5y + 20
2x - 5
30
x + 25
4y + 35
30

Question

Accepted Answer

To solve this problem, we need to use the properties of triangles and transversals. The figure shows two triangles, $ \triangle ABC $ and $ \triangle CDE $, with a transversal line $ AC $.

### Given:
- $ AB = 5y + 20 $
- $ BC = 30 $
- $ AC = 2x - 5 $
- $ CD = 30 $
- $ DE = x + 25 $
- $ CE = 4y + 35 $

Since $ BC = CD = 30 $, triangles $ \triangle ABC $ and $ \triangle CDE $ share a common side $ BC $ and are isosceles.

### To find $ x $ and $ y $:

1. **Using the property of isosceles triangles:**
   - In $ \triangle ABC $, since $ BC = 30 $ and $ AB = 5y + 20 $, set them equal:
     $$
     5y + 20 = 30
     $$
     $$
     5y = 10 \implies y = 2
     $$

2. **Similarly, for $ \triangle CDE $:**
   - Since $ CD = 30 $ and $ DE = x + 25 $, set them equal:
     $$
     x + 25 = 30
     $$
     $$
     x = 5
     $$

Thus, the values are $ x = 5 $ and $ y = 2 $.

5y + 20 2x - 5 30 x + 25 4y + 35 30

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