Question 470220
You can think of {{{ y = abs( 2x + 3 ) }}} as 
the same as {{{ y = 2x + 3 }}}, except that
whenever {{{y}}} is negative, you just ignore the
minus sign and make {{{y}}} positive.
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Note that {{{y = 0}}} when {{{ x = -3/2 }}}, since
{{{ y = abs(2*(-3/2) + 3) }}}
{{{ y = abs(-3 + 3) }}}
{{{ y = 0 }}}
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When {{{ x < -3/2 }}}, then {{{ y = 2x + 3 }}}
becomes negative, but you just ignore the minus sign.
Graphically, that amounts to reflecting the line
about the y-axis so that {{{y}}} ends up positive
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Here's the graph so far:
{{{ graph( 400, 400, -10, 10, -10, 10, abs(2x + 3) ) }}}
and, adding {{{ y = 1 - x }}},
{{{ graph( 400, 400, -10, 10, -10, 10, abs(2x + 3),  1 - x ) }}}
Now I want all {{{x}}} for which {{{ x + abs(2x + 3) = 1 }}}
Now subtract {{{x}}} from both sides
{{{ abs(2x + 3) = 1 - x }}}
This is just the solutions, or intersections of the graphs.
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The reflected part of {{{ y = 2x + 3 }}} will have slope = {{{-2}}}
and the y-intercept is {{{ y = -3 }}} so the equation is {{{y = -2x - 3 }}}
So I want the solutions to:
(1) {{{ y = 2x + 3 }}}
(2) {{{ y = -2x - 3 }}}
(3) {{{ y = 1 - x }}}
(3) and (1):
{{{ 1 - x = 2x + 3 }}}
{{{ 3x = -2 }}}
{{{ x = -2/3 }}}
and
{{{ y = 1 - x }}}
{{{ y = 1 -(-2/3) }}}
{{{ y = 5/3 }}}
(-2/3, 5/3) is a solution
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(3) and (2):
{{{ 1 - x = -2x - 3 }}}
{{{ x = -4 }}}
and
{{{ y = 1 - x }}}
{{{ y = 1 -(-4) }}}
{{{ y = 5 }}}
(-4, 5) is a solution
The answer to (II) is {{{x = -2/3}}}
and {{{x = -4}}}