1. **State the problem:** We need to classify each system of linear equations (A, B, C) as consistent independent, consistent dependent, or inconsistent, and find the solution if it exists. 2. **Recall definitions:** - Consistent independent: exactly one unique solution (lines intersect at one point). - Consistent dependent: infinitely many solutions (lines coincide). - Inconsistent: no solution (lines are parallel and distinct). --- ### System A: Equations: $$y = -\frac{1}{2}x + 1$$ $$y = -\frac{1}{2}x + 2$$ Both lines have the same slope $-\frac{1}{2}$ but different y-intercepts (1 and 2), so they are parallel and never intersect. **Classification:** Inconsistent system. **Solution:** No solution. --- ### System B: Equations: $$y = -2x + 5$$ $$y = x - 1$$ Slopes are $-2$ and $1$, which are different, so lines intersect at exactly one point. To find the solution, set the right sides equal: $$-2x + 5 = x - 1$$ Move terms: $$-2x - x = -1 - 5$$ $$-3x = -6$$ Divide both sides by $-3$: $$x = \frac{\cancel{-6}}{\cancel{-3}} = 2$$ Substitute $x=2$ into $y = x - 1$: $$y = 2 - 1 = 1$$ **Classification:** Consistent independent. **Solution:** $(2, 1)$ --- ### System C: Equations: $$y = -\frac{2}{3}x + 4$$ $$2x + 3y = 12$$ Rewrite second equation to slope-intercept form: $$3y = 12 - 2x$$ $$y = \frac{12 - 2x}{3} = -\frac{2}{3}x + 4$$ Both equations represent the same line. **Classification:** Consistent dependent. **Solution:** Infinitely many solutions (all points on the line). --- **Final answers:** - System A: Inconsistent, no solution. - System B: Consistent independent, unique solution $(2,1)$. - System C: Consistent dependent, infinitely many solutions.