which statements are true about the graph of y ≤ 3x + 1 and y ≥ -x + 2? check all that apply. the slope of…

Explanation:

Step1: Identify boundary - line equations and slopes

The boundary - line equations for $y\leq3x + 1$ and $y\geq - x+2$ are $y = 3x + 1$ and $y=-x + 2$. The slopes are 3 and - 1 respectively, so the statement "The slope of one boundary line is 2" is false.

Step2: Determine line type

Since the inequalities are $\leq$ and $\geq$, the boundary lines $y = 3x + 1$ and $y=-x + 2$ are solid. So the statement "Both boundary lines are solid" is true.

Step3: Check if (1,3) is a solution

For $y\leq3x + 1$, when $x = 1$ and $y = 3$, $3\leq3\times1+1$ (i.e., $3\leq4$) is true. For $y\geq - x+2$, when $x = 1$ and $y = 3$, $3\geq-1 + 2$ (i.e., $3\geq1$) is true. So the statement "A solution to the system is (1,3)" is true.

Step4: Analyze shading direction

For $y\leq3x + 1$, we shade below the line $y = 3x + 1$. For $y\geq - x+2$, we shade above the line $y=-x + 2$. So the statement "Both inequalities are shaded below the boundary lines" is false.

Step5: Check for intersection

Set $3x + 1=-x + 2$. Solving for $x$: $$3x+x=2 - 1$$ $$4x=1$$ $$x=\frac{1}{4}$$ Substitute $x=\frac{1}{4}$ into $y=-x + 2$, we get $y=-\frac{1}{4}+2=\frac{7}{4}$. So the boundary lines intersect, and the statement "The boundary lines intersect" is true.

Answer:

• Both boundary lines are solid.
• A solution to the system is (1,3).
• The boundary lines intersect.