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

If $\text{AB}=\mathbf{15}$, $\text{BC}=\mathbf{2}\text{x}-\text{y}$and $\text{CD}=\text{x}+\text{y}$, 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 CD are $2x-y$and $x+y$ respectively. Length of AB is 15 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 BC is:

$BC=AB$

Then,

$2x-y=15$ …… (1)

The length of CD is:

$CD=AB$

Then,

$x+y=15$ …… (2)

Solve equations (1) and (2) as follows:

$\frac{\begin{array}{c}x+y=15\\ 2x-y=15\end{array}}{3x=10}$

Then,

$x=10$.

Substitute 10 for x in equation (2) as follows:

$\begin{array}{c}10+y=15\\ y=5\end{array}$

Thus, the length of x is 10 units and y is 5 units.

Therefore, the length of x and y is 10 units and 5 units respectively.