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 shapes: top-right graph is a coordinate plane with two intersecting lines; the rising line passes through (0,-3) and (5,-2), and the falling line passes through (0,-3) and (3,-7). position_hint ∈ {top-right, center}
Graph shapes: bottom-right graph is a blank coordinate plane with no plotted line. position_hint ∈ {bottom-right}User: Solve it with graph
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:**
Start with $$3y - 2x - 9 = 0$$
Add $$2x + 9$$ to both sides:
$$3y = 2x + 9$$
Divide both sides by 3:
$$y = \frac{2}{3}x + 3$$
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 the lines:**
- $$\ell_1$$ passes through (0, -2) and has slope -1 (down 1 unit, right 1 unit).
- $$\ell_2$$ passes through (0, 3) and has slope $$\frac{2}{3}$$ (up 2 units, right 3 units).
5. **Find the intersection point:**
Set $$y$$ from both equations equal:
$$-x - 2 = \frac{2}{3}x + 3$$
Add $$x$$ to both sides:
$$-2 = \frac{2}{3}x + x + 3$$
Combine $$x$$ terms:
$$-2 = \frac{2}{3}x + \frac{3}{3}x + 3 = \frac{5}{3}x + 3$$
Subtract 3 from both sides:
$$-2 - 3 = \frac{5}{3}x$$
$$-5 = \frac{5}{3}x$$
Multiply both sides by $$\cancel{\frac{3}{5}}$$ to isolate $$x$$:
$$x = -5 \times \cancel{\frac{3}{5}} = -3$$
6. **Find $$y$$ coordinate:**
Substitute $$x = -3$$ into $$y = -x - 2$$:
$$y = -(-3) - 2 = 3 - 2 = 1$$
7. **Solution:**
The lines intersect at $$(-3, 1)$$.
**Final answer:** $$\boxed{(-3, 1)}$$