19. A boy is one-sixth the age of his father, and five years older than his sister; the united ages of all three being 51, how old is each ?

20. Find the continued product of x2 + ax + α2, x2 x4 - a2x2 + aa.

21. Show without actual division that x pression x3 2x2 — 5x + 6.

-

ax + a2,

3 is a factor of the ex

by the usual method. Is the work shortened by proceeding as in Art. 119? Show that the square of the H. C. F. is contained in the second expression.

26. Find the factors of (i.) 14 a2—11 a−15; (ii.) aa+5 a2b2+9ba.

27. Of a party 5 more than one-third are Americans, 7 less than one-half are Englishmen, and the remainder, 8 in number, are Germans: find the number in the party.

30. Find three numbers whose sum is 21, and of which the greatest exceeds the least by 4, and the middle one is half the sum of the other two.

31. Employ the Factor Theorem in finding the H. C. F. of

a3 - 2 a2 + 1 and 2 a3 + a2 + 4 a − 7.

33. Two trains go from P to Q by different routes, one of which is 15 miles longer than the other. A train on the shorter route takes 6 hours, and a train on the longer, travelling 10 miles less per hour, takes 8 hours. Find length of each route.

L

34. Find the factors of

(i.) 4x24 xy-15 y2; (ii.) 9 x 82 x2y2+9 y.

36. The number of months in the age of a man on his birthday in 1875 was exactly half of the number denoting the year in which he was born. In what year was he born?

38. Divide 28+x3y2+x1y1+x2у6+y3 by x1 — x3y+x2y2 —xy3+y1 and find the value of the quotient when x = 0 and y = 1.

40. A regiment has sufficient food for m days; but if it were reinforced by p men, would have food enough for n days only. Find the number of men in the regiment.

CHAPTER XVII.

SIMULTANEOUS EQUATIONS.

167. Consider the equation 2x+5y = 23, which contains

two unknown quantities.

23-2x

From this we get y =

5

(1).

Now for every value we give to x there will be one corresponding value of y. Thus we shall be able to find as many pairs of values as we please which satisfy the given equation. Such an equation is called indeterminate.

For instance, if x 1, then from (1) y= 21.

=

Again, if x=-2, then y=27; and so on.

But if also we have a second equation of the same kind expressing a different relation between x and y, such as

If now we seek values of x and y which satisfy both equations, the values of y in (1) and (2) must be identical.

Multiplying across, 92-8x=120 – 15 x ;

7x=28; ..x= 4.

Substituting this value in equation (1), we have

Thus, if both equations are to be satisfied by the same values of x and y, there is only one solution possible.

168. DEFINITION. When two or more equations are satisfied by the same values of the unknown quantities, they are called simultaneous equations.

169. In the example already worked, we have used the method of solution which best illustrates the meaning of the term simultaneous equations; but in practice it will be found that this is rarely the readiest mode of solution. It must be borne in mind that since the two equations are simultaneously true, any equation formed by combining them will be satisfied by the values of x and y which satisfy the original equations. Our object will always be to obtain an equation which involves one only of the unknown quantities.

170. The process by which we cause either of the unknown quantities to disappear is called elimination. It may be effected in different ways, but three methods are in general use: (1) by Addition or Subtraction; (2) by Sub stitution; and (3) by Comparison.

Here it will be more convenient to eliminate y.

(1), (2)

To find y, substitute this value of x in either of the given equations. Thus from (1)

and

35+2y=47;
.. y = 6,

x = 5.

To eliminate x we multiply (1) by 5 and (2) by 3, so as to make the coefficients of x in both equations equal. This gives

To find x, substitute this value of y in either of the given equations.

Thus from (1)

3x+2127;

and

.*. x = 2,
y = 3.

In this solution we eliminated x by subtraction.

Rule.

Multiply, when necessary, in such a manner as to make the coefficients of the unknown quantity to be eliminated equal in both equations. Add the resulting equations if these coefficients are unlike in sign; subtract if like in sign.

Transposing - 5 y in (1), and dividing by 2, we obtain

(1),

(2).

This value substituted in either (1) or (2) gives

x = 3.

Rule. From one of the equations, find the value of the unknown quantity to be eliminated in terms of the other and known quantities; then substitute this value for that quantity in the other equation, and reduce.