which statements are true about the graph of the system of linear inequalities? select two options.\ny > 3x…

which statements are true about the graph of the system of linear inequalities? select two options.\ny > 3x - 4\ny ≤ 1/2x + 1\n□ the graph of y > 3x - 4 has shading above a dashed line.\n□ the graph of y ≤ 1/2x + 1 has shading below a dashed line.\n□ the graphs of the inequalities will intersect.\n□ there are no solutions to the system.\n□ the graphs of the two inequalities intersect the y - axis at (0, 1) and (0, 4).
Answer
Explanation:
Step1: Analyze the inequality (y>3x - 4)
For a linear inequality (y>mx + b) ((m = 3), (b=-4)), the boundary line (y = 3x-4) is dashed (because the inequality is strict, (y>3x - 4) not (y\geq3x - 4)), and the shading is above the line.
Step2: Analyze the inequality (y\leq\frac{1}{2}x+1)
For a linear inequality (y\leq mx + b) ((m=\frac{1}{2}), (b = 1)), the boundary line (y=\frac{1}{2}x + 1) is solid (because the inequality is non - strict, (y\leq\frac{1}{2}x+1)), and the shading is below the line.
Step3: Check for intersection of the boundary lines
Set (3x-4=\frac{1}{2}x + 1). [ \begin{align*} 3x-\frac{1}{2}x&=1 + 4\ \frac{6x-x}{2}&=5\ \frac{5x}{2}&=5\ x&=2 \end{align*} ] Substitute (x = 2) into (y=\frac{1}{2}x+1), (y=\frac{1}{2}\times2+1=2). The boundary lines (y = 3x-4) and (y=\frac{1}{2}x + 1) intersect at ((2,2)). Since (y>3x - 4) (shading above (y = 3x-4)) and (y\leq\frac{1}{2}x+1) (shading below (y=\frac{1}{2}x + 1)), there are no common points (because (3x-4>\frac{1}{2}x + 1) for (x>2) and the direction of shading is opposite).
Step4: Check (y) - intercepts
For (y=3x - 4), when (x = 0), (y=-4). For (y=\frac{1}{2}x+1), when (x = 0), (y = 1).
Answer:
The graph of (y>3x - 4) has shading above a dashed line. ; There are no solutions to the system.