and taking the square root, (3.) + (1.), then, 2 x2

x2 - 2x2y2 + 3*
y*
4 x2y2

x2 + 2 x2y2 + y*
x2 + y2

=

= +

8, or x2 9 or

=

and .. x = ± 3 or ± 2√√√1.

(3.) - (1.), then, 2 y2 = 8 or 18, or y = 4 or 9,

-

andy2 or ± 3√

I.

NOTE. The student will see that the pairs of values which satisfy the given equations are, x = 3, y = 2; x = - 3, y = −2; x = 2 √ − 1, y = 3√ = 1; x = 2√ - 1, y =

Now (xy) is a factor of each side, and hence, striking it

Equations (3.) and (4.) may not be used as simultaneous equations, but each of them may be used in turn with either of the given equations.

Thus, taking equations (3.) and (1.), we have—

and hence from (3.), by substitution, we easily get

Hence, the pairs of values satisfying the given equations

NOTE. It is worth while remarking that when each of the terms of the given equations contain at least one of the unknown quantities, the values x = 0, y O will always satisfy.

=

or, transposing, 34 x or, (2 x

275 xy + 261 xy + 5 y) (17 x

=

440 y2 21 22
440 y2 = 0;
88 y)

14 xy;

= 0, 88 y.

=

from which 2 x = 5 y, and 17 x =

Each of these equations taken in turn with either (1.) or (2) will easily give the required values of x and y.

Ex. 5. x2 + y2

=

337....
7

.(1.),

(2.)

From (2), raising each side to the fourth power, we have—

x2 + 4x3y + 6 x2y2 + 4 xy3 + y2 (3)(1), then 4x3y + 6xy + 4 xy3

or, 2 x3y + 3x2y2 + 2 xy3

From (2) and (4), x - y may now be easily obtained, and hence also the required values of x and y.

30. x4 - 2x22 + y2 = 1 + 4 xy, x2(x + 1) + y2(y + 1) 24

A

48. Given R = 1 + r, P = 4, (1 − R−"), M = PR",

Problems Producing Quadratic Equations.

7. We shall now discuss one or two problems whose solutions depend upon quadratic equations.

Ex. 1. A person raised his goods a certain rate per cent., and found that to bring them back to the original price he must lower them 31 less per cent. than he had raised them. Find the original rise per cent.

Let x the original rise per cent.,

=

The value x = 20 is alone applicable to the problem. Remembering, however, the algebraical meaning of the negative sign, it is easy to see that the second value, x gives us the solution of the following problem:

=

16,

A person lowered his goods a certain rate per cent., and found that to bring them back to the original price he must raise them 3 more per cent. than he had lowered them. Find the original fall per cent.

The above solution tells us that the fall required is 163 per cent.

=

Had we worked the latter problem first, we should have obtained x 16% or 20, the value x - 20 indicating the solution of the former problem.

=