First, we must find the slope of the equation in the problem by transforming the equation to the slope-intercept form. The slope-intercept form of a linear equation is: #y = color(red)(m)x + color(blue)(b)#

Where #color(red)(m)# is the slope and #color(blue)(b)# is the y-intercept value. Solving for #y# gives:

#y - 2 + color(blue)(2) = (2/3)x + color(blue)(2)#

#y - 0 = (2/3)x + 2#

#y = color(red)(2/3)x + color(blue)(2)# Therefore:

The slope is #color(red)(m = 2/3)#

A parallel line will have a slope the same as the line it is parallel to, for this problem #color(red)(m = 2/3)# We can now use the point-slope formula to find an equation for the parallel line. The point-slope formula states: #(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))#

Where #color(blue)(m)# is the slope and #color(red)(((x_1, y_1)))# is a point the line passes through. Substituting the slope we determined and the point from the problem gives:

#(y - color(red)(2)) = color(blue)(2/3)(x - color(red)(3))#

Or, we can solve for #y# to put the equation in slope-intercept form:

#y - color(red)(2) = (color(blue)(2/3) xx x) - (color(blue)(2/3) xx color(red)(3))#

#y - color(red)(2) = 2/3x - 6/3#

#y - color(red)(2) = 2/3x - 2#

#y - color(red)(2) + 2 = 2/3x - 2 + 2#

#y - 0 = 2/3x - 0#

#y = color(red)(2/3)x + color(blue)(0)#

Or

#y = 2/3x#

Three formulas solving this problem are:

#(y - color(red)(2)) = color(blue)(2/3)(x - color(red)(3))#

#y = color(red)(2/3)x + color(blue)(0)#

#y = 2/3x#