Question 1075580
{{{ x+2y+3=6 }}}   (1)
{{{ 2x+3y+4+6-x+y=4 }}}   (2)
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First, simplify each:
{{{ x+2y = 3 }}}     (1')
{{{ x+4y = -6 }}}    (2')

Ordinarily, you'd now multiply one equation by some constant so that either the coefficient of 'x' (or 'y') would match in the two equations, but we already have '1x' in each equation, so we can eliminate x by subtracting (2') from (1'):
  {{{ x - x + 2y - 4y = 3-(-6) }}}
  {{{            -2y = 9 }}}
  {{{   y = -9/2 }}}

Find 'x' using (1') (or (2') if you wish):
{{{x + 2y = 3 }}}  —> {{{ x + 2(-9/2) = 3 }}} —>  {{{ x = 3 + 9 = 12 }}}

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Ans:  {{{x=12}}}  and {{{y=-9/2}}}
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Check using (2'):  12+4(-9/2) = 12 + (-18) = -6   (ok)
Check using (1') (we used (1') to find 'x' so this better check!):   12 + 2(-9/2) = 12 -9 = 3 (ok)