Let us solve this system, first graphically, then algebraically. The equation x-y=1 is linear and is represented by a straight line. The graph of x2+ y2= 25 is a circle. Fig. 20 shows the graph of x2 + y2 = 25, (1), and x line (2) intersects the circle in two points (-3, — 4) and (4, 3). The coördinates of these points satisfy both equations and are therefore the values sought; namely, 140. Observe that x and y, when subject to the one condition x2 + y2 = 25, are variables, because they are capable of taking on successively a never ending series of different values. For the same reason x and y are also variables when subject to the one condition x y = 1. But x and y can no longer assume successively a never ending series of different values, when they are subject at the same time to both conditions, Hence x and y are now no longer variables, x2 + y2 = 25 y = : 1 but constants. It is the purpose of the solution to find the values of these constants. In drawing graphs we are dealing with variables; in finding the points of intersection of these graphs, we are finding the values of constants which, at the outset, are unknown. 141. The algebraic solution of x2 + y2 = 25 x-y=1 can be effected easiest by the method of substitution. (1) (2) In checking, substitute the values of x and y in both of the given equations; the answers might be wrong, yet might satisfy one of the two equations. For instance, x = 3, y = 2 will satisfy (2), but not (1). If in Fig. 20 the line should move, parallel to its present position, until it were tangent to the circle, we would have just one set of values satisfying the pair of equations. Show that this is the case, if the equations are x2 + y2 = 25 and x - y = 5√2. If the line should not intersect the circle, nor be tangent to it, we would have a set of imaginary values satisfying both equations. Show that this is the case, if the equations are x2 + y2 = 25, x − y = 10. 142. Instead of elimination by substitution, some special device is frequently used. |