Question 952013
{{{3x + y = 12}}}...eq.1
{{{y = 3x + 12}}}...eq.2
--------------------------substitute {{{y}}} from eq.2 in eq.1

{{{3x + 3x + 12= 12}}}...eq.1...solve for {{{x}}}

{{{6x= 12-12}}}

{{{6x= 0}}}

{{{highlight(x=0)}}}

{{{y = 3x + 12}}}...eq.2...substitute {{{0}}} for {{{x}}}

{{{y = 3*0 + 12}}}

{{{highlight(y = 12)}}}

the linear system has one solution, and intersection point is: ({{{0}}},{{{12}}})


{{{ graph( 600, 600, -10, 10, -10, 15, 3x + 12, -3x + 12) }}}