We have a figure with two triangles, $\triangle TPQ$ and $\triangle SRQ$. We are given that $PT$ and $RS$ are parallel. We are also given the lengths of the sides $PQ = 6-x$, $SQ = 3+x$, $TQ = 3$, and $RQ = 6+x$. We need to find the value of $x$.

1. Problem Description

We have a figure with two triangles, $\triangle TPQ$ and $\triangle SRQ$. We are given that $PT$ and $RS$ are parallel. We are also given the lengths of the sides $PQ = 6-x$, $SQ = 3+x$, $TQ = 3$, and $RQ = 6+x$. We need to find the value of $x$.

2. Solution Steps

Since $PT$ and $RS$ are parallel, we have that $\angle TPQ = \angle RSQ$ and $\angle PTQ = \angle SRQ$. Also, $\angle PQT = \angle SQR$ because they are vertical angles. Therefore, $\triangle TPQ \sim \triangle RSQ$ by Angle-Angle-Angle (AAA) similarity.

Since the triangles are similar, the ratios of corresponding sides are equal. We have:

$\frac{PQ}{RQ} = \frac{TQ}{SQ}$

Substituting the given values:

$\frac{6-x}{6+x} = \frac{3}{3+x}$

Cross-multiply:

$(6-x)(3+x) = 3(6+x)$

$18 + 6x - 3x - x^2 = 18 + 3x$

$18 + 3x - x^2 = 18 + 3x$

$-x^2 = 0$

$x^2 = 0$

$x = 0$

3. Final Answer

$x = 0$