Question 1154829
{{{y = (1/2)x - 3}}}.....eq.1
{{{y = (3/2)x - 1}}}.....eq.2
---------------------------------subtract eq.2 from eq.1 to eliminate {{{y}}}

{{{y-y = (1/2)x - 3-((3/2)x - 1)}}}

{{{0= (1/2)x - 3-(3/2)x +1}}}

{{{0= (1/2)x -(3/2)x- 2 }}}

{{{0= -(2/2)x- 2 }}}

{{{0= -x- 2 }}}

{{{0+x= x-x- 2 }}}

{{{x=- 2 }}}

go to

{{{y = (1/2)x - 3}}}.....eq.1, substitute {{{x}}}

{{{y = (1/2)(-2) - 3}}}

{{{y = -2/2 - 3}}}

{{{y = -1 - 3}}}

{{{y = -4}}}


solution: 

{{{x=- 2 }}}
{{{y = -4}}}


{{{drawing( 600, 600, -10,10, -10, 10,

circle(-2,-4,.12), locate(-2,-4,p(-2,-4)),
 
 graph( 600, 600, -10,10, -10, 10,(3/2)x - 1 ,(1/2)x - 3)) }}}