Question

solve the system. enter the smallest x-coordinate first. \

$$\begin{cases} x^2 - 4y^2 = 16 \\\\ x - 6y = -4 \\end{cases}$$

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Explanation:

Step1: Solve for \( x \) from the linear equation

From \( x - 6y = -4 \), we can express \( x \) as \( x = 6y - 4 \).

Step2: Substitute \( x = 6y - 4 \) into the hyperbola equation

Substitute \( x = 6y - 4 \) into \( x^2 - 4y^2 = 16 \):

$$ (6y - 4)^2 - 4y^2 = 16 $$

Expand \( (6y - 4)^2 \): \( 36y^2 - 48y + 16 - 4y^2 = 16 \)
Combine like terms: \( 32y^2 - 48y + 16 - 16 = 0 \)
Simplify: \( 32y^2 - 48y = 0 \)
Factor out \( 16y \): \( 16y(2y - 3) = 0 \)

Step3: Solve for \( y \)

Set each factor equal to zero:

• \( 16y = 0 \) gives \( y = 0 \)
• \( 2y - 3 = 0 \) gives \( y = \frac{3}{2} \)

Step4: Find corresponding \( x \) values

• When \( y = 0 \), \( x = 6(0) - 4 = -4 \)
• When \( y = \frac{3}{2} \), \( x = 6(\frac{3}{2}) - 4 = 9 - 4 = 5 \)

Answer:

\((-4, 0)\) and \((5, \frac{3}{2})\)