Examples of Equations of the Second Degree.

242. When the first member of one of the equations, reduced as in art. 118, is homogeneous in regard to two unknown quantities, the solution is often simplified by substituting for the two unknown quantities, two other unknown quantities, one of which is their quotient.

The same method of simplification can also be employed when such a homogeneous equation is readily obtained from the given equations.

x2y+6x y2+8 y3+ (x−2y) (y2—5y+4=0.

Solution. Retaining the unknown quantity y, introduce instead of x, the unknown quantity q, such that

from which the given equations become

q2 y2 — 6 q y2+8 y2 = 0,

.

q2 y3 — 6 q y3 + 8 y3+(qy—2y) (y2—5y+4)=0.

Both these equations are satisfied by the value of y,

But if we divide the first of these equations by y2, and the second by y, we have

q26q8=0,

q2 y2 — 6 q y2+8 y2+(q−2) (y2 —5y+4)=0;

being substituted in the other equation, reduces the first member to zero, and therefore y is indeterminate; that is, x and y may have any values whatever, with the limitation that x is the double of y.

Solution. 13 times the first equation, diminished by the

second equation, is

12x513 x3 y2 — 4 x y1 = 0;

and, if we make

we have

x = qy,

12 q5 y5 + 13 q3 y5 — 4 q y5 — 0.

Which is satisfied by the value of y,

y= = 0;

and this value of y, being substituted in the given equations produces

x5

5,

x5 = 65;

which are evident impossibilities, and therefore the value y=0 is impossible.

Examples of Substitution of Unknown Quantities.

Dividing, then, by y5, we have

12 q5 13 q3-4q=0;

Now the first of the given equations becomes, by the substitution of

x= qy,

q5 y5+q3 y5—5;

hence, by the substitution of the above values of q, we have

(y2—y)2+(3 x y+2y)2—9 x2(2y+3)—12 y (x+2y)=0.

or x=

(-5-5,) and y = 1±v-5;

or x=-5,

4. Solve the two equations

x3 + 2 x y2 = 3,

and y = 1.

x y2 + 2 x2 y = 3.

Ans. x 1, and y = 1.

Examples of Substitution of Unknown Quantities.

5. What two numbers are they, twice the sum of whose squares is 5 times their product, and the sum of whose sixth powers is 65. Ans. 2 and 1, or -2 and -1.

6. What two numbers are they, the difference of whose fourth powers is 65, and the square of the sum of whose squares is 169. Ans. 2, and ± 3.

To find the last Term.

CHAPTER VII.

PROGRESSIONS.

SECTION I.

Arithmetical Progression.

244. An Arithmetical Progression, or a progression by differences, is a series of terms or quantities which continually increase or decrease by a constant quantity.

This constant increment or decrement is called the common difference of the progression.

Throughout this section the following notation will be retained. We shall use

a = the first term of the progression,

7 the last term,

r = the common difference,

n = the number of terms,

S the sum of all the terms.

245. Problem. To find the las. ierm of an arithmetical progression when its first term, common difference, and number of terms are known.

Solution. In this case a, r, and n, are supposed to be known, and 7 is to be found. Now the successive terms of the series if it is increasing are

a, a+r, a+2r, a+3r, a+4r, &c.;