Question #6b8f6

1 Answer
Mar 21, 2017

Answer:

Slope is #2/3#
#y_("intercept")" "=" "4/3#

#x_("intercept")" "=-2#

Explanation:

#color(brown)("Extensive explanation given - using first principles method")#

#color(purple)("Note that the shortcut method is based on the outcome of first principle method")#

Target: Manipulate the given equation into the form of #y=mx+c#
#" "#where m is the gradient and c is the y-intercept.

The #3/2y# is positive so we need to keep the #y# on that side of the equation.

To get rid of the #x# on the left we change it to 0
What we do to one side of the equation we also do to the other.

Add #color(red)(x)# to both sides.

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

But #-x+x=0#

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

Note that #x+2# has the same value as #2+x#

#color(green)(3/2y=x+2#

To 'get rid' of the #3/2# change it to 1 as #1xxy=y#

Multiply both sides by #color(red)(2/3)#

#color(green)(3/2color(red)(xx2/3)xxy=color(red)(2/3)(x+2)#

#color(green)(y=2/3x+4/3 rarr" "2/3" is the gradient (slope)"#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

The y-axis crosses the x-axis at #x=0# so substitute 0 for #x#

#y_("intercept")=2/3(0)+4/3" "=" "4/3#

The x-axis cross the y-axis at y=0 so substitute 0 for y

#0=2/3x+4/3#

Subtract from both sides #4/3#

#-4/3=2/3x#

Multiply both sides by #3/2#

#x=3/2xx(-4/3)#

#x=- (3/3xx4/2)#

#x_("intercept")=-2#