Question 927106
An equation that looks like this typically
has a solution set of 2 values for {{{ x }}}
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{{{ abs( 1 - x ) = abs( 2x + 3 ) }}}
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The way to think of this is to simplify it
Suppose you have:
{{{ abs( j ) = abs( k ) }}}
This looks like {{{ j = k }}}, but {{{ j = -k }}}
also works ( definition of absolute value )
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Now go back to your problem and do the same thing
(1) {{{ 1 - x = 2x + 3 }}} works, but
(2) {{{ 1 - x = -( 2x + 3 ) }}} also works
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Solve:
(1) {{{ 1 - x = 2x + 3 }}}
(1) {{{ 3x = -2 }}}
(1) {{{ x = -2/3 }}}
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Solve:
(2) {{{ 1 - x = -( 2x + 3 ) }}}
(2) {{{ 1 - x = -2x - 3 }}}
(2) {{{ x = -4 }}}
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The solutions are:
{{{ x = -2/3 }}}
{{{ x = -4 }}}
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check:
{{{ abs( 1 - x ) = abs( 2x + 3 ) }}}
{{{ abs( 1 - (-2/3) ) = abs( 2*(-2/3) + 3 ) }}}
{{{ abs( 5/3 ) = abs( -4/3 + 9/3 ) }}}
{{{ abs( 5/3 ) = abs( 5/3 ) }}}
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{{{ abs( 1 - x ) = abs( 2x + 3 ) }}}
{{{ abs( 1 - (-4) ) = abs( 2*(-4) + 3 ) }}}
{{{ abs( 5 ) = abs( -8 + 3 ) }}}
{{{ abs( 5 ) = abs( -5 ) }}}
OK