Together, Michael and Sarah can mow their lawn in 2 hours. Michael alone can mow the lawn in x hours.

a) Write an equation to find how long it takes Sarah to mow the lawn by herself (in terms of x).

b)  If it takes Sarah 3 hours longer than Michael to mow the lawn by herself, solve the equation you wrote in part a) to find how long it takes each of them to mow the lawn by themselves.

Let's start by analyzing the given information and formulating the equations step-by-step.

### Part a) Writing the equation

We know the following:

1. Together, Michael and Sarah can mow the lawn in 2 hours.
2. Michael alone can mow the lawn in \( x \) hours.
3. We need to find how long it takes Sarah to mow the lawn by herself, in terms of \( x \).

First, let's define some terms:
- Let \( x \) be the number of hours it takes Michael to mow the lawn by himself.
- Let \( y \) be the number of hours it takes Sarah to mow the lawn by herself.

The rate at which Michael mows the lawn is \( \frac{1}{x} \) lawns per hour.
The rate at which Sarah mows the lawn is \( \frac{1}{y} \) lawns per hour.

Together, their combined rate is \( \frac{1}{2} \) lawns per hour.

The equation representing their combined rates is:
\[ \frac{1}{x} + \frac{1}{y} = \frac{1}{2} \]

### Part b) Solving the equation

We are also given that it takes Sarah 3 hours longer than Michael to mow the lawn by herself. Therefore, we can express \( y \) in terms of \( x \):
\[ y = x + 3 \]

Now, substitute \( y \) in the combined rate equation:
\[ \frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2} \]

To solve this equation, we first find a common denominator for the left-hand side:
\[ \frac{(x + 3) + x}{x(x + 3)} = \frac{1}{2} \]
\[ \frac{2x + 3}{x(x + 3)} = \frac{1}{2} \]

Next, cross-multiply to eliminate the fraction:
\[ 2(2x + 3) = x(x + 3) \]
\[ 4x + 6 = x^2 + 3x \]

Rearrange the equation to form a standard quadratic equation:
\[ x^2 + 3x - 4x - 6 = 0 \]
\[ x^2 - x - 6 = 0 \]

Solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -6 \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 + 24}}{2} \]
\[ x = \frac{1 \pm \sqrt{25}}{2} \]
\[ x = \frac{1 \pm 5}{2} \]

This gives us two solutions:
\[ x = \frac{6}{2} = 3 \]
\[ x = \frac{-4}{2} = -2 \]

Since time cannot be negative, we discard \( x = -2 \) and accept \( x = 3 \).

Therefore, it takes Michael 3 hours to mow the lawn by himself.

Now, substituting \( x = 3 \) into \( y = x + 3 \):
\[ y = 3 + 3 = 6 \]

Thus, it takes Sarah 6 hours to mow the lawn by herself.

### Summary
- Michael takes 3 hours to mow the lawn by himself.
- Sarah takes 6 hours to mow the lawn by herself.

Do you want more details or have any questions?

Here are 5 questions you might want to ask:
1. How would the equation change if they could mow the lawn together in a different time?
2. What if Sarah could mow the lawn in 4 hours more than Michael?
3. How can we verify the solution by substituting back into the original equations?
4. What are some real-world applications of these types of rate problems?
5. Can you show how to solve this problem graphically?

**Tip:** When solving word problems, carefully define each variable and write down what they represent to avoid confusion later on.

Explore how Michael and Sarah's lawn mowing rates are analyzed and solved mathematically. Discover how their combined efficiency and individual rates are calculated with detailed steps and explanations. Suitable for Grades 7-9 students learning about rates and quadratic equations.

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