Question 254911
{{{x^2+2y^2=22}}} Start with the second equation.
 


{{{x^2+2(1+2x^2)=22}}} Plug in {{{y^2=1+2x^2}}}



{{{x^2+2+4x^2=22}}} Distribute



{{{x^2+2+4x^2-22=0}}} Subtract 22 from both sides.



{{{5x^2-20=0}}} Combine like terms.



Notice that the quadratic {{{5x^2-20}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=5}}}, {{{B=0}}}, and {{{C=-20}}}



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



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



{{{x = (0 +- sqrt( (0)^2-4(5)(-20) ))/(2(5))}}} Plug in  {{{A=5}}}, {{{B=0}}}, and {{{C=-20}}}



{{{x = (0 +- sqrt( 0-4(5)(-20) ))/(2(5))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (0 +- sqrt( 0--400 ))/(2(5))}}} Multiply {{{4(5)(-20)}}} to get {{{-400}}}



{{{x = (0 +- sqrt( 0+400 ))/(2(5))}}} Rewrite {{{sqrt(0--400)}}} as {{{sqrt(0+400)}}}



{{{x = (0 +- sqrt( 400 ))/(2(5))}}} Add {{{0}}} to {{{400}}} to get {{{400}}}



{{{x = (0 +- sqrt( 400 ))/(10)}}} Multiply {{{2}}} and {{{5}}} to get {{{10}}}. 



{{{x = (0 +- 20)/(10)}}} Take the square root of {{{400}}} to get {{{20}}}. 



{{{x = (0 + 20)/(10)}}} or {{{x = (-0 - 20)/(10)}}} Break up the expression. 



{{{x = (20)/(10)}}} or {{{x =  (-20)/(10)}}} Combine like terms. 



{{{x = 2}}} or {{{x = -2}}} Simplify. 



So the solutions for 'x' are {{{x = 2}}} or {{{x = -2}}} 



Now take each solution for 'x' and plug them into {{{y^2=1+2x^2}}} to find the corresponding 'y' solutions.



Let's find the corresponding solutions for y when {{{x=2}}}



{{{y^2=1+2x^2}}} Start with the first equation.



{{{y^2=1+2(2)^2}}} Plug in {{{x=2}}} (the first 'x' solution)



{{{y^2=1+2(4)}}} Square 2 to get 4.



{{{y^2=1+8}}} Multiply



{{{y^2=9}}} Add



{{{y=""+-sqrt(9)}}} Take the square root of both sides (don't forget the plus/minus)



{{{y=sqrt(9)}}} or {{{y=-sqrt(9)}}} Break up the plus/minus



{{{y=3}}} or {{{y=-3}}} Take the square root of 9 to get 3.



So when {{{x=2}}}, we have two solutions for 'y' which are {{{y=3}}} or {{{y=-3}}}. So we have two ordered pairs (2,3) and (2,-3)


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Now let's find the corresponding solutions for y when {{{x=-2}}}



{{{y^2=1+2x^2}}} Start with the first equation.



{{{y^2=1+2(2)^2}}} Plug in {{{x=-2}}} (the second 'x' solution)



{{{y^2=1+2(4)}}} Square 2 to get 4.



{{{y^2=1+8}}} Multiply



{{{y^2=9}}} Add



{{{y=""+-sqrt(9)}}} Take the square root of both sides (don't forget the plus/minus)



{{{y=sqrt(9)}}} or {{{y=-sqrt(9)}}} Break up the plus/minus



{{{y=3}}} or {{{y=-3}}} Take the square root of 9 to get 3.



So when {{{x=-2}}}, we have two solutions for 'y' which are {{{y=3}}} or {{{y=-3}}}. So we have two more ordered pairs (-2,3) and (-2,-3)


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Answer:



So the four ordered pair solutions are (2,3), (2,-3), (-2,3), and (-2,-3) which means that the answer is B)