Question 945147

{{{y = x + 4}}}......eq.1
{{{y = 3x}}}......eq.2
------------------------substitute {{{3x}}} for {{{y}}} in eq.1

{{{3x = x + 4}}}......eq.1...solve for {{{x}}}

{{{3x -x = 4}}}

{{{2x = 4}}}

{{{x = 4/2}}}

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

go to {{{y = 3x}}}......eq.2 substitute {{{2}}} for {{{x}}} and solve for {{{y}}}


{{{y = 3*2}}}

{{{highlight(y = 6)}}}


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