Question #d842e

First, we need to find the slope of the equation given in the problem by converting it to the familiar slope-intercept form by solving for #y#:

#2x - color(red)(2x) - 3y = -color(red)(2x) + 8#

#0 - 3y = -2x + 8#

#-3y = -2x + 8#

#(-3y)/color(red)(-3) = (-2x + 8)/color(red)(-3)#

#(color(red)(cancel(color(black)(-3)))y)/cancel(color(red)(-3)) = 2/3x - 8/3#

#y = 2/3x - 8/3#

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.

Therefore we know the slope of this line and a line parallel to this line is #color(red)(m = 2/3)#

We can now use this slope and the point and the point-slope formula to find an equation for the line requested in the problem:

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 values from the problem and the calculate gives:

#(y - color(red)(-1)) = color(blue)(2/3)(x - color(red)(3))#

#(y + color(red)(1)) = color(blue)(2/3)(x - color(red)(3))#

Or converting to the slope-intercept form:

#y + color(red)(1) = 2/3x - (2/3 xx color(red)(3))#

#y + 1 = 2/3x - 2#

#y + 1 - 1 = 2/3x - 2 - 1#

#y = 2/3x - 3#

#2x-3y=8#
we change to #y=mx+c#, where #m#=gradient.

#2x-8=3y#
#y=(2x-8)/3#
#y=2/3x-8/3#

since it is a parallel line, they have a same value of gradient,where #m=2/3#.

use #y-y_1=m(x-x_1)#
#y-(-1)=2/3(x-3)#
#y+1=2/3x-2#
#y=2/3x-2-1#
#y=2/3x-3#
#3y=2x-9#
#9=2x-3y#