Answer
(a) nonlinear
(b) $(\sqrt {10}, 3\sqrt {10})$ and $(-\sqrt {10}, -3\sqrt {10})$
Work Step by Step
(a) The system of equations is nonlinear because two of the terms are greater than first degree.
(b) We will use the substitution method.
The second equation has the $y$ term already isolated, so we will use the expression to substitute into the first equation:
$x^{2} + y^{2} = 100$
$x^{2} + (3x)^{2} = 100$
$x^{2} + 9x^{2} = 100$
Combine like terms:
$10x^{2} = 100$
Divide both sides by $10$:
$x^{2} = 10$
Take the square root of both sides:
$x = \sqrt {10}$ or $x = -\sqrt {10}$
Now that we have the values for $x$, we can substitute this value into the second equation to get the values for $y$:
$y = 3(\sqrt {10})$ or $y = 3(-\sqrt {10})$
$y = 3\sqrt {10}$ or $y = -3\sqrt {10}$
The points where these two equations intersect are $(\sqrt {10}, 3\sqrt {10})$ and $(-\sqrt {10}, -3\sqrt {10})$.
