Answer
$10x+6y+5z=30$
The graph is a plane.
Work Step by Step
Consider the following linear equation in three variables:
$$10x+6y+5z=30.$$
The graph of this equation is a plane. In order to draw it we calculate its intercepts.
Find the $x$-intercept by setting $y=0$ and $z=0$ and solving the resulting equation:
$$10x=30\Rightarrow x=3$$
The $x$-intercept is $3$, so we will plot the point $(3,0,0)$.
Find the $y$-intercept by setting $x=0$ and $z=0$ and solving the resulting equation:
$$6y=30\Rightarrow y=5$$
The $y$-intercept is $5$, so we will plot the point $(0,5,0)$.
Find the $z$-intercept by setting $x=0$ and $y=0$ and solving the resulting equation:
$$5z=30\Rightarrow z=6$$
The $z$-intercept is $6$, so we will plot the point $(0,0,6)$.
We join the $3$ points and obtain a triangular region which is a part of the plane described by the given equation.
