Question 992474
Using: 

{{{highlight(Elimination)}}} 

1) {{{2x+3y=18}}}
     {{{4x+6y=12}}}

*[invoke solving_linear_system_by_elimination 2, 3, 18, 4, 6, 12] 


2) {{{4x+y=6}}}
   {{{-8x-2y=21}}}

*[invoke solving_linear_system_by_elimination 4, 1, 6, -8, -2, 21] 
     



{{{highlight(Substitution)}}}


1) {{{2x+3y=18}}}
     {{{4x+6y=12}}}

*[invoke linear_substitution "x", "y", 2, 3, 18, 4, 6, 12]


2) {{{4x+y=6}}}
   {{{-8x-2y=21}}}

*[invoke linear_substitution "x", "y", 4, 1, 6, -8,-2,21]




{{{highlight(Determinants )}}}



1) {{{2x+3y=18}}}
     {{{4x+6y=12}}}

{{{2}}} by {{{2}}} systems of linear equations are of the form

{{{a x + b y = c}}}

{{{d x + e y = f}}


 Cramer's rule state that

{{{x = (c e - f b) / D}}} and {{{y = (a f - d c) / D}}} where {{{D}}} is the coefficient determinant given by {{{D = a e - b d}}}. 

in your case {{{a=2}}},{{{b=3}}},{{{c=18}}},{{{d=4}}},{{{e=6}}},and {{{f=12}}}

then you have:

determinant  {{{D = a e - b d}}}=>{{{D = 2*6 -3* 4}}}=>{{{D = 12 -12}}}=>{{{D = 0}}}

since determinant equal to zero, this system has no solution


2) {{{4x+y=6}}}
   {{{-8x-2y=21}}}

in this case {{{a=4}}},{{{b=1}}},{{{c=6}}},{{{d=-8}}},{{{e=-2}}},and {{{f=21}}}

then you have:

determinant  {{{D = a e - b d}}}=>{{{D = 4(-2) - 1(-8)}}}=>{{{D =-8 - (-8)}}}=>
{{{D =-8+8}}}=>{{{D =0}}}

so, determinant equal to zero, this system has no solution



{{{highlight(Graphical)}}}


1) {{{2x+3y=18}}}
     {{{4x+6y=12}}}

*[invoke solve_by_graphing 2, 3, 18, 4, 6, 12] 


2) {{{4x+y=6}}}
   {{{-8x-2y=21}}}

*[invoke solve_by_graphing 4, 1, 6, -8, -2, 21]