Question

1. \\(\

$$\begin{cases}2y = x - 4 \\\\ 2y + 4 = - 3x\\end{cases}$$

\\)\

1. \\(\

$$\begin{cases}- 3y = x - 15 \\\\ 3y - 9 = - x\\end{cases}$$

\\)

Explanation:

Problem 7

Step1: Substitute \(2y\) from first equation

From \(2y = x - 4\), substitute into \(2y + 4 = -3x\):
\((x - 4) + 4 = -3x\)

Step2: Simplify and solve for \(x\)

Simplify: \(x = -3x\)
Add \(3x\): \(4x = 0\) → \(x = 0\)

Step3: Substitute \(x\) to find \(y\)

Substitute \(x = 0\) into \(2y = x - 4\):
\(2y = 0 - 4\) → \(2y = -4\) → \(y = -2\)

Problem 8

Step1: Rewrite second equation

From \(3y - 9 = -x\), rewrite as \(x = 9 - 3y\)

Step2: Substitute \(x\) into first equation

Substitute \(x = 9 - 3y\) into \(-3y = x - 15\):
\(-3y = (9 - 3y) - 15\)

Step3: Simplify and solve for \(y\)

Simplify: \(-3y = -3y - 6\)
Add \(3y\): \(0 = -6\) → Contradiction? Wait, re - check:
Wait, first equation: \(-3y = x - 15\) → \(x = -3y + 15\)
Second equation: \(3y - 9 = -x\) → \(x = 9 - 3y\)
Set equal: \(-3y + 15 = 9 - 3y\)
Add \(3y\): \(15 = 9\) → No solution? Wait, maybe miscalculation.
Wait, let's use elimination.
First equation: \(x + 3y = 15\)
Second equation: \(x + 3y = 9\)
Parallel lines (same slope, different intercepts) → No solution. Wait, but maybe I made a mistake.
Wait, original equations:
\(-3y = x - 15\) → \(x + 3y = 15\)
\(3y - 9 = -x\) → \(x + 3y = 9\)
Yes, two lines \(x + 3y = 15\) and \(x + 3y = 9\) are parallel (same \(x\) and \(y\) coefficients), so no solution. But wait, maybe the problem expects a solution. Wait, re - check the equations:
Wait, the second equation: \(3y - 9 = -x\) → \(x = 9 - 3y\)
First equation: \(-3y = x - 15\) → substitute \(x\):
\(-3y = (9 - 3y) - 15\) → \(-3y = -3y - 6\) → \(0 = -6\), which is false. So no solution. But maybe the problem has a typo, or I misread.
Wait, maybe the second equation is \(3y - 9 = x\)? No, as per the image. So the system has no solution (parallel lines).

Answer:

Problem 7: \(x = 0\), \(y = -2\)
Problem 8: No solution (parallel lines)