Question 200162
{{{x^2-y^2=1/2}}} Start with the second equation.



{{{x^2=1/2+y^2}}} Add {{{y^2}}} to both sides.



{{{2x^2+4y=11}}} Move back to the first equation.



{{{2(1/2+y^2)+4y=11}}} Plug in {{{x^2=1/2+y^2}}}



{{{2+2y^2+4y=11}}} Distribute



{{{2+2y^2+4y-11=0}}} Subtract 11 from both sides.



{{{2y^2+4y-9=0}}} Combine like terms.



Notice that the quadratic {{{2y^2+4y-9}}} is in the form of {{{Ay^2+By+C}}} where {{{A=2}}}, {{{B=4}}}, and {{{C=-9}}}



Let's use the quadratic formula to solve for "y":



{{{y = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{y = (-(4) +- sqrt( (4)^2-4(2)(-9) ))/(2(2))}}} Plug in  {{{A=2}}}, {{{B=4}}}, and {{{C=-9}}}



{{{y = (-4 +- sqrt( 16-4(2)(-9) ))/(2(2))}}} Square {{{4}}} to get {{{16}}}. 



{{{y = (-4 +- sqrt( 16--72 ))/(2(2))}}} Multiply {{{4(2)(-9)}}} to get {{{-72}}}



{{{y = (-4 +- sqrt( 16+72 ))/(2(2))}}} Rewrite {{{sqrt(16--72)}}} as {{{sqrt(16+72)}}}



{{{y = (-4 +- sqrt( 88 ))/(2(2))}}} Add {{{16}}} to {{{72}}} to get {{{88}}}



{{{y = (-4 +- sqrt( 88 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{y = (-4 +- 2*sqrt(22))/(4)}}} Simplify the square root 



{{{y = (-4+2*sqrt(22))/(4)}}} or {{{y = (-4-2*sqrt(22))/(4)}}} Break up the expression.  



{{{y = (-2+sqrt(22))/(2)}}} or {{{y = (-2-sqrt(22))/(2)}}} Reduce



So the solutions (for "y") are {{{y = (-2+sqrt(22))/(2)}}} or {{{y = (-2-sqrt(22))/(2)}}}



which approximate to {{{y=1.35}}} or {{{y=-3.35}}} 



Now plug these values into {{{x^2=1/2+y^2}}} to find the corresponding "x" values. 


Note: You should get four ordered pair solutions (there are two x values for each y value you just found).