Question

regents exam questions
a.rei.d.12: graphing systems of linear inequalities i
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9 on the set of axes below, graph the following system of inequalities:
2y + 3x ≤ 14
4x - y < 2

Explanation:

Step1: Rewrite inequalities to slope-intercept form

First inequality:
$2y + 3x \leq 14$
Subtract $3x$: $2y \leq -3x + 14$
Divide by 2: $y \leq -\frac{3}{2}x + 7$

Second inequality:
$4x - y < 2$
Subtract $4x$: $-y < -4x + 2$
Multiply by -1 (reverse inequality): $y > 4x - 2$

Step2: Graph boundary lines

For $y \leq -\frac{3}{2}x + 7$:

• Boundary line: $y = -\frac{3}{2}x + 7$ (solid line, since $\leq$ includes equality)
• Y-intercept: $(0,7)$, slope: $-\frac{3}{2}$

For $y > 4x - 2$:

• Boundary line: $y = 4x - 2$ (dashed line, since $<$ excludes equality)
• Y-intercept: $(0,-2)$, slope: $4$

Step3: Shade solution regions

For $y \leq -\frac{3}{2}x + 7$: Shade below the solid line.
For $y > 4x - 2$: Shade above the dashed line.
The overlapping shaded area is the solution set.

Answer:

1. Solid boundary line: $y = -\frac{3}{2}x + 7$, shade below it.
2. Dashed boundary line: $y = 4x - 2$, shade above it.
3. The overlapping shaded area is the solution to the system.