Question

What are the solutions to the nonlinear system of equations below?
Check all that apply.

Answer 1

(-4,0) and (4,0)

Option C and Option F are the right options.

solution,

`{x}^(2) + {y}^(2) = 16......(equation \: i) \\ \frac{ {x}^(2) }{ {4}^(2) } - \frac{ {y}^(2) }{ {4}^(2) } = 16......( \: equation \: ii) \\ from \: equation \: ii \\ \frac{ {x}^(2) }{ {4}^(2) } - \frac{ {y}^(2) }{ {4}^(2) } = 1 \\ \frac{ {x}^(2) - {y}^(2) }{ {4}^(2) } = 1 \\ \frac{ {x}^(2) - {y}^(2) }{16} = 1 \\ {x}^(2) - {y}^(2) = 16......... \: equation \: iii \\ now \: adding \: equation \: i \: and \: ii \\ {x}^(2) + {y}^(2) = 16 \\ {x}^(2) - {y}^(2) = 16 \\ ...................... \\ 2 {x}^(2) = 32 \\ or \: {x}^(2) = (32)/(2) \\ or \: {x}^(2) = 16 \\ or \: x = √(16) \\ or \: x = \sqrt{ {(4)}^(2) } \: \: \: or \: \: \sqrt{ {( - 4)}^(2) } \\ x = + 4 \: or \: - 4`

Now

When X=4,

`{y}^(2) = 16 - {x}^(2) \\ {y}^(2) = 16 - {4}^(2) \\ {y}^(2) = 16 - 16 \\ {y = 0}^(2) \\ y = \sqrt{ {(0)}^(2) } \\ y = 0`

When X=-4,

`{y}^(2) = 16 - {x}^(2) \\ {y}^(2) = 16 - {( - 4)}^(2) \\ {y}^(2) = 16 - 16 \\ {y}^(2) = 0 \\ y = \sqrt{ {(0)}^(2) } \\ y = 0`

The solutions are (4,0) and (-4,0)

Hope this helps...

good luck on your assignment..

Answer 2

C and F

Explanation:

x² + y² = 16 ------------------(I)

`(x^(2))/(4^(2))-(y^(2))/(4^(2))=1\\\\x^(2)-y^(2)=4^(2)`

x² - y² = 16 ---------------(I)

(I) x² + y² = 16

(I) x² - y² = 16 {add & y² will be eliminated}

2x² = 32

x² = 32/2

x² = 16

x = √16

x = ±4

When x = 4,

4² + y² = 16

16 + y² = 16

y² = 16 - 16

y² = 0

y = 0

(4,0) is a solution

When x = -4,

(-4)² + y² = 16

16 + y² = 16

y² = 16 - 16

y² = 0

y = 0

(-4, 0 ) is a solution