Question 1111881
Solve the simultaneous equation<b>s</b>:
{{{2x+y-1=0}}}
2 to the power of x+y =32   (Doesn't format correctly)
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{{{ 2x+y-1 = 0 }}}     (1)
{{{ 2^(x+y) = 32 }}}     (2)

(2) implies {{{ x+y = 5}}}  (2')   because {{{2^5 = 32 }}}

(1) can be re-written  {{{ 2x + y = 1 }}}  (1')

Subtract (1') from (2') to eliminate y   (there are other methods, this just looked easiest to me):

   {{{ x + y = 5 }}}
-  {{{ 2x + y = 1 }}}
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       -x  = 4   

x = -4  —>  y = 9

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Ans:   {{{highlight( x = -4) }}}    and  {{{ highlight(y = 9) }}}

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Check (using original equations):

                 1.  2x + y - 1 = 2(-4)+9-1 = -8+9-1 = 0  (ok)
                 2. {{{ 2^(x+y) = 2^(-4+9) = 2^5 = 32 }}}  (ok)