Question 258767
Break up the number line into 3 distinct regions using the absolute value arguments.
3x+1=0
x=-1/3
When {{{x<-1/3}}}, {{{abs(3x + 1)=-(3x+1)}}}
When {{{x>-1/3}}}, {{{abs(3x + 1)=3x+1}}}
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3-2x=0
3=2x
x=3/2
When {{{x<3/2}}}, {{{abs(3-2x)=3-2x}}}
When {{{x>3/2}}},{{{abs(3-2x)=-(3-2x)}}}
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Here are the 3 regions.
({{{ -infinity}}},{{{-1/3}}})
{{{-(3x+1)+(3-2x)=11}}}
{{{-3x-1+3-2x=11}}}
{{{-5x+2=11}}}
{{{x=-9/5}}}
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({{{-1/3}}},{{{3/2}}})
{{{(3x+1)+(3-2x)=11}}}
{{{x+4=11}}}
{{{x=7}}}
Not in the valid region, no solution.
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({{{3/2}}},{{{infinity}}})
{{{(3x+1)-(3-2x)=11}}}
{{{3x+1-3+2x=11}}}
{{{5x-2=11}}}
{{{5x=13}}}
{{{x=13/5}}}
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Two solutions: x=-9/5 and x=13/5.