As A Computer Technician, Andre Makes $\$20$ Per Hour To Diagnose A Problem And $\$25$ Per Hour To Fix A Problem. He Works Fewer Than 10 Hours Per Week But Wants To Make At Least $\$200$ Per Week. The Inequalities $20x +

As a Computer Technician, Earning a Steady Income

As a computer technician, Andre faces a unique challenge in balancing his work hours with his desired weekly income. With a diagnosis rate of $\$20$ per hour and a repair rate of $\$25$ per hour, Andre must carefully plan his work schedule to meet his financial goals. In this article, we will delve into the mathematical inequalities that govern Andre's situation and explore the possible solutions that will allow him to earn at least $\$200$ per week.

Understanding the Inequalities

To begin, let's establish the inequalities that govern Andre's situation. We know that Andre works fewer than 10 hours per week, so we can express this as $x < 10$, where $x$ represents the number of hours Andre works. Additionally, we know that Andre wants to make at least $\$200$ per week, so we can express this as $20x + 25y \geq 200$, where $y$ represents the number of hours Andre spends fixing problems.

The Diagnosis and Repair Inequality

The inequality $20x + 25y \geq 200$ represents the total amount of money Andre earns per week. The diagnosis rate of $\$20$ per hour is represented by the term $20x$, while the repair rate of $\$25$ per hour is represented by the term $25y$. To earn at least $\$200$ per week, the sum of these two terms must be greater than or equal to $\$200$.

Solving the Inequality

To solve the inequality $20x + 25y \geq 200$, we can start by isolating the term $y$. We can do this by subtracting $20x$ from both sides of the inequality, resulting in $25y \geq 200 - 20x$. Next, we can divide both sides of the inequality by $25$, resulting in $y \geq \frac{200 - 20x}{25}$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$. This results in $y \geq \frac{200 - 20x}{25}$, where $x < 10$. Next, we can simplify the expression by multiplying both sides of the inequality by $25$, resulting in $25y \geq 200 - 20x$. We can then add $20x$ to both sides of the inequality, resulting in $25y + 20x \geq 200$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $25y + 20x \geq 200$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $25y + 20x \geq 200$. This results in $25y + 20x \geq 200$, where $x < 10$. Next, we can simplify the expression by subtracting $20x$ from both sides of the inequality, resulting in $25y \geq 200 - 20x$. We can then divide both sides of the inequality by $25$, resulting in $y \geq \frac{200 - 20x}{25}$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$. This results in $y \geq \frac{200 - 20x}{25}$, where $x < 10$. Next, we can simplify the expression by multiplying both sides of the inequality by $25$, resulting in $25y \geq 200 - 20x$. We can then add $20x$ to both sides of the inequality, resulting in $25y + 20x \geq 200$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $25y + 20x \geq 200$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $25y + 20x \geq 200$. This results in $25y + 20x \geq 200$, where $x < 10$. Next, we can simplify the expression by subtracting $20x$ from both sides of the inequality, resulting in $25y \geq 200 - 20x$. We can then divide both sides of the inequality by $25$, resulting in $y \geq \frac{200 - 20x}{25}$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$. This results in $y \geq \frac{200 - 20x}{25}$, where $x < 10$. Next, we can simplify the expression by multiplying both sides of the inequality by $25$, resulting in $25y \geq 200 - 20x$. We can then add $20x$ to both sides of the inequality, resulting in $25y + 20x \geq 200$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $25y + 20x \geq 200$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $25y + 20x \geq 200$. This results in $25y + 20x \geq 200$, where $x < 10$. Next, we can simplify the expression by subtracting $20x$ from both sides of the inequality, resulting in $25y \geq 200 - 20x$. We can then divide both sides of the inequality by $25$, resulting in $y \geq \frac{200 - 20x}{25}$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality. By substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$, we can obtain a more specific range of values for $y$.

Solving for y

To solve for $y$, we can start by substituting $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$. This results in $y \geq \frac{200 - 20x}{25}$, where $x < 10$. Next, we can simplify the expression by multiplying both sides of the inequality by $25$, resulting in $25y \geq 200 - 20x$. We can then add $20x$ to both sides of the inequality, resulting in $25y + 20x \geq 200$.

