Linear Inequalities Eed391

1. **Stating the problem:** We are given four systems of inequalities labeled A, B, C, and D. Each system contains two linear inequalities. 2. **Goal:** For each system, we want to understand the solution region defined by the inequalities. Since the user did not specify a particular question, we will analyze system A completely as the first problem. 3. **System A:** $$3x + 2y < 11$$ $$2x - y < 9$$ 4. **Rewrite inequalities to express $y$ in terms of $x$:** - From $$3x + 2y < 11$$, subtract $$3x$$: $$2y < 11 - 3x$$ Divide both sides by 2: $$y < \frac{11 - 3x}{2}$$ - From $$2x - y < 9$$, subtract $$2x$$: $$-y < 9 - 2x$$ Multiply both sides by $$-1$$ and reverse inequality: $$y > 2x - 9$$ 5. **Summary of inequalities for system A:** $$y < \frac{11 - 3x}{2}$$ $$y > 2x - 9$$ 6. **Interpretation:** The solution region is the set of points $(x,y)$ where $y$ is simultaneously less than the line $$y = \frac{11 - 3x}{2}$$ and greater than the line $$y = 2x - 9$$. 7. **Finding intersection point of boundary lines:** Set $$\frac{11 - 3x}{2} = 2x - 9$$ Multiply both sides by 2: $$11 - 3x = 4x - 18$$ Add $$3x$$ to both sides: $$11 = 7x - 18$$ Add $$18$$: $$29 = 7x$$ Divide both sides by 7: $$x = \frac{29}{7}$$ Substitute back to find $y$: $$y = 2\left(\frac{29}{7}\right) - 9 = \frac{58}{7} - 9 = \frac{58}{7} - \frac{63}{7} = -\frac{5}{7}$$ 8. **Final answer:** The solution region for system A is all points $(x,y)$ such that $$2x - 9 < y < \frac{11 - 3x}{2}$$ with the boundary lines intersecting at $$\left(\frac{29}{7}, -\frac{5}{7}\right)$$. This completes the solution for the first system (A).