Question

1. solve the system of linear equations: \\(\

$$\begin{cases} 3x - 2y = -5 \\\\ x + 6y = -1 \\end{cases}$$

\\) \\( (-1, -1) \\) \\( (-1, 1) \\) \\( (1, -1) \\) \\( (1, 1) \\) 10) an ellipse has the equation \\( \frac{(x - 3)^2}{4^2} + \frac{(y + 6)^2}{6^2} = 1 \\). if the ellipse is shifted two spaces to the left and five spaces up, what is the new equation? \\( \frac{x^2}{4^2} + \frac{y^2}{6^2} = 1 \\) \\( \frac{(x + 3)^2}{4^2} + \frac{(y - 6)^2}{6^2} = 1 \\) \\( \frac{(x + 1)^2}{4^2} + \frac{(y + 11)^2}{6^2} = 1 \\) \\( \frac{(x - 1)^2}{4^2} + \frac{(y + 10)^2}{6^2} = 1 \\) 11) find the vertex of the parabola: \\( y = 3x^2 - 6x + 7 \\) \\( (3, -6) \\) \\( (1, 4) \\) \\( (0, 7) \\) \\( (-6, 7) \\)

Explanation:

Problem 9

Step1: Write given equations

Equation 1: $3x - 2y = 5$
Equation 2: $x + 2y = -1$

Step2: Eliminate $y$ via addition

Add Equation 1 and Equation 2:
$$3x - 2y + x + 2y = 5 + (-1)$$
$$4x = 4$$

Step3: Solve for $x$

$$x = \frac{4}{4} = 1$$

Step4: Substitute $x=1$ into Equation 2

$$1 + 2y = -1$$
$$2y = -1 - 1 = -2$$
$$y = \frac{-2}{2} = -1$$

Problem 10

Step1: Recall ellipse shift rules

For an ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$:

• Shift left $h'=h-2$, shift up $k'=k+5$

Step2: Identify original $h,k$

Original $h=2$, $k=-6$

Step3: Calculate new $h',k'$

$h'=2-2=0$, $k'=-6+5=-1$

Step4: Write new equation

$$\frac{(x-0)^2}{3^2}+\frac{(y-(-1))^2}{6^2}=1 \implies \frac{x^2}{9}+\frac{(y+1)^2}{36}=1$$

Problem 11

Step1: Recall vertex formula for parabola

For $y=ax^2+bx+c$, vertex $x=-\frac{b}{2a}$

Step2: Identify $a,b$

$a=3$, $b=-6$

Step3: Calculate vertex $x$-coordinate

$$x = -\frac{-6}{2\times3} = \frac{6}{6} = 1$$

Step4: Calculate vertex $y$-coordinate

Substitute $x=1$ into $y=3x^2-6x+7$:
$$y=3(1)^2-6(1)+7=3-6+7=4$$

Answer:

Problem 9: $(1, -1)$
Problem 10: $\frac{x^2}{9}+\frac{(y+1)^2}{36}=1$
Problem 11: $(1, 4)$