Question 378107
1)  Add corresponding sides of the equations: {{{2x^2 = 10}}}, or {{{x^2 = 5}}}.  Substituting this into the top equation, we get {{{y^2 = 4}}}.  Thus there are 4 solutions: ({{{-sqrt(5)}}}, -2), ({{{-sqrt(5)}}}, 2), ({{{sqrt(5)}}}, -2), ({{{sqrt(5)}}}, 2).  (They represent the intersection points of a circle centered at (0,0) with radius 3 and a hyperbola with transverse axis  corresponding to the x-axis.)


2)Multiply the top eq'n by 2, and then add corresponding sides: {{{8x^2 = 72}}}, or {{{x^2 = 9}}}.  Substituting into the lower eq'n, we get {{{16y^3 = -2}}}, or {{{y^3 = -1/8}}}, or y = -1/2.  Thus there are 2 solutions: (-3, -1/2) and (3, -1/2).


3)  Multiply the top eq'n by 2, and then add corresponding sides: 1/y = 3, or y = 1/3.  Substituting into the top eq'n, we get: {{{2/x -9 = 1}}}, or {{{2/x = 10}}}, or x = 1/5.