Answer
(a) nonlinear
(b) $(1,−2)$ and $(\frac{5}{3},0)$
Work Step by Step
(a) This system of equations is nonlinear because one of the terms is greater than first degree.
(b) We will use the substitution method to isolate one of the variables, $y$, so we can substitute it into the other equation. We will use the second equation.
We subtract $3x$ from each side:
$3x - y = 5$
$3x - 3x - y = 5 - 3x$
$-y = 5 - 3x$
Divide both sides by $-1$:
$\frac{-y}{-1} = \frac{5 - 3x}{-1}$
$y = 3x - 5$
Now that we have an expression for $y$, we can substitute this expression into the other equation to solve for the other variable, $x$:
$6x + y^{2} = 10$
$6x + (3x - 5)^{2} = 10$
We use the foil method to simplify the squared term:
$6x + (3x - 5)(3x - 5) = 10$
$6x + 9x^{2} - 15x - 15x + 25 = 10$
Simplify the equation by combining like terms and setting the equation equal to 0.
$9x^{2} - 30x + 6x + 25 = 10$
$9x^{2} - 24x + 25 = 10$
$9x^{2} - 24x + 15 = 0$
We can simplify this equation even further by dividing all terms by 3:
$3(3x^{2} - 8x + 5) = 0$
Divide each side by 3 to get:
$3x^{2} - 8x + 5 = 0$
Now, we factor:
$(3x - 5)(x - 1) = 0$
Set each term equal to 0 and solve for $x$:
$3x - 5 = 0$
$3x = 5$
$x = \frac{5}{3}$
$x - 1 = 0$
$x = 1$
Now we can plug in the value we found for $x$ into the second equation to get the values for $y$:
$3(\frac{5}{3}) - y = 5$
$\frac{15}{3} - y = 5$
$5 - y = 5$
$-y = 0$
$y = 0$
$3(1) - y = 5$
$3 - y = 5$
$-y = 2$
$y = -2$
The points where these two graphs intersect are $(1, -2)$ and $(\frac{5}{3}, 0)$.
