To create a real-world problem that models the equation \( 0.08x = 0.06x + 140 \), we can consider a scenario involving savings accounts with different interest rates.
Real-World Problem:
Scenario: Imagine you are deciding between two different savings accounts to put your money into.
- Account A offers an interest rate of 8% per year.
- Account B offers an interest rate of 6% per year.
You plan to invest some amount of money, denoted by \( x \), in these accounts.
After one year, you find yourself in a situation where the interest earned from Account A exceeds the interest earned from Account B by $140.
Question: How much money \( x \) do you need to invest in each account for the interest from Account A to be $140 more than the interest from Account B after one year?
Equation Explanation:
- The interest from Account A after one year can be expressed as \( 0.08x \).
- The interest from Account B after one year can be expressed as \( 0.06x \).
- The problem states that the interest from Account A is $140 more than the interest from Account B, leading to the equation:
\[ 0.08x = 0.06x + 140 \]
This equation sets up the relationship between the investments in the two accounts based on their respective interest rates and the additional income from Account A.
Solution Steps:
To find the value of \( x \):
- Subtract \( 0.06x \) from both sides:
\[ 0.08x - 0.06x = 140 \]
- This simplifies to:
\[ 0.02x = 140 \]
- To solve for \( x \), divide both sides by \( 0.02 \):
\[ x = \frac{140}{0.02} = 7000 \]
Thus, you would need to invest $7000 in each account to achieve the desired interest difference.