The value of x will be easily found to be ±2, or 14. Find the values of x and y in the Equations 3y2x2-39, and x2+4xy=256-4y2. Substituting vy for x in these Equations 3y2-v2y2-39, and v2y2+4vy2=256-4y2. Finding the values of y2 from these Equations, and equating them, we have = 3-v2 v2+4v+4 Clearing this Equation of fractions, 39v2+156v+156-768-256v2. 15. Find the values of x and y in the Equations Assuming xv+z, and y=v-z, we shall have (x+y)=(v+2)+(v-z)=2v=12. Substituting these values of x and y in the third Equation, we have (6+z)3+(6—2)3—18(6+z) (6—z). Developing both sides of this Equation, and reducing, 54z2-216; hence z=±2. Then since x=6+z, we have x=6±2=8, or 4. 16. Find the values of x and y in the Equations Assuming xv+z, and y=v—z, we have x+y=2v=10. Hence v=5. Then x=5+z, and y=5-z. Substituting these values of x and y in the second Equation, (5+)3+(5-2)3=280. Developing the first member of this Equation, and reducing, 302230; hence z=±1. Since x=6+z, we have x=5±1=6, or 4. Since y=6-2, we have y=51=4, or 6. 17. Find the values of x and y in the Equations x+y=11, and x1+y1=2657. Substituting v+z for x, and v-z for y in the first Equation, and uniting similar terms, we find Substituting these values of x and y in the second Equation, Developing each term of the first member, of this Equation, by the method of the Binomial Theorem, we have Adding these two Equations together, and recurring to the preceding Equation, we find This Equation, by developing the two binomials in the first member, becomes Clearing this Equation of fractions, 29282+580822+32442512. From this Equation, omitting imaginary roots, we shall have 18. Find the values of x and y in the Equations, x+y=10, and x5+y5=17050. Substituting v+z for x, and v-z for y in the first Equation, and Therefore x5+2, and y=5—z. Substituting these values of x and y in the second Equation, have (5+2)+(5-2)=17050. Developing each binomial in the first member of this Equation, we (5+z)5=3125+3125z+1250z2+250z3+25z4+25. Adding these Equations together, (5+2)+(5-2)56250+2500z2+5024. Hence we shall have 6250+2500z2+50z4-17050. From this Equation, omitting imaginary values, we shall finu z=±2. Hence x=5±2=7, or 3; and y=52=3, or 7. Miscellaneous Solutions and Exercises. 19. Find the values of x and y in the Equations xy=6, and x2+x=18-y2—y. From the second Equation, we have x2+y2+x+y=18. Adding twice the first Equation, x2+2xy+y2+x+y=30. This Equation may be expressed by (x+y)2+(x+y)=30; which is quadratic with reference to x+y, and from which we shall find x+y=5; or x+y=-6; hence x=5—y; or x——6—y. Substituting these values of x, successively, in the first Equation, we have The first of these Equations will give y=3, or 2; and the second will give y=−3±√√3. Since x=5-y; or x= -6—y; we have x=5-3=2, x=5—2 =3, or x=—6—(−3±√/3)=−3+√/3. 20. Find the values of x and y in the Equations x+y=6, and x2y2+4xy=96. The second Equation may be expressed by (xy)2+4(xy)=96. This Equation is quadratic with reference to xy, and will give Substituting the first value of x in the first Equation, By substituting each of these values of y for y in the first Equation we shall find the corresponding values of x to be 2, 4, and 3 √21. 21. Find the values of x and y in the Equations x3ya—2y2, and 8x*—y*14. Dividing the first Equation by y3, we find x+2y1; whence y1 = Substituting this value of y in the second Equation, Substituting each of the values of 23 in the second Equation, 112-y=14, and 16-y-14. From the first of these Equations, 98; hence y=9604. |