Question 1203939
{{{x = 5y - 1}}}.....eq.1
{{{x + 2y = 13}}}.....eq.2
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{{{x + 2y = 13}}}.....eq.2, substitute {{{x}}} from eq.1

{{{5y - 1 + 2y = 13}}}

{{{7y = 13+1}}}

{{{7y = 14}}}

{{{y = 14/7}}}

{{{y = 2}}}


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{{{x = 5y - 1}}}.....eq.1, substitute {{{y}}}

{{{x = 5*2 - 1}}}

{{{x = 9}}}



solution:

{{{x = 9}}}

{{{y = 2}}}


{{{ drawing( 600, 600, -10, 10, -10, 10,circle(9,2,.12),locate(8.5,2,p(9,2)),
graph( 600, 600, -10, 10, -10, 10, (x+1)/5, (13-x)/2)) }}}