#### Answer

(a) Standard form
$2x-y=8$
(b) Slope-intercept form
$y=2x-5$

#### Work Step by Step

RECALL:
(1) The standard form of a line's equation is $Ax+By=C$.
(2) THe slope-intercept form of a line's equation is $y=mx+b$ where $m$=slope and $(0, b)$ is the line's y-intercept.
(3) Parallel lines have the same slope.
(4) Perpendicular lines have slopes whose product is $-1$ (negative reciprocals of each other).
The line is parallel to $2x-y=5$. Solve the equation for $y$ to obtain:
$2x-y+y=5+y
\\2x=5+y
\\2x-5=5+y-5
\\2x-5=y
\\y=2x-5$
The slope of the line above is $2$. Thus, the slope of the line parallel to it is also $2$.
This means that the tentative equation of the line we are looking for is $y=2x+b$.
To find the value of $b$, substitute the $x$ and $y$ values of the point $(3, -2)$ to obtain:
$y=2x+b
\\-2=2(3)+b
\\-2=6+b
\\-2-6=b
\\-8=b$
Therefore, the equation of the line is $y=2x-8$.
(a) Standard form
Add $8$ to both sides of the equation to obtain:
$y+8=2x-8+8
y+8=2x$
Subtract $y$ to both sides of the equation to obtain:
$y+8-y=2x-y
\\8=2x-y
\\2x-y=8$
(b) Slope-intercept form
$y=2x-5$