Calculus of a Single Variable, Chapter 10, 10.2, Section 10.2, Problem 2

x = 5-4t
y=2+5t
To graph a parametric equation, assign values to t. Since there is no given interval for t, let's consider the values from t=-3 to t=3.
t=-3
x=5-4(-3) = 17
y= 2+5(-3) = -13
t=-2
x=5-4(-2)=13
y=2+5(-2)=-8
t=-1
x=5-4(-1)=9
y=2+5(-1)=-3
t=0
x=5-4(0)=5
y=2+5(0)=2
t=1
x=5-4(1)=1
y=2+5(1)=7
t=2
x=5-4(2)=-3
y=2+5(2)=12
t=3
x=5-4(3)=-7
y=2+5(3)=17
And, plot the points (x,y) in the xy-plane.

(Please see attachment for the orientation of the curve.)
Take note that the graph of a parametric equation has direction. For this equation, as the value of t increases, the points (x,y) are going to upward to the left.
To convert a parametric equation to rectangular form, isolate the t in one of the equation. Let's consider the equation for x.
x= 5-4t
x-5=-4t
-(x-5)/4=t
Then, plug-in this to the other equation.
y=2+5(t)
y=2+5(-(x-5)/4)
y=2-(5(x-5))/4
y=2-(5(x-5))/4
y=2-(5x - 25)/4
y=8/4-(5x-25)/4
y=(33 - 5x)/4
y=-(5x)/4 + 33/4
Therefore, the rectangular form of the given parametric equation is y=-(5x)/4 + 33/4 .