If 8 pens and 4 rulers cost $116 and 2 pens and 6 rulers cost$54. Find the cost of 30 pens and 30 rulers.

Let's solve this problem step-by-step.

### Step 1: Define Variables
Let:
- $ x $ be the cost of one pen.
- $ y $ be the cost of one ruler.

### Step 2: Set Up Equations
From the problem statement:
1. The cost of 8 pens and 4 rulers is $116:
   \[
   8x + 4y = 116
   \]
2. The cost of 2 pens and 6 rulers is $54:
   \[
   2x + 6y = 54
   \]

### Step 3: Solve the System of Equations
#### Simplify the equations:
1. Divide the first equation by 4:
   \[
   2x + y = 29
   \]
2. Keep the second equation as it is:
   \[
   2x + 6y = 54
   \]

#### Subtract the first simplified equation from the second:
\[
(2x + 6y) - (2x + y) = 54 - 29
\]
\[
5y = 25
\]
\[
y = 5
\]

#### Substitute $ y = 5 $ back into the first simplified equation:
\[
2x + 5 = 29
\]
\[
2x = 24
\]
\[
x = 12
\]

### Step 4: Calculate the Cost of 30 Pens and 30 Rulers
Now that we know $ x = 12 $ and $ y = 5 $:
- The cost of 30 pens: $ 30 \times 12 = 360 $
- The cost of 30 rulers: $ 30 \times 5 = 150 $

Total cost:
\[
360 + 150 = 510
\]

### Answer
The cost of 30 pens and 30 rulers is $510.

Would you like a more detailed explanation or have any questions?

Here are five related questions to deepen your understanding:
1. What if the number of pens and rulers were different, how would you set up the equations?
2. How would the solution change if the price of the items was given in another currency?
3. Can you find the cost of 10 pens and 15 rulers using the same equations?
4. How do you determine if a system of equations is consistent and has a unique solution?
5. What are other methods to solve a system of linear equations besides subtraction?

**Tip:** In problems with two unknowns, elimination and substitution are both effective methods for solving systems of equations. Choose the one that simplifies calculations.

If 8 pens and 4 rulers cost $116 and 2 pens and 6 rulers cost $54, find the cost of 30 pens and 30 rulers.

Explore how to find the cost of 30 pens and 30 rulers by solving a system of linear equations. Given that 8 pens and 4 rulers cost $116, and 2 pens and 6 rulers cost $54, this problem demonstrates step-by-step methods for solving linear equations. Suitable for Grades 7-9.

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