I guess those 'L' are 't'.
Solve for t in both parametric equations:
[tex]x(t) = 2 - 3t \iff t = \dfrac{x-2}{-3}[/tex]
[tex]y(t)=4+t \iff t = y-4[/tex]
Now, you have two expressions for t. The must be equal to each other:
[tex]\dfrac{x-2}{-3} = y - 4[/tex]
Solve for y in this last equation:
[tex]y = \dfrac{x-2}{-3} + 4 = -\dfrac{1}{3}x+\dfrac{14}{3}[/tex]
And you've got the slope-intercept form.
Yo could also find from the beginning the slope: it's the quotient of the coefficients of the parameter:
[tex]m = \dfrac{1}{-3}[/tex]
And, then find the intercept by plugging the point (2,4) into the equation:
[tex]y=-\dfrac{1}{3}x+n[/tex]
