Question 945142
{{{y = 2x + 7}}}......eq.1
{{{y = 5x + 4}}}......eq.2
-----------------------------substitute {{{5x + 4}}} for {{{y}}} in eq.1

{{{5x + 4 = 2x + 7}}}......eq.1...solve for {{{x}}}

{{{5x-2x = 7-4}}}

{{{3x = 3}}}

{{{x = 3/3}}}

{{{highlight(x = 1)}}}

go to {{{y = 5x + 4}}}......eq.2 substitute {{{1}}} for {{{x}}} and solve for {{{y}}}

 {{{y = 5*1 + 4}}}

 {{{highlight(y = 9)}}}


{{{drawing( 600, 600, -10, 10, -10, 10,circle(1,9,.12),locate(1,9,p(1,9)), graph( 600, 600, -10, 10, -10, 10, 5x + 4, 2x + 7)) }}}