$\overleftrightarrow{\text{AE}},\overleftrightarrow{\text{BF}},\overleftrightarrow{\text{CG}}$and $\overleftrightarrow{\text{DH}}$are parallel, with $\text{EF}=\text{FG}=\text{GH}$. Complete.

If $\text{AB}=\mathbf{12}$, $\text{BC}=\mathbf{2}\text{x}+\mathbf{3}\text{y}$and $\text{BD}=\mathbf{8}\text{x}$, then $\text{x}=$ ? and $\text{y}=$ ? .

The parallel lines are shown as:

AE, BF, CG and DH are parallel to each other and $EF=FG=GH$. Length of BC and BD are $2x+3y$and $8x$ respectively. Length of AB is 12 units.

Here, ADHE is the trapezoid as AE is parallel to DH.

As $EF=FG=GH$ and AE, BF, CG and DH are parallel to each other, then, the lines BF and CG are the midsegment of the trapezoid.

Hence,

$AB=BC=CD$

The length of BD is:

$\begin{array}{c}BD=BC+CD\\ =BC+BC\\ BD=2BC\end{array}$

The length of BD is:

$\begin{array}{c}BD=2BC\\ 8x=2\left(2x+3y\right)\\ 8x=4x+6y\\ 4x=6y\end{array}$

Then,

$2x=3y$ …… (1)

The length of BC is:

$\begin{array}{c}BC=AB\\ 2x+3y=12\\ 3y+3y=12\left(\text{Fromeq.}\left(\text{1}\right)\right)\\ 6y=12\\ y=2\text{units}\end{array}$

Substitute 12 for y in equation (1) as follows:

$\begin{array}{c}2x=3y\\ 2x=3\left(2\right)\\ 2x=6\\ x=3\text{units}\end{array}$

Thus, the length of x is 3 units and y is 2 units.

Therefore, the length of x and y is 3 units and 2 units respectively.