CHAPTER XX. SIMULTANEOUS QUADRATIC EQUATIONS. 258. Quadratic equations involving two unknown numbers require different methods for their solution, according to the form of the equations. CASE I. 259. When from One of the Equations the Value of One of the Unknown Numbers can be found in Terms of the Other, and this Value substituted in the Other Equation. The solution of which gives x = 1, or x = − 5. y=1-2= -1; (1) (2) The original equations are both satisfied by either pair of values. But the values x=1, y == - 1. the values x = — 5, y = 7, will not satisfy the equations; nor will The student must be careful to join to each value of x the corresponding value of y. CASE II. 260. When the Left Side of Each of the Two Equations is Homogeneous and of the Second Degree. Solve: 2y2-4xy+3x2 = 17) Let y = vx, and substitute vx for y in both equations. From (1), From (2), 2 v2x2 - 4vx2 + 3x2 = 17. Equate the values of x2, 2 v2 - 4v+3 32 v2 - 64 v + 48: 15 v2 - 64 v 17 2 v2 - 4v+3 16. 16 v2 1 v2 17 v2 - 17, = — 65, (1) (2) CASE III. 261. When the Two Equations are Symmetrical with Respect to x and y; that is, when x and y are similarly involved. Thus, the expressions 2x3+3x2y2+2y3, 2xy−3x-3y+1, x1−3x2y—3xy2+y*, are symmetrical expressions. In this case the general rule is to combine the equations in such a manner as to remove the highest powers of x and y. To remove x and y1, raise (2) to the fourth power, Divide by 2, x2 + 2x3y + 3 x2y2 + 2 xy3 + y2 = 1369. Extract the square root, Subtract (3) from (2)2, (1) (2) We now have to solve the two pairs of equations, 262. The preceding cases are general methods for the solution of equations which belong to the kinds referred to; often, however, in the solution of these and other kinds of simultaneous equations involving quadratics, a little ingenuity will suggest some step by which the roots may be found more easily than by the general method. |