Question 159543
I didn't get to the same place.
{{{sqrt(2-x) = 1 + sqrt(3x+1)}}}
{{{(2-x) = 1 + 2*sqrt(3x+1)+(3x+1)}}}
{{{(2-x)-1-(3x+1) = 2*sqrt(3x+1)}}}
{{{(2-1-1)-x-3x=2*sqrt(3x+1)}}}
{{{-4x=2*sqrt(3x+1)}}}
{{{16x^2=4(3x+1)}}}
{{{16x^2=12x+4}}}
I have the additional 4. 
Use the quadratic formula or factor for the solution.
{{{16x^2-12x-4=0}}}
{{{(4x+1)(4x-4)=0}}}
{{{4x+1=0}}}
{{{x=-1/4}}}
Check the answer
{{{sqrt(2-x) - sqrt(3x+1)=1}}}
{{{sqrt(2+1/4)-sqrt(3(-1/4)+1)=1}}}
{{{1.5-0.5=1}}}
{{{1=1}}}
True statement.
Good answer.
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{{{4x-4=0}}}
{{{x=1}}}
Check the answer
{{{sqrt(2-x) - sqrt(3x+1)=1}}}
{{{sqrt(2-1)-sqrt(3(1)+1)=1}}}
{{{1-2=1}}}
{{{-1=1}}}
False statement.
Not a good answer.
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One solution, 
{{{x=-1/4}}}
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To your question, if you were at that point,
{{{16x^2=12x}}}
Yes, divide both sides by 16x and you get,
{{{x=12/16=3/4}}}
But that is not a solution.