Question 972391

{{{3x +y = 6}}}..........eq.1
x+ y + 2 I guess you have {{{x+ y = 2}}}..........eq.2

start with {{{x+ y = 2}}}..........eq.2 and solve for {{{x}}}

{{{x = 2-y}}}........substitute in eq.1

{{{3(2-y) +y = 6}}}..........eq.1 and solve for {{{y}}}

{{{6-3y +y = 6}}}

{{{6-2y = 6}}}

{{{6-6 = 2y}}}

{{{0 = 2y}}}

{{{highlight(y=0)}}}.....substitute in {{{x = 2-y}}}


{{{x = 2-0}}}

{{{highlight(x = 2)}}}

so, intersection point for given lines is: ({{{2}}},{{{0}}})


{{{ graph( 600,600, -10, 10, -10, 10,  2-x, -3x + 6) }}}