Eliminate the parameter.

[tex]\[
\begin{array}{c}
\left\{\begin{array}{l}
x = 2 - 3t \\
y = 5 + t
\end{array}\right. \\
y = -\frac{1}{3} x + \frac{11}{3}
\end{array}
\][/tex]



Answer :

Sure, let's eliminate the parameter [tex]\( t \)[/tex] from the given parametric equations and express [tex]\( y \)[/tex] as a function of [tex]\( x \)[/tex].

### Step 1: Express [tex]\( t \)[/tex] in terms of [tex]\( x \)[/tex] from the first equation
The first equation is:
[tex]\[ x = 2 - 3t \][/tex]

We can solve for [tex]\( t \)[/tex]:
[tex]\[ x = 2 - 3t \][/tex]
[tex]\[ 3t = 2 - x \][/tex]
[tex]\[ t = \frac{2 - x}{3} \][/tex]

### Step 2: Substitute [tex]\( t \)[/tex] into the second equation
The second equation is:
[tex]\[ y = 5 + t \][/tex]

Substitute the expression we found for [tex]\( t \)[/tex]:
[tex]\[ y = 5 + \frac{2 - x}{3} \][/tex]

### Step 3: Simplify the expression
To make this simpler, let's separate the fraction:
[tex]\[ y = 5 + \frac{2}{3} - \frac{x}{3} \][/tex]
[tex]\[ y = 5 + \frac{2}{3} - \frac{x}{3} \][/tex]
[tex]\[ y = 5\frac{3}{3} + \frac{2}{3} - \frac{x}{3} \][/tex]
[tex]\[ y = \frac{15}{3} + \frac{2}{3} - \frac{x}{3} \][/tex]
[tex]\[ y = \frac{15 + 2 - x}{3} \][/tex]
[tex]\[ y = \frac{17 - x}{3} \][/tex]

### Step 4: Express [tex]\( y \)[/tex] in the form of [tex]\( y = mx + b \)[/tex]
[tex]\[ y = -\frac{x}{3} + \frac{17}{3} \][/tex]

Therefore, the slope [tex]\( m \)[/tex] is [tex]\(-\frac{1}{3}\)[/tex] and the [tex]\( y \)[/tex]-intercept [tex]\( b \)[/tex] is [tex]\(\frac{17}{3}\)[/tex].

So, the expression for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex] is:
[tex]\[ y = -\frac{1}{3} x + \frac{17}{3} \][/tex]

Thus, placing the constants into the required form, [tex]\( y = -\frac{\boxed{1}}{\boxed{3}} x + \boxed{17/3} \)[/tex].