Question

sketch the solution to this system of inequalities:
$x - y \leq 3$
$3x + y \leq 3$
\\(\bigcirc\\) a)
image of graph a
\\(\boldsymbol{\bigcirc}\\) b)
image of graph b
\\(\bigcirc\\) c)
image of graph c
\\(\bigcirc\\) d)
image of graph d

Explanation:

Step1: Analyze the first inequality \(x - y \leq 3\)

Rewrite it as \(y \geq x - 3\). The boundary line is \(y = x - 3\) (a straight line with slope 1 and y - intercept - 3), and we shade above the line (since \(y\geq x - 3\)).

Step2: Analyze the second inequality \(3x + y \leq 3\)

Rewrite it as \(y \leq - 3x+3\). The boundary line is \(y=-3x + 3\) (a straight line with slope - 3 and y - intercept 3), and we shade below the line (since \(y\leq - 3x + 3\)).

Step3: Find the intersection of the two shaded regions

We need to find the region that is shaded above \(y=x - 3\) and below \(y=-3x + 3\).

Now let's analyze the graphs:

• For option A: The shaded region does not match the intersection of the two inequalities.
• For option B: The shaded region is below \(y =- 3x+3\) and above \(y=x - 3\), which matches the intersection of the two inequalities.
• For option C: The shaded region does not match the intersection of the two inequalities.
• For option D: The shaded region does not match the intersection of the two inequalities.

Answer:

B. The graph in option B