SOLUTION: This one really has me stumped. Find the exact solution(s)of the system: x^2/4-y^2=1 and x=y^2+1 a. (4,√3),(4,-√3),(-4,√3),(-4,-√3) b. (4,√3),

Algebra ->  Equations -> SOLUTION: This one really has me stumped. Find the exact solution(s)of the system: x^2/4-y^2=1 and x=y^2+1 a. (4,√3),(4,-√3),(-4,√3),(-4,-√3) b. (4,√3),      Log On


   

Question 99123: This one really has me stumped.
Find the exact solution(s)of the system: x^2/4-y^2=1 and x=y^2+1
a. (4,√3),(4,-√3),(-4,√3),(-4,-√3)
b. (4,√3),(-4,√3)
c. (2,1),(2,-1),(4,√3),(4,-√3)
d. (4,√3),(4,-√3)
Thank you.
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
x=y%5E2%2B1 Start with the second equation


x-1=y%5E2 Subtract 1 from both sides


x%5E2%2F4-y%5E2=1+ Now move onto the second equation


x%5E2%2F4-%28x-1%29=1+ Replace y%5E2 with x-1

x%5E2%2F4=1%2B%28x-1%29+ Add x-1 to both sides


x%5E2%2F4=x+ Combine like terms


x%5E2=4x+ Multiply both sides by 4


x%5E2-4x=0+ Subtract 4x from both sides



x%28x-4%29=0 Factor the left side (note: if you need help with factoring, check out this solver)



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%5E2%2B1 Start with the second equation


0=y%5E2%2B1 Plug in x=0


-1=y%5E2 Subtract 1 from both sides


sqrt%28-1%29=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



---------------------------------------------------------------------------

Let's find y when x=4
x=y%5E2%2B1 Start with the second equation


4=y%5E2%2B1 Plug in x=4


3=y%5E2 Subtract 1 from both sides


0%2B-sqrt%283%29=y Take the square root of both sides. So when x=4, y=0%2B-sqrt%283%29 which means y=sqrt%283%29 or y=-sqrt%283%29


So our solution is



(4,sqrt%283%29), (4,-sqrt%283%29)


which means the answer is D