Question

1. $8x - \frac{1}{3}y = 0$

$12x + 3 = y$

Explanation:

Step1: Substitute \( y = 12x + 3 \) into \( 8x-\frac{1}{3}y = 0 \)

We have the system of equations:
\[

$$\begin{cases} 8x-\frac{1}{3}y = 0 \\ y = 12x + 3 \end{cases}$$

\]
Substitute \( y = 12x + 3 \) into the first equation:
\( 8x-\frac{1}{3}(12x + 3)=0 \)

Step2: Simplify the equation

First, distribute \( -\frac{1}{3} \) in \( -\frac{1}{3}(12x + 3) \):
\( 8x - 4x - 1 = 0 \)
Combine like terms:
\( 4x - 1 = 0 \)

Step3: Solve for \( x \)

Add 1 to both sides:
\( 4x = 1 \)
Divide both sides by 4:
\( x=\frac{1}{4} \)

Step4: Solve for \( y \)

Substitute \( x = \frac{1}{4} \) into \( y = 12x + 3 \):
\( y = 12\times\frac{1}{4}+3 \)
\( y = 3 + 3 \)
\( y = 6 \)

Answer:

The solution to the system is \( x = \frac{1}{4} \), \( y = 6 \)