Question

persevere find the values of x and y if $\triangle pqs \cong \triangle rqs$.
(there is a triangle diagram with points p, q, r, s. s is on pr, qs is perpendicular to pr. in $\triangle pqs$, pq is labeled $2x$, ps is labeled $x$. in $\triangle rqs$, rq is labeled $3y + 8$, rs is labeled $2y$. there are input boxes for x and y: $x = \square$, $y = \square$)

Explanation:

Step1: Use congruent triangles' sides

Since $\triangle PQS \cong \triangle RQS$, their corresponding sides are equal. So, $PQ = RQ$ and $PS = RS$.
For $PQ = RQ$: $2x = 3y + 8$.
For $PS = RS$: $x = 2y$.

Step2: Substitute $x = 2y$ into $2x = 3y + 8$

Substitute $x$ in the first equation: $2(2y) = 3y + 8$.
Simplify: $4y = 3y + 8$.
Subtract $3y$: $y = 8$.

Step3: Find $x$ using $x = 2y$

Substitute $y = 8$: $x = 2\times8 = 16$.

Answer:

$x = 16$, $y = 8$