By operating in the same way on equations IV the value of y can be found, and this value of y will be the result found by changing, in the value of x, a into b and b into a, a' into b' and b' into a'. This determination gives By a discussion in Algebra is meant the examination of the various cases about which any question can be raised. The discussion now arises of the principal circumstances presented by the solution of equations (1) and (2) of the first degree in two unknown quantities. 232. Case when ab' ba' is different from 0. -It is clear that every system of values of x and y which satisfies the two equations (1) and (2) of system I satisfies also the two equations (3) and (4) of system II, which were derived as shown in 228; but, these equations (3) and (4) are satisfied in accordance with 2212, 2, only by the values of x and y given by the formulae (5) and (6) of system III; therefore the proposed equations have but one solution. To show that these values satisfy the two given equations, substitute for x and y the values given by formulae (5) and (6) of system III in the first member ax + by of (1) and arrange the result with respect to c and c', thus: Since the coefficient of c' is zero, and that of c is equal to the denominator, the above quantity is equal to the second member c and the equation is satisfied. The first member a'x+b'y of the second equation becomes, in like manner, The coefficient of c is zero and the coefficient of c' is equal to the denominator, and this quantity is equal to the second member c' of equation (2); and the equation is therefore satisfied. THEOREM. When the denominator ab' — a'b is different from zero, the two given equations have one solution and only one. Since ab'a'b 5 11-7 (-3)=760, the two equations have the solutions x = 3 and y = 2 given by the formulae (5) and (6). 233. Case when ab' - a'b = 0. Suppose now that the denominator ab' a'b = 0 and that one of the numerators, for example, cb' — bc'=0. If the two given equations are satisfied by the values of x and y, then the equation, x (ab' — a'b) = cb' — be', is satisfied, but, since ab'a'b 0,x; therefore the given system has no solution and the equations are said to be incompatible. EXAMPLE. Apply the theory to equations (1) and (5) of 209: But it is easy to show from another point of view that the equations are impossible. Since cb'-be'=/=0, then one at least of the coefficients, and b', is different from 0. from 0; hence from the condition ab' Suppose that b is different a'b = 0, it follows that ; this value being substituted in equation (2), the equation If the coefficient b' is zero, the first member is zero and the second member c'b is different from zero, since cl' c'b=0 by hypothesis. This is impossible. If b' is not zero it is possible to write b' (ax+by) = c'b but the first member of equation (1) is also ax + by, which would which is not allowed. Hence the equations are incompatible. Put ab'a'b = 0 in formulae (5) and (6) of III; then in case a = 0. But, if a = = 0, since the binomial ab' — a'b it follows that 0 y= which is indeterminate. 234. The case when ab' - a'b 0, cb' - bc' = 0, ac' — ca'=0. It has been seen that the second equation becomes Suppose that at least one of the four coefficients a, b, a', b', is not zero, say b; consider the equation in the form be b then, from the relation cb' — be' = 0, c = and equation (2) becomes identical with equation (1). If the coefficient b' 0, then also c' = 0, since cb'be' = 0; and b = 0, and equation becomes an identity b' (ax+by) = be' The system is then reduced to the first equation, which can be satisfied by an infinite number of values of x and y. If the four coefficients, a, b, a', b', were zero, the given equa tions would become 0: = c, 0 = c', which would be impossible unless c and c' were zero; and in that case, the equations would become 0 = 0, an equation which is absolutely indeterminate. Any values whatever might be assigned to x and y and the equations would be satisfied. 235. Résumé of the Discussion. For the sake of brevity put D = ab' — a'b, Nx = cb' — c'b, Ny = ac' - a'c. The following table gives a résumé of the discussion of the system. y = D incompatible equations ( indeterminateness: one arbitrary 236. When the second members of system I become zero, i. e., c and c' equal zero, the system becomes ax + by Since all the terms in both equations are of the first degree in r and y, they are called homogeneous equations. If ab'a'b is not zero, then they have the solution, x = 0, y = 0, but if ab'-a'b=0, then they have an infinity of solutions, since by (5) and (6), III, Hence, the two equations are satisfied by a single infinite set of values of x and y. If a and a' are zero, there is still an infinity of solutions, since the values of x and y are indeterminate. From the equations, by = 0, b'y = 0, b =⇒ 0, l' ÷ 0, y can take only the value zero, but x is arbitrary; if b and b' are zero, and a and a' are not both zero, then x can take only the value y is arbitrary. zero, but THE CONDITION THAT Two EQUATIONS OF HAVE A COMMON ROOT THE FIRST DEGREE 237. Let the two equations in one unknown quantity be Suppose that one at least, say a, of the coefficients is different from zero. Then the first equation has the unique root, x = b a In order that the two equations may have a common root it is b necessary and sufficient that the second equation have the root a If both coefficients a and a' are zero at the same time, then ab' — a'b = 0; but, if b or b' is not zero, then one of the equations could not have a finite root; and therefore, in this case, the given equations could not have a finite root in common. |