Examples of Substitution of Unknown Quantities. 6. Solve the two equations 22 y + xy = 6, Ans. X= 1, or = y=2, or = 1. 7. Solve the two equations 96 — qum ya, *+y=6. Ans. 2 = 2, or 4, or 3 + ✓ 21; y=4, or 2, or 3 F 21. 8. Find two numbers such, that their sum and product may together be 34, and the sum of their squares may exceed the sum of the numbers themselves by 42. Ans. 4 and 6; or 3 (-11+v - 59,) and 1 (-11-V-59). 9. What two numbers are they, whose sum is 3, and the sum of whose fourth powers is 17 ? Ans. 2 and 1 ; or 1 (3+V-55), and } (3–V - 55). 10. What two numbers are they, whose product is 3, and the sum of whose fourth powers is 82? Ans. £ 1, and £ 3; or + - 1, and FV-- 9. 240. Corollary. In many cases, in which two unknown quantities enter into the given equations symmetrically except in regard to their signs, the solution is simplified by substituting for them two other unknown quantities, one of which is their difference, and the other is their sum or their product. Examples of Substitution of Unknown Quantities. 241. EXAMPLES. 1. Solve the two equations (x - y) (x2 + y2) = 13, (x - y) x y 6. Solution. These equations become, by the substitution of 2. Solve the two equations 22 - y2 =7, 23 + y2 = 91 (2 — y). x+y=s, 1 (53 +3 st)=91 t. Hence, by the elimination of t, we have $4 - 2401 Examples of Equations of the Second Degree. 4 and S = V 2401 =+7, or = +7V-1; y=+3, or = +4v . 1. 3. Solve the two equations 23 - y3 = 7, Ans. x = 2, or = - - 1; y=1, or = - 2. 4. Solve the two equations 23 - y3 = 215, Ans. x=6, or = =-1; y=1, or=-6. 5. Solve the two equations 23 y - 2 y2 + x y = 156, y=3, or = +4, or = (-12V-51). or a=+fv(-1572+2V(624+1574)) — 781, y=+iv(-1572+27(624+1574))+784. 6. What two numbers are they, whose difference is 1, and the difference of whose third powers is 7 ? Ans. 1 and 2, or -2 and -1. 7. What two numbers are they, whose difference is 3, and the sum of whose fourth powers is 257 ? Ans. 4 and 1, or 4 and -1, or i (+v(-79) + 3) and 1(+v(-79)-3). Examples of Equations of the Second Degree. 242. When the first member of one of the equations, reduced as in art. 118, is homogeneous in regard to two unknown quantities, the solution is often simplified by substituting for the two unknown quantities, two other unknown quantities, one of which is their quotient. The same method of simplification can also be employed when such a homogeneous equation is readily obtained from the given equations. 243. EXAMPLES. 1. Solve the two equations 32 - 6xy + 8y = 0, x2y 62y2 + 8y3 +(-2y) (y2 _5y+4=0. Solution. Retaining the unknown quantity y, introduce instead of t, the unknown quantity 9, such that 2 9 = or 2 =qY; from which the given equations become 69 y2 + 8y2 = 0, q2y3 - 69 y3 + 8y3+(ay-2y)(y2–5y+4)=0. Both these equations are satisfied by the value of y, y=0, whence But if we divide the first of these equations by ys, and the second by y, we have q? – 69+8=0, Examples of Substitution of Unknown Quantities. the first of which gives q= 2, or = 4. The value of q, q=2, being substituted in the other equation, reduces the first member to zero, and therefore y is indeterminate ; that is, J and y may have any values whatever, with the limitation that x is the double of y. The value of q, 9 = 4. being substituted in the other equation, gives 2 (y2 — 5y + 4)=0; whence y=1, or = 4, and = 4, or = 16. 2. Solve the two equations 25 + x3y2 = 5, 25 + 4 x y* = 65. Solution. 13 times the first equation, diminished by the second equation, is 12 25 + 13 x y — 4 x y* = 0; and, if we make 2 =9Y, we have 12 q” y + 13 93 y - 4975 = 0. Which is satisfied by the value of y, y = 0; and this value of y, being substituted in the given equations, produces 25 = 5, 25 = 65; which are evident impossibilities, and therefore the value y=0 is impossible. |