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Let's break down the two questions shown in the image.

### Question 3: Chef's Purchase of Ribs and Chicken
We are given the equation:
\[ 0.9c + 1.89r = 17.01 \]
Where:
- $ c $ is the pounds of chicken bought (at $0.90 per pound),
- $ r $ is the pounds of ribs bought (at $1.89 per pound),
- $17.01 is the total cost for both the ribs and the chicken.

We are asked to show that the following two equations are equivalent to the original one:
- (a) $ c = 18.9 - 2.1r $
- (b) $ r = \frac{-10}{21}c + 9 $

#### (a) Verify $ c = 18.9 - 2.1r $
We need to start from the original equation and isolate $ c $:
\[
0.9c + 1.89r = 17.01
\]
Subtract $ 1.89r $ from both sides:
\[
0.9c = 17.01 - 1.89r
\]
Now, divide both sides by 0.9:
\[
c = \frac{17.01 - 1.89r}{0.9} = \frac{17.01}{0.9} - \frac{1.89r}{0.9}
\]
Simplify the fractions:
\[
c = 18.9 - 2.1r
\]
So, $ c = 18.9 - 2.1r $ is equivalent to the original equation.

#### (b) Verify $ r = \frac{-10}{21}c + 9 $
Start again from the original equation:
\[
0.9c + 1.89r = 17.01
\]
Subtract $ 0.9c $ from both sides:
\[
1.89r = 17.01 - 0.9c
\]
Now, divide both sides by 1.89:
\[
r = \frac{17.01 - 0.9c}{1.89}
\]
Simplify the terms:
\[
r = \frac{17.01}{1.89} - \frac{0.9c}{1.89} = 9 - \frac{0.9}{1.89}c
\]
Now simplify $ \frac{0.9}{1.89} $, which equals $ \frac{10}{21} $:
\[
r = 9 - \frac{10}{21}c
\]
So, $ r = 9 - \frac{10}{21}c $, which is equivalent to $ r = \frac{-10}{21}c + 9 $.

#### Explanation of Equivalent Forms:
- Equation (a) isolates $ c $, making it easier to calculate how much chicken the chef bought for different quantities of ribs.
- Equation (b) isolates $ r $, making it easier to calculate how much ribs the chef bought for different quantities of chicken.

---

### Question 4: Car Traveling at a Constant Rate

#### (a) Complete the table:
The car traveled 180 miles, so the rate of travel $ r $ is calculated as:
\[
r = \frac{\text{distance}}{\text{time}} = \frac{180}{t}
\]
We need to fill out the table:

| travel time (hours) | rate of travel (miles per hour) |
|---------------------|--------------------------------|
| 5                   | $ \frac{180}{5} = 36 $       |
| 4.5                 | $