5. Bao is considering getting a part-time job. Instead of taking money out of his account each month, he would put money in. His account still has \( \$ 175 \), and his goal is to have at least \( \$ 1,000 \) in the account a year from now. a Write an inequality where \( x \) represents the amount of money Bao should put in each month to reach his goal. b Solve the inequality you wrote and explain what the solutions mean in Bao's situation.

To solve Bao's problem, we will follow the steps outlined: ### Step 1: Extract Known Conditions - Current amount in the account: \( \$ 175 \) - Goal amount in the account: \( \$ 1,000 \) - Time frame: 1 year (which is 12 months) - Let \( x \) represent the amount of money Bao puts in each month. ### Step 2: Write the Inequality To find out how much Bao needs to put in each month, we can set up the following inequality: \[ 175 + 12x \geq 1000 \] This inequality states that the current amount in the account plus the total amount he will deposit over 12 months must be at least \( \$ 1,000 \). ### Step 3: Solve the Inequality Now, we will solve the inequality \( 175 + 12x \geq 1000 \). 1. Subtract \( 175 \) from both sides: \[ 12x \geq 1000 - 175 \] \[ 12x \geq 825 \] 2. Divide both sides by \( 12 \): \[ x \geq \frac{825}{12} \] Now, we will calculate \( \frac{825}{12} \). ### Step 4: Calculate the Value Let's perform the calculation: \[ \frac{825}{12} = 68.75 \] ### Step 5: Interpret the Solution The solution to the inequality is: \[ x \geq 68.75 \] This means that Bao needs to put in at least \( \$ 68.75 \) each month to reach his goal of having at least \( \$ 1,000 \) in his account after one year. ### Conclusion In summary: - **Inequality**: \( 175 + 12x \geq 1000 \) - **Solution**: \( x \geq 68.75 \) This indicates that if Bao deposits \( \$ 68.75 \) or more each month, he will achieve his goal of \( \$ 1,000 \) in his account by the end of the year.