Question

1. $4x - 2y = 14$

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

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

$12x + 3 = y$

Explanation:

Problem 4: Solve the system of equations \( 4x - 2y = 14 \) and \( y = \frac{1}{2}x - 1 \)

Step 1: Substitute \( y \) into the first equation

We know \( y = \frac{1}{2}x - 1 \), so substitute this into \( 4x - 2y = 14 \).
\( 4x - 2(\frac{1}{2}x - 1) = 14 \)

Step 2: Simplify the equation

First, distribute the -2: \( 4x - x + 2 = 14 \)
Then combine like terms: \( 3x + 2 = 14 \)

Step 3: Solve for \( x \)

Subtract 2 from both sides: \( 3x = 14 - 2 = 12 \)
Divide both sides by 3: \( x = \frac{12}{3} = 4 \)

Step 4: Solve for \( y \)

Substitute \( x = 4 \) into \( y = \frac{1}{2}x - 1 \)
\( y = \frac{1}{2}(4) - 1 = 2 - 1 = 1 \)

Step 1: Substitute \( y \) into the first equation

We know \( y = 12x + 3 \), so substitute this into \( 8x - \frac{1}{3}y = 0 \)
\( 8x - \frac{1}{3}(12x + 3) = 0 \)

Step 2: Simplify the equation

Distribute the \( -\frac{1}{3} \): \( 8x - 4x - 1 = 0 \)
Combine like terms: \( 4x - 1 = 0 \)

Step 3: Solve for \( x \)

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

Step 4: Solve for \( y \)

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

Answer:

\( x = 4, y = 1 \)

Problem 6: Solve the system of equations \( 8x - \frac{1}{3}y = 0 \) and \( y = 12x + 3 \)