Question 983956
{{{ x*y = 1 }}}
{{{ y = 1/x }}}
by substitution:
{{{ 2x + 3*(1/x) = 5 }}}
{{{ 2x^2 + 3 = 5x }}}
{{{ 2x^2 - 5x + 3 = 0 }}}
Use quadratic formula
{{{ x = ( -b +- sqrt( b^2-4*a*c )) / (2*a) }}}
{{{ a = 2 }}}
{{{ b = -5 }}}
{{{ c = 3 }}}
{{{ x = ( -(-5) +- sqrt( (-5)^2 - 4*2*3 )) / (2*2) }}}
{{{ x = ( 5 +- sqrt( 25 - 24 )) / 4 }}}
{{{ x = ( 5 + 1 ) / 4 }}}
{{{ x = 3/2 }}}
and
{{{ x = ( 5 - 1 ) / 4 }}}
{{{ x = 1 }}}
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and
{{{ y = 1/x }}}
{{{ y = 2/3 }}}
and
{{{ y = 1 }}}
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The solutions are:
( 3/2, 2/3 ) and ( 1,1 )
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check:
{{{ 2x + 3y = 5 }}}
{{{ 2*(3/2) + 3*(2/3) = 5 }}}
{{{ 3 + 2 = 5 }}}
{{{ 5 = 5 }}}
OK
{{{ 2x + 3y = 5 }}}
{{{ 2*1 + 3*1 = 5 }}}
{{{ 2 + 3 = 5 }}}
{{{ 5 = 5 }}}
OK