Question 204074
It's probably a good idea to draw yourself a diagram of this situation.
Draw a larger rectangle to represent the garden with the gravel path in place, then inside of that rectangle, draw a smaller similar rectangle to represent the 11ft. by 15ft. garden.
Let the width of the gravel path be x ft.
You need to write the area of the gravel path in terms of x, its width. 
Think...write an expression for the whole thing, garden plus gravel path, then subtract the area of the garden alone. This will give you an expression for the area of the gravel path.
Step 1. Area of the whole thing:
{{{A = L*W}}}
The length of the whole thing can be expressed as L = 15+2x
The width can be expressed as W = 11+2x
The area is:
{{{A[w] = (15+2x)(11+2x)}}} Use the FOIL method to multiply.
{{{A[w] = 165+52x+4x^2}}} or {{{highlight(A[w] = 4x^2+52x+165)}}} Now we subtract the area of the garden alone and this is:
Step 2. Area of the gravel path.
{{{A[g] = L*W}}} Substitute L = 15 and W = 11
{{{A[g] = (15)*(11)}}}
{{{highlight_green(A[g] = 165)}}}
So the area of the gravel path alone is:
{{{A[p] = highlight(A[w])-highlight_green(A[g])}}} or:
{{{A[p] = highlight(4x^2+52x+165)-highlight_green(165)}}} and this is equal to 192 sq.ft., so...
{{{4x^2+52x = 192}}} Subtract 192 from both sides.
{{{4x^2+52x-192 = 0}}} Factor out a 4 to simplify the calculations a bit.
{{{4(x^2+13x+48) = 0}}} so we can write:
{{{x^2+13x+48 = 0}}} Factor this trinomial.
{{{(x+16)(x-3) = 0}}} Apply the zero product rule:
{{{x+16 = 0}}} or {{{x-3 = 0}}} which means that...
{{{cross(x = -16)}}} or {{{x = 3}}} Discard the negative solution as the width of the gravel path must be a positive value.
The width of the gravel path is 3 feet.