Solve the following system of inequalities graphically:
$4x + 3y \leq 60, y \geq 2x, x \geq 3, x, y \geq 0$

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(N/A) The given system of inequalities is:
$4x + 3y \leq 60$ .....$(1)$
$y \geq 2x$ .....$(2)$
$x \geq 3$ .....$(3)$
To solve this graphically,we first draw the corresponding lines:
$1$. For $4x + 3y = 60$: If $x=0, y=20$; if $y=0, x=15$. The line passes through $(0, 20)$ and $(15, 0)$.
$2$. For $y = 2x$: The line passes through $(0, 0)$ and $(3, 6)$.
$3$. For $x = 3$: This is a vertical line passing through $(3, 0)$.
Analysis of regions:
- Inequality $(1)$ represents the region below the line $4x + 3y = 60$.
- Inequality $(2)$ represents the region above the line $y = 2x$.
- Inequality $(3)$ represents the region to the right of the line $x = 3$.
The common shaded region in the graph represents the solution set of the given system of linear inequalities.

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