The problem gives the supply and demand equations for a product: Supply (S): $20p - 3q = 50$ Demand (D): $10p + 2q = 200$ The goal is to find the equilibrium point, which is the point (p, q) where the supply and demand equations are satisfied simultaneously. Here, $p$ represents the price and $q$ represents the quantity.

1. Problem Description

The problem gives the supply and demand equations for a product:

Supply (S): $20p - 3q = 50$

Demand (D): $10p + 2q = 200$

The goal is to find the equilibrium point, which is the point (p, q) where the supply and demand equations are satisfied simultaneously. Here, $p$ represents the price and $q$ represents the quantity.

2. Solution Steps

To find the equilibrium point, we need to solve the system of equations:

$20p - 3q = 50$

$10p + 2q = 200$

We can use substitution or elimination method. Let's use the elimination method. Multiply the second equation by 2:

$2(10p + 2q) = 2(200)$

$20p + 4q = 400$

Now we have the following system:

$20p - 3q = 50$

$20p + 4q = 400$

Subtract the first equation from the second equation to eliminate $p$:

$(20p + 4q) - (20p - 3q) = 400 - 50$

$20p + 4q - 20p + 3q = 350$

$7q = 350$

$q = \frac{350}{7}$

$q = 50$

Substitute $q = 50$ into either of the original equations. Let's use the second equation:

$10p + 2(50) = 200$

$10p + 100 = 200$

$10p = 200 - 100$

$10p = 100$

$p = \frac{100}{10}$

$p = 10$

Therefore, the equilibrium point is $(p, q) = (10, 50)$.

3. Final Answer

The equilibrium point is $(10, 50)$.