Answer:
Solutions are (6,3) and (-2,3).
Step-by-step explanation:
From the first equation given we can write:
[tex](x-2)^{2} = 16-(y-3)^{2}\\[/tex]
substituting for [tex](x-2)^{2}[/tex] in the second equation given we get
[tex]\frac{16-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} = 1[/tex]
[tex]\frac{-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} =0[/tex]
∴ y=3
Putting y=3 in the first equation we get
[tex](x-2)^{2} = 16[/tex]
x-2 = ±4
Hence x=6 or x=-2.
