Question 139463
Start with the given system

{{{y=x^2-4}}}
{{{x-y=-2}}}



{{{x=y-2}}} Solve for x in the second equation



{{{y=(y-2)^2-4}}} Plug in {{{x=y-2}}}



{{{y=y^2-4y+4-4}}} Foil



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



{{{0=y^2-5y}}} Combine like terms



{{{0=y(y-5)}}} Factor the right side





Now set each factor equal to zero:

{{{y=0}}} or  {{{y-5=0}}} 


{{{y=0}}} or  {{{y=5}}}    Now solve for y in each case



So our y-values are


 {{{y=0}}} or  {{{y=5}}} 



Let's find x when {{{y=0}}}


{{{x=y-2}}} Start with the second equation



{{{x=0-2}}} Plug in {{{y=0}}}



{{{x=-2}}} Subtract



So when {{{x=-2}}} then {{{y=0}}}




Let's find x when {{{y=5}}}


{{{x=y-2}}} Start with the second equation



{{{x=5-2}}} Plug in {{{y=5}}}



{{{x=3}}} Subtract



So when {{{x=3}}} then {{{y=5}}}





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Answer:


So the solutions are


(-2,0) or (3,5)