Part 24

21. The glass of a mirror is 18 inches by 12 inches, and it has a frame of uniform width whose area is equal to that of the glass. Find the width of the frame.

OUTLINE OF SIMULTANEOUS QUADRATICS

~Simultaneous Quadratics~

CASE I.

One equation linear. The other quadratic. 2x + y = 7, x^2 + 2y^2 = 22.

METHOD: Solve for x as in terms of y, or vice versa, in the linear and substitute in the quadratic.

CASE II.

Both equations homogeneous and of the second degree. x^2 - xy + y^2 = 39, 2x^2 - 3xy + 2y^2 = 43.

METHOD: Let y = vx, and substitute in both equations.

ALTERNATE METHOD: Solve for x in terms of y in one equation and substitute in the other.

CASE III.

Any two of the quantities x + y x^2 + y^2 xy x - y x^3 + y^3 x^3 - y^3 x^2 + xy + y^2 x^2 - xy + y^2 given.

x + y = 5, x^2 - xy + y^2 = 7.

METHOD: Solve for x + y and x - y; then add to get x, subtract to get y.

CASE IV.

Both equations symmetrical or symmetrical except for sign. Usually one equation of high degree, the other of the first degree. x^5 + y^5 = 242, x + y = 2.

METHOD: Let x = u + v and y = u - v, and substitute in both equations.

~Special Devices~

I. Consider some compound quantity like xy, [x - y]^(1/2), ^(1/2), x/y, etc., as the unknown, at first. Solve for the compound unknown, and combine the resulting equation with the simpler original equation.

x^2 y^2 + xy = 6, x + 2y = -5.

II. Divide the equations member by member. Then solve by Case I, II, or III.

x^3 - y^3 = 152, x - y = 2.

III. Eliminate the quadratic terms. Then solve by Case I, II, or III.

xy + x = 15, xy + y = 16.

SIMULTANEOUS QUADRATICS

Solve:

1. x + y = 7, x^2 + 4xy = 57.

2. 2x^2 = 46 + y^2, xy + y^2 = 14.