Question 774225
You can do this by substitution
{{{ x + 2y = 6 }}}
{{{ x = 6 - 2y }}}
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{{{ x^2 + y^2 = 17 }}}
{{{ ( 6 - 2y )^2 + y^2 = 17 }}}
{{{ 36 - 24y + 4y^2 + y^2 = 17 }}}
{{{ 5y^2 - 24y + 19 = 0 }}}
{{{ y = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 5 }}}
{{{ b = -24 }}}
{{{ c = 19 }}}
{{{ y = (-(-24) +- sqrt( (-24)^2 - 4*5*19 )) / (2*5) }}}
{{{ y = ( 24 +- sqrt( 576 - 380 )) / 10 }}}
{{{ y = ( 24 +- sqrt( 196 )) / 10 }}}
{{{ y = ( 24 +- 14) / 10 }}}
{{{ y = 38/10 }}}
{{{ y = 3.8 }}}
and
{{{ y = 10/10 }}}
{{{ y = 1 }}}
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Plug these results back into either equation
{{{ x + 2y = 6 }}}
{{{ x + 2*3.8 = 6 }}}
{{{ x + 7.6 = 6 }}}
{{{ x = -1.6 }}}
So you have the point ( -1.6, 3.8 )
and, plugging in  {{{ y = 1 }}}
{{{ x + 2y = 6 }}}
{{{ x + 2*1 = 6 }}}
{{{ x = 4 }}}
This is the point ( 4,1 )
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Here's the plot:
{{{ graph( 400, 400, -8, 8, -8, 8,  -(1/2)*x + 3, sqrt( 17 - x^2 ), -sqrt( 17 - x^2 ) ) }}}