Question: Solve each linear system using the method of GRAPHING [6]
ℓ1: 4x + 5y = -15
ℓ2: y = 4/5 x - 3
b) ℓ1: y = -x - 2
ℓ2: 3y - 2x - 9 = 0
Graph shape: bottom-right blank coordinate graph with x- and y-axes, no lines drawn.
1. **State the problem:** Solve the system of linear equations by graphing:
$$\ell_1: y = -x - 2$$
$$\ell_2: 3y - 2x - 9 = 0$$
2. **Rewrite the second equation in slope-intercept form $y = mx + b$:**
$$3y - 2x - 9 = 0$$
$$3y = 2x + 9$$
$$y = \frac{2}{3}x + 3$$
3. **Identify slopes and intercepts:**
- For $$\ell_1$$: slope $$m_1 = -1$$, y-intercept $$b_1 = -2$$
- For $$\ell_2$$: slope $$m_2 = \frac{2}{3}$$, y-intercept $$b_2 = 3$$
4. **Plot points for $$\ell_1$$:**
- Start at y-intercept (0, -2)
- Use slope $$-1$$: from (0, -2), go down 1 and right 1 to (1, -3)
- Another point: (2, -4)
5. **Plot points for $$\ell_2$$:**
- Start at y-intercept (0, 3)
- Use slope $$\frac{2}{3}$$: from (0, 3), go up 2 and right 3 to (3, 5)
- Another point: (6, 7)
6. **Graph the two lines and find their intersection:**
- Draw line through points (0, -2), (1, -3), (2, -4)
- Draw line through points (0, 3), (3, 5), (6, 7)
7. **Find intersection algebraically to confirm:**
Set $$-x - 2 = \frac{2}{3}x + 3$$
Multiply both sides by 3 to clear denominator:
$$3(-x - 2) = 3\left(\frac{2}{3}x + 3\right)$$
$$-3x - 6 = 2x + 9$$
Bring all terms to one side:
$$-3x - 6 - 2x - 9 = 0$$
$$-5x - 15 = 0$$
Add 15 to both sides:
$$-5x = 15$$
Divide both sides by -5:
$$x = \frac{\cancel{-5}x}{\cancel{-5}} = \frac{15}{-5} = -3$$
Substitute $$x = -3$$ into $$y = -x - 2$$:
$$y = -(-3) - 2 = 3 - 2 = 1$$
8. **Final answer:**
The lines intersect at $$\boxed{(-3, 1)}$$.
This is the solution to the system by graphing and algebraic confirmation.