Linear cost and revenue equations are shown below. Find the break-even quantity.
C(x) = 13x + 11800,   R(x) = 213x

To find the break-even quantity, we need to determine the value of \( x \) where the cost \( C(x) \) equals the revenue \( R(x) \). In other words, we solve the equation:

\[
C(x) = R(x)
\]

Given:
\[
C(x) = 13x + 11800
\]
\[
R(x) = 213x
\]

Setting these equal:
\[
13x + 11800 = 213x
\]

Now, let's isolate \( x \):
1. Subtract \( 13x \) from both sides:
   \[
   11800 = 200x
   \]

2. Divide both sides by 200 to solve for \( x \):
   \[
   x = \frac{11800}{200} = 59
   \]

Therefore, the break-even quantity is:
\[
x = 59
\]

### Answer:
The break-even quantity is \( x = 59 \).

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. What is the total cost at the break-even point?
2. What is the revenue at the break-even point?
3. How would the break-even quantity change if the fixed cost increased?
4. What happens to the break-even quantity if the revenue per unit decreases?
5. Can there be multiple break-even points for non-linear cost and revenue functions?

**Tip:** The break-even point is found where total cost equals total revenue, which can reveal the minimum sales needed to avoid a loss.

Linear cost and revenue equations are shown below. Find the break-even quantity. C(x) = 13x + 11800, R(x) = 213x

Learn how to find the break-even quantity using linear cost and revenue equations. Given the cost function C(x) = 13x + 11800 and revenue function R(x) = 213x, we solve for the point where cost equals revenue. This problem is suitable for students in Grades 8-10 and involves algebraic solutions.

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