Question 200163
Start with the given system

{{{system(x^2-y^2=25,2x^2-y^2=x+55)}}}



Multiply both sides of the first equation by -1


{{{system(-1(x^2-y^2)=-1(25),2x^2-y^2=x+55)}}}



Distribute and multiply


{{{system(-x^2+y^2=-25,2x^2-y^2=x+55)}}}




Now add the two equations



{{{(-x^2+y^2)+(2x^2-y^2)=-25+x+55}}}



{{{(-x^2+2x^2)+(y^2-y^2)=x+(-25+55)}}} Group like terms.



{{{x^2=x+30}}} Combine like terms.



{{{x^2-x-30=0}}} Get all terms to the left side.



Notice that the quadratic {{{x^2-x-30}}} is in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-1}}}, and {{{C=-30}}}



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



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



{{{x = (-(-1) +- sqrt( (-1)^2-4(1)(-30) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-1}}}, and {{{C=-30}}}



{{{x = (1 +- sqrt( (-1)^2-4(1)(-30) ))/(2(1))}}} Negate {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1-4(1)(-30) ))/(2(1))}}} Square {{{-1}}} to get {{{1}}}. 



{{{x = (1 +- sqrt( 1--120 ))/(2(1))}}} Multiply {{{4(1)(-30)}}} to get {{{-120}}}



{{{x = (1 +- sqrt( 1+120 ))/(2(1))}}} Rewrite {{{sqrt(1--120)}}} as {{{sqrt(1+120)}}}



{{{x = (1 +- sqrt( 121 ))/(2(1))}}} Add {{{1}}} to {{{120}}} to get {{{121}}}



{{{x = (1 +- sqrt( 121 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (1 +- 11)/(2)}}} Take the square root of {{{121}}} to get {{{11}}}. 



{{{x = (1 + 11)/(2)}}} or {{{x = (1 - 11)/(2)}}} Break up the expression. 



{{{x = (12)/(2)}}} or {{{x =  (-10)/(2)}}} Combine like terms. 



{{{x = 6}}} or {{{x = -5}}} Simplify. 



So the solutions (for the x values) are {{{x = 6}}} or {{{x = -5}}}



Now take each x value and plug them into either equation to solve for "y". Let me know if you need more help from here.



Note: In this case, you should get 4 ordered. However, one solution is repeated twice. So there are really only 3 ordered pair solutions.