CHAPTER XII. MISCELLANEOUS THEOREMS AND EXAMPLES. 153. Elimination. When more equations are given than are necessary to determine the values of the unknown quantities, the constants in the equations must be connected by one or more relations, and it is often of importance to determine these relations. Since the relations required are not to contain any of the unknown quantities, what we have to do is to eliminate all the unknown quantities from the given system. The following are some examples of Elimination: Ex. 1. Eliminate x from the equations ax+b=0, a'x+b'=0. b From the first equation we have x=- and from the second a From the first two equations we have [Art. 143] These values of x and y must satisfy the third equation; х hence or a" (bc' - b'c)+b" (ca′ – c'a)+c" (ab' – a′b) =0, the required result. The general case of the elimination of n-1 unknown quantities from n equations of the first degree will be considered in the Chapter on Determinants. It should be remarked that the above condition is also the condition that the two expressions ax2+ bx + c and a'x2 + b'x+c' may have a common factor of the form x-a; for if the expressions have a common factor of the form x-a they must both vanish for the same value of x. Ex. 5. Eliminate x from the equations ax2+ bx+c=0...... a'x3 + b2x2+ c'x+d' = 0 Multiply (i) by a'x, (ii) by a, and subtract; then, (ab' - ba') x2+(ac' - ca') x + ad' = 0........... We can now eliminate x from (i) and (iii) as in Ex. 3. .(i), .(ii). .(iii). By continued multiplication of the corresponding members of the three equations, we have =2abc+a (4f2 - 2bc) +b (4g2 – 2ca) +c (4h2 – 2ab). a2 y2 Hence abc+2fgh- af 2 - bg2 - ch2=0, the required relation. 154. Equations in which there is some restriction on the values of the letters. A single equation which contains two or more unknown quantities can be satisfied by an indefinite number of values of the unknown quantities, provided that these values are not in any way restricted. If however the values of the unknown quantities are subject to any restriction, a single equation may be sufficient to determine more than one unknown quantity. For example, if we have the single equation 2x+5y=7, and restrict both x and y to positive integral values, the equation can only be satisfied by one set of values, namely by the values = 1, y = 1. Again, from the single equation 3 (x − a)2 + 4 (y — b)2 = 0, with the restriction that all the quantities must be real, we can conclude both that x a = 0, and that y-b=0; for the squares of real quantities must be positive, and the sum of two or more positive quantities cannot be zero unless they are all zero. Ex. 1. If (a+b+c)2 = 3 (bc+ca+ab), then a=b=c. We have a2+b2+c2-bc - ca - ab=0, that is } { (b − c)2 + (c − a)2 + (a − b)2} = 0. Whence b-c, c − a and a − b must all be zero. Ex. 2. If x, x', y, y' be all real, and 2 (x2 + x22 — xx') (y2+y'2 — yy') = x2y2+x12y'2 ; then will x=x' and y=y'. We have `x2 (y2+2y12 — 2yy') — 2xx' (y2+ y22 — yy')+x22 (2y2+y'2 — 2yy') = 0 ; that is .. (x2 – 2xx' +x'2) (y2 − 2yy' +y'2) + x2y'2 — 2xx'yy' + x'2y2=0, (x-x') (y-y')2+(xy' x'y)2=0. From the second relation x=x' or y=y'; and either of these combined with the first relation shews that both x=x' and y=y'. Multiply the equations in order by q2, p2 and -2pq respectively, and add; we then have (qa1-pb1)2 + (qa, − pb2)2+(qa, − pb3)2+ ......=0. Hence qα-pb1=0=qa2-pb2=qaz - pbz = ......&c. 155. We have already proved that a3 +b3 + c3 − 3abc = (a + b + c) (a2 + b2 + c2 − bc − ca — ab) = (a+b+c) {(b − c)2 + (c − a)2+(a−b)2} = (a+b+c) (a + wb + w3c) (a + w2b+wc), |