3. $x-y>1$ \n$x+4y>20$
3. $x-y>1$ \n$x+4y>20$
Answer
Explanation:
For the system $\boldsymbol{7x + 3y \leq 18}$ and $\boldsymbol{2x - 3y \leq 9}$
Step1: Rewrite first inequality
Solve $7x+3y\leq18$ for $y$: $3y \leq -7x + 18$ $y \leq -\frac{7}{3}x + 6$
Step2: Rewrite second inequality
Solve $2x-3y\leq9$ for $y$: $-3y \leq -2x + 9$ $y \geq \frac{2}{3}x - 3$
Step3: Graph boundary lines
- For $y = -\frac{7}{3}x + 6$: Slope $-\frac{7}{3}$, y-intercept $(0,6)$, x-intercept $(\frac{18}{7}, 0 \approx 2.57)$. Draw as a solid line (inequality is $\leq$).
- For $y = \frac{2}{3}x - 3$: Slope $\frac{2}{3}$, y-intercept $(0,-3)$, x-intercept $(\frac{9}{2}, 0 = 4.5)$. Draw as a solid line (inequality is $\geq$).
Step4: Shade solution regions
- For $y \leq -\frac{7}{3}x + 6$: Shade below the line.
- For $y \geq \frac{2}{3}x - 3$: Shade above the line.
Step5: Identify overlap
The solution is the double-shaded area satisfying both inequalities.
For the system $\boldsymbol{4x - 5y > -5}$ and $\boldsymbol{y < -1}$
Step1: Rewrite first inequality
Solve $4x-5y>-5$ for $y$: $-5y > -4x - 5$ $y < \frac{4}{5}x + 1$
Step2: Graph boundary lines
- For $y = \frac{4}{5}x + 1$: Slope $\frac{4}{5}$, y-intercept $(0,1)$, x-intercept $(-\frac{5}{4}, 0 = -1.25)$. Draw as a dashed line (inequality is $<$).
- For $y = -1$: Horizontal line through $(0,-1)$. Draw as a dashed line (inequality is $<$).
Step3: Shade solution regions
- For $y < \frac{4}{5}x + 1$: Shade below the line.
- For $y < -1$: Shade below the line $y=-1$.
Step4: Identify overlap
The solution is the double-shaded area satisfying both inequalities.
Answer:
- For $\boldsymbol{7x + 3y \leq 18}$ and $\boldsymbol{2x - 3y \leq 9}$: The solution is the overlapping shaded region bounded by the solid lines $y = -\frac{7}{3}x + 6$ (shade below) and $y = \frac{2}{3}x - 3$ (shade above).
- For $\boldsymbol{4x - 5y > -5}$ and $\boldsymbol{y < -1}$: The solution is the overlapping shaded region bounded by the dashed lines $y = \frac{4}{5}x + 1$ (shade below) and $y=-1$ (shade below).