Question

d) 3x + y - 9 = 0
e) y = 2x - 4
g) y = 3(x + 3)
h) 2x - y = -3
f) 3x + 5y - 15 = 0
j) 3y = 3 - x

Explanation:

Since the problem is not clearly stated (e.g., what to do with these equations like solving for a variable, finding intersections, etc.), I'll assume we need to solve each equation for \( y \) (a common task with linear equations). Let's handle each one:

Equation (d): \( 3x + y - 9 = 0 \)

Step 1: Isolate \( y \)

Subtract \( 3x \) and add \( 9 \) to both sides.
\( y = -3x + 9 \)

Equation (e): \( y = 2x - 4 \)

This is already solved for \( y \), so no further steps.
\( y = 2x - 4 \)

Equation (g): \( y = 3(x + 3) \)

Step 1: Distribute the 3

Multiply 3 with \( x \) and 3.
\( y = 3x + 9 \)

Equation (h): \( 2x - y = -3 \)

Answer:

Step 1: Simplify

\( y = 2 \)

If you meant a different operation (like finding intersection points between pairs of lines), please clarify which equations to pair. For now, here are the equations solved for \( y \):

• (d): \( \boldsymbol{y = -3x + 9} \)
• (e): \( \boldsymbol{y = 2x - 4} \)
• (g): \( \boldsymbol{y = 3x + 9} \)
• (h): \( \boldsymbol{y = 2x + 3} \)
• (f) (assumed): \( \boldsymbol{y = -\frac{3}{5}x + 3} \)
• (i) (assumed): \( \boldsymbol{y = -\frac{1}{3}x + 1} \)
• (the \( y = 5 - 3 \) one): \( \boldsymbol{y = 2} \)