Question 250930
1.{{{x^2 + xy = 7}}}
2.{{{x + 2y = 5}}}
From 2,
{{{2y=5-x}}}
{{{y=(1/2)(5-x)}}}
{{{xy=(1/2)(5x-x^2)}}}
Substitute into 1,
1.{{{x^2 + xy = 7}}}
{{{x^2 + (1/2)(5x-x^2) = 7}}}
{{{2x^2 + 5x-x^2 = 14}}}
{{{x^2+5x-14=0}}}
{{{(x+7)(x-2)=0}}}
Two solutions:
{{{ x+7=0 }}}
{{{ x=-7}}}
{{{ 2y=5-x=5-(-7)=12}}}
{{{ y=6}}}
({{{ -7}}},{{{ 6}}})
.
.
.
{{{ x-2=0}}}
{{{ x=2}}}
{{{ 2y=5-x=5-2=3}}}
{{{ y=3/2}}}
({{{ 2}}},{{{ 3/2}}})
Graphically, here is the solution also,
{{{drawing( 300, 300, -10, 10, -10, 10,grid( 1 ),circle( -7, 6, .3 ),
circle( 2, 1.5, .3 ),
graph( 300,300,-10, 10, -10, 10, (7-x^2)/x, (1/2)*(5-x))) }}}