1. **Problem:** Determine which system of equations has exactly one solution. 2. **Recall:** A system of two linear equations has exactly one solution if the lines intersect at a single point, meaning their slopes are different. 3. **Check each system:** - System 1: \begin{cases} -3x + 2y = 8 \\ 6x + 2y = 8 \end{cases} Subtract the first from the second: $$\cancel{6x} + 2y - (-3x + 2y) = 8 - 8$$ $$6x + 2y + 3x - 2y = 0$$ $$9x = 0 \Rightarrow x = 0$$ Substitute back to find y: $$-3(0) + 2y = 8 \Rightarrow 2y = 8 \Rightarrow y = 4$$ So solution is $(0,4)$, exactly one solution. - System 2: $$3x = 0 \Rightarrow x = 0$$ Only one equation, so infinite solutions for y. Not a system. - System 3: $$2x + 10 = 2x - 10$$ Subtract $2x$: $$\cancel{2x} + 10 = \cancel{2x} - 10$$ $$10 = -10$$ Contradiction, no solution. - System 4: \begin{cases} y = 2x + 10 \\ y = 2x - 10 \end{cases} Both lines have slope 2 but different intercepts, so parallel lines, no solution. - System 5: \begin{cases} y = \frac{1}{3}x + 2 \\ y = -\frac{1}{3}x + 1 \end{cases} Set equal: $$\frac{1}{3}x + 2 = -\frac{1}{3}x + 1$$ Add $\frac{1}{3}x$ and subtract 2: $$\frac{1}{3}x + \frac{1}{3}x = 1 - 2$$ $$\frac{2}{3}x = -1$$ Multiply both sides by $\frac{3}{2}$: $$x = -1 \times \frac{3}{2} = -\frac{3}{2} = -1.5$$ Substitute back: $$y = \frac{1}{3}(-1.5) + 2 = -0.5 + 2 = 1.5$$ So solution is $(-1.5, 1.5)$, exactly one solution. - System 6: \begin{cases} -3x + 6x + 6y = -15 \\ 3x + 2y = 5 \\ -9x - 6y = 15 \end{cases} Simplify first equation: $$3x + 6y = -15$$ Compare with second: $$3x + 2y = 5$$ Multiply second by 3: $$9x + 6y = 15$$ Add to third: $$-9x - 6y = 15$$ Adding: $$0 = 30$$ Contradiction, no solution. **Answer:** Systems 1 and 5 have exactly one solution. --- **Slug:** one-solution **Subject:** algebra **Desmos:** {"latex":"y=\frac{1}{3}x+2","features":{"intercepts":true,"extrema":true}} **q_count:** 2