1. **State the problem:** We are given the system of linear equations: $$10x + 3y = -3$$ $$-18x - 19y = -18$$ We need to determine whether this system has no solution, one solution, or infinitely many solutions. 2. **Recall the method:** To analyze the system, we can use the method of comparing ratios of coefficients. For two equations: $$a_1x + b_1y = c_1$$ $$a_2x + b_2y = c_2$$ - If $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$, the system has **one unique solution** (lines intersect). - If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$$, the system has **no solution** (parallel lines). - If $$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$$, the system has **infinitely many solutions** (same line). 3. **Calculate the ratios:** $$\frac{a_1}{a_2} = \frac{10}{-18} = -\frac{5}{9}$$ $$\frac{b_1}{b_2} = \frac{3}{-19} = -\frac{3}{19}$$ $$\frac{c_1}{c_2} = \frac{-3}{-18} = \frac{1}{6}$$ 4. **Compare the ratios:** $$-\frac{5}{9} \neq -\frac{3}{19}$$ Since $$\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$$, the system has **one unique solution**. 5. **Conclusion:** The two lines intersect at exactly one point, so the system has one solution. **Final answer:** one solution