Question 99123
{{{x=y^2+1}}} Start with the second equation



{{{x-1=y^2}}} Subtract 1 from both sides



{{{x^2/4-y^2=1 }}} Now move onto the second equation



{{{x^2/4-(x-1)=1 }}} Replace {{{y^2}}} with {{{x-1}}}


{{{x^2/4=1+(x-1) }}} Add {{{x-1}}} to both sides



{{{x^2/4=x }}} Combine like terms



{{{x^2=4x }}} Multiply both sides by 4



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




{{{x(x-4)=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=0}}} or {{{x-4=0}}}



{{{x=0}}} or {{{x=4}}} Now solve for x in each case



Now let's use these two x-values to find the y-values



Let's find y when x=0

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



{{{0=y^2+1}}} Plug in x=0



{{{-1=y^2}}} Subtract 1 from both sides



{{{sqrt(-1)=y}}} Take the square root of both sides. Since you cannot take the square root of a negative number, x=0 is not in the solution




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Let's find y when x=4

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



{{{4=y^2+1}}} Plug in x=4



{{{3=y^2}}} Subtract 1 from both sides



{{{0+-sqrt(3)=y}}} Take the square root of both sides. So when x=4, {{{y=0+-sqrt(3)}}} which means {{{y=sqrt(3)}}} or {{{y=-sqrt(3)}}} 



So our solution is 




(4,{{{sqrt(3)}}}), (4,{{{-sqrt(3)}}})



which means the answer is D