Question 949492
Suppose the line through points ({{{n}}},{{{6}}}) and ({{{1}}},{{{2}}}) is parallel to the graph of {{{2x+y=3}}}. Find the value of {{{n}}}.

you are looking for {{{y=mx+b}}} parallel to the graph of {{{2x+y=3}}}
parallel lines have same slope, so find slope of the line {{{2x+y=3}}} 

{{{y=-2x+3}}}-> slope {{{m=-2}}}

so far we have {{{y=-2x+b}}}

if the line passes through point({{{1}}},{{{2}}}), then we can use it to find y-intercept {{{b}}}

{{{2=-2*1+b}}}

{{{2=-2+b}}}

{{{2+2=b}}}

{{{b=4}}}

so, our line is {{{y=-2x+4}}}

if the line passes through point({{{n}}},{{{6}}}), then {{{x=n}}} and {{{y=6}}}, so plug it in  {{{y=-2x+4}}} and solve for {{{n}}}

 {{{6=-2n+4}}}

{{{2n=-6+4}}}

{{{2n=-2}}}

{{{n=-1}}}

so, the line passes through point({{{-1}}},{{{6}}})

check it on a graph:


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,6,.12),circle(1,2,.12),
locate(-1,6,p(-1,6)),locate(1,2,p(1,2)),
locate(-4,7,y=(-2x+3)),locate(-2,8,y=(-2x+4)),
 graph( 600, 600, -10, 10, -10, 10, -2x+3, -2x+4)) }}}