where r and t are also subject to the limitations, 7>r>5, 12 >t> 10. Explain the formula in words. What is the "rise" when the tread is 101"? 11"?

FIG. 8.

12. Engineers have determined that the velocity (v feet per minute) of a stream at the bottom of a river is connected with the velocity (V feet per minute) at the surface, by the formula v=V+1-2VV. Find the velocity at the bottom when the velocity at the surface is 5 ft. per minute, 8 ft. per minute, 95 ft. per minute.

G