4 years ago

PLEASE HELPPPP WITH MATHH

Mathematics

1 answer:

Svetllana [295]4 years ago

6 0

Answer:

im pretty sure it is A

Step-by-step explanation:

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Answer:

Solutions are (6,3) and (-2,3).

Step-by-step explanation:

From the first equation given we can write:

$(x-2)^{2} = 16-(y-3)^{2}\\$

substituting for $(x-2)^{2}$ in the second equation given we get

$\frac{16-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} = 1$

$\frac{-(y-3)^{2}}{16} + \frac{(y-3)^{2}}{64} =0$

∴ y=3

Putting y=3 in the first equation we get

$(x-2)^{2} = 16$

x-2 = ±4

Hence x=6 or x=-2.

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