Question 1095283
"... he increased the length by the same amount."


Not clear if that was to one dimension or both dimensions.


Assuming x distance added to BOTH dimensions,
{{{(16+x)(12+x)}}}, the new area of the garden.


{{{(x+16)(x+12)-(16*12)=165}}}
Solve for x and evaluate {{{x+16}}} and {{{x+12}}}.
{{{x^2+16x+12x+(16*12)-16*12=165}}}
{{{x^2+28x-165=0}}}
{{{(x+33)(x-5)=0}}}
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{{{highlight_green(x=5)}}}


New dimensions are 21 feet and 17 feet.