Question 1124476
{{{1/(4x+3)^2=16/49}}}


{{{1/(4x+3)^2=(4/7)^2}}}


{{{(4x+3)^2=(7/4)^2}}}    *

{{{4x+3=0+- 7/4}}}

{{{4x=-3+- 7/4}}}

{{{x=-3/4+- 7/16}}}

{{{x=-12/16+- 7/16}}}

{{{system(highlight_green(x=-19/16),or,highlight_green(x=-5/16))}}}





*
--------mistake made here---------------------------------

{{{cross(4x+3=0+- sqrt(7)/2)}}}

{{{4x=-3+- sqrt(7)/2}}}

{{{cross(x=-3/4+- sqrt(7)/8)}}}
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