Answer
The dimensions of each plot should have $10$ ft $\times 8$ ft or $12$ ft $\times \frac{20}{3}$ ft.
Work Step by Step
Let $x$ ft and $y$ ft be the length and width of each plot.
Since the area is $80$ ft$^2$, we have
$x\cdot y=80$
Meanwhile, she has 88 ft of fencing material, then
$4x+6y=88$
Now, we have the system:
$xy=80$
$4x+6y=88$
Multiplying the second equation by $\frac{1}{2}x$,
$2x^2+3xy=44x$
Substituting the first equation,
$2x^2+240=44x$
$2x^2-44x+240=0$ $(\div 2$)
$x^2-22x+120=0$ (Factorize)
$(x-10)(x-12)=0$
$x=10$ or $x=12$
If $x=10$, then $y=\frac{80}{x}=\frac{80}{10}=8$.
If $x=12$, then $y=\frac{80}{x}=\frac{80}{12}=\frac{20}{3}$.
Therefore, the dimensions of each plot should have $10$ ft $\times 8$ ft or $12$ ft $\times \frac{20}{3}$ ft.
