Hello, Catherine!

The first three can be solved with the same trick . . .

1) $\displaystyle ABC$ is a right triangle (with $\displaystyle A=90^o$). .$\displaystyle AB$ is 27 ft.

The perimeter of this triangle is 108 feet.

What is the length of $\displaystyle AC$ and $\displaystyle BC$? First, make a sketch . . . Code:

C *
| *
| * a
b | *
| *
| *
A * - - - - - - - - * B
27

Let $\displaystyle a = BC,\;b = AC$

The perimeter is 108: . $\displaystyle a + b + 27 \:=\:108\quad\Rightarrow\quad a + b \:=\:81$

Pythagorus tells us: .$\displaystyle a^2 \:=\:b^2 + 27^2\quad\Rightarrow\quad a^2-b^2 \:=\:729$

We have a system of equations: .$\displaystyle \begin{array}{cccc}a^2-b^2 &=&729 & {\color{blue}[1]}\\ a+b &=& 81 & {\color{blue}[2]}\end{array}$

Equation [1] is: .$\displaystyle (a-b)(a+b) \:=\:729$

Divide by [2]: .$\displaystyle \frac{(a-b)(a+b)}{a+b} \:=\:\frac{729}{81}\quad\Rightarrow\quad a-b\:=\:9\;\;{\color{blue}[3]}$

Add [2] and [3]: .$\displaystyle \begin{array}{ccc}a + b &=&81 \\ a-b &=&9\end{array}\quad\Rightarrow\quad 2a \:=\:90\quad\Rightarrow\quad a \:=\:45$

Substitute into [2]: .$\displaystyle 45 + b \:=\:81\quad\Rightarrow\quad b \:=\:36$

Therefore: .$\displaystyle AC = 36,\;BC = 45$ feet

2) If $\displaystyle x-y\:=\:-3$ and $\displaystyle x^2-y^2\:=\:6$, find $\displaystyle x$ and $\displaystyle y.$ Can you see that this system is identical to the system in #1?

We have: .$\displaystyle \begin{array}{cccc}x - y &=& \text{-}3 & {\color{blue}[1]}\\ (x-y)(x+y) &=& 6 & {\color{blue}[2]}\end{array}$

$\displaystyle \begin{array}{cccc}\text{Divide {\color{blue}[2]} by {\color{blue}[1]}:} & x + y &=&-2 \\

\text{Add {\color{blue}[1]}:} & x - y & =& -3\end{array}\quad\Rightarrow \quad 2x\:=\:-5\quad\Rightarrow\quad\boxed{x \:=\:-\frac{5}{2}}$

Substitute into [1]: .$\displaystyle -\frac{5}{2} - y \:=\:-3\quad\Rightarrow\quad\boxed{y \:=\:\frac{1}{2}}$

4) A garden has a shape of a rectangle.

If we remove 3 feet from the length of a garden, and add 6 feet to the width,

the garden gets the shape of a square and the total area gains 78 sq ft.

What is the length and width of the garden? I approached this one "backwards".

Let $\displaystyle x$ = side of the square garden. .Its area is: $\displaystyle x^2$ ft²

The original garden was 3 feet longer: .$\displaystyle \text{Length} \:=\:x+3$

. . and was 6 feet narrower: .$\displaystyle \text{Width} \:=\:x-6$

The area of the original garden was: .$\displaystyle (x+3)(x-6)$ ft²

The square garden has 78 ft² more than the original garden.

. . . $\displaystyle x^2 \;=\;(x+3)(x-6) + 78\quad\Rightarrow\quad x \:=\:20$

Therefore, the original garden was $\displaystyle 23 \times 14$ ft.