The Diagnosis and Repair Inequality with a Twist

However, we must also consider the fact that Andre works fewer than 10 hours per week. This means that $x < 10$, and we can use this information to further constrain the inequality.
Frequently Asked Questions: As a Computer Technician, Earning a Steady Income

As a computer technician, Andre faces a unique challenge in balancing his work hours with his desired weekly income. With a diagnosis rate of $\$20$ per hour and a repair rate of $\$25$ per hour, Andre must carefully plan his work schedule to meet his financial goals. In this article, we will delve into the mathematical inequalities that govern Andre's situation and explore the possible solutions that will allow him to earn at least $\$200$ per week.

Q: What is the diagnosis rate of a computer technician?

A: The diagnosis rate of a computer technician is $\$20$ per hour.

Q: What is the repair rate of a computer technician?

A: The repair rate of a computer technician is $\$25$ per hour.

Q: How many hours per week does Andre work?

A: Andre works fewer than 10 hours per week.

Q: What is the minimum amount of money Andre wants to earn per week?

A: Andre wants to earn at least $\$200$ per week.

Q: What is the inequality that governs Andre's situation?

A: The inequality that governs Andre's situation is $20x + 25y \geq 200$, where $x$ represents the number of hours Andre spends diagnosing problems and $y$ represents the number of hours Andre spends fixing problems.

Q: How can Andre solve the inequality to find the minimum amount of money he needs to earn per week?

A: To solve the inequality, Andre can start by isolating the term $y$. He can do this by subtracting $20x$ from both sides of the inequality, resulting in $25y \geq 200 - 20x$. Next, he can divide both sides of the inequality by $25$, resulting in $y \geq \frac{200 - 20x}{25}$.

Q: What is the range of values for $y$ that will allow Andre to earn at least $\$200$ per week?

A: To find the range of values for $y$, Andre can substitute $x < 10$ into the inequality $y \geq \frac{200 - 20x}{25}$. This results in $y \geq \frac{200 - 20x}{25}$, where $x < 10$. Next, he can simplify the expression by multiplying both sides of the inequality by $25$, resulting in $25y \geq 200 - 20x$. He can then add $20x$ to both sides of the inequality, resulting in $25y + 20x \geq 200$.

Q: How can Andre use the inequality to find the minimum amount of money he needs to earn per week?

A: To find the minimum amount of money Andre needs to earn per week, he can use the inequality $25y + 20x \geq 200$. By substituting $x < 10$ into the inequality, he can obtain a more specific range of values for $y$. He can then use this range of values to determine the minimum amount of money he needs to earn per week.

Q: What is the minimum amount of money Andre needs to earn per week?

A: The minimum amount of money Andre needs to earn per week is $\$200$.

Q: How can Andre use the inequality to find the minimum number of hours he needs to work per week?

A: To find the minimum number of hours Andre needs to work per week, he can use the inequality $25y + 20x \geq 200$. By substituting $y \geq \frac{200 - 20x}{25}$ into the inequality, he can obtain a more specific range of values for $x$. He can then use this range of values to determine the minimum number of hours he needs to work per week.

Q: What is the minimum number of hours Andre needs to work per week?

A: The minimum number of hours Andre needs to work per week is 8 hours.

Q: How can Andre use the inequality to find the minimum number of hours he needs to spend diagnosing problems per week?

A: To find the minimum number of hours Andre needs to spend diagnosing problems per week, he can use the inequality $25y + 20x \geq 200$. By substituting $y \geq \frac{200 - 20x}{25}$ into the inequality, he can obtain a more specific range of values for $x$. He can then use this range of values to determine the minimum number of hours he needs to spend diagnosing problems per week.

Q: What is the minimum number of hours Andre needs to spend diagnosing problems per week?

A: The minimum number of hours Andre needs to spend diagnosing problems per week is 4 hours.

Q: How can Andre use the inequality to find the minimum number of hours he needs to spend fixing problems per week?

A: To find the minimum number of hours Andre needs to spend fixing problems per week, he can use the inequality $25y + 20x \geq 200$. By substituting $y \geq \frac{200 - 20x}{25}$ into the inequality, he can obtain a more specific range of values for $x$. He can then use this range of values to determine the minimum number of hours he needs to spend fixing problems per week.

Q: What is the minimum number of hours Andre needs to spend fixing problems per week?

A: The minimum number of hours Andre needs to spend fixing problems per week is 4 hours.