Question 135530
Start with the given system

{{{y=x^2-4}}}
{{{y=6x-13}}}



{{{x^2-4=6x-13}}} Plug in {{{y=x^2-4}}} into the second equation



{{{x^2-4-6x+13=0}}} Get all terms to left side



{{{x^2-6x+9=0}}} Combine like terms



{{{(x-3)(x-3)=0}}} Factor the left side (note: if you need help with factoring, check out this <a href=http://www.algebra.com/algebra/homework/playground/change-this-name4450.solver>solver</a>)




Now set each factor equal to zero:

{{{x-3=0}}} or  {{{x-3=0}}} 


{{{x=3}}} or  {{{x=3}}}    Now solve for x in each case



Since we have a repeating answer, our only answer is {{{x=3}}}



{{{y=6x-13}}} Go back to the second equation



{{{y=6(3)-13}}} Plug in {{{x=3}}} 



{{{y=5}}} Simplify



Answer:


So the solution is (3,5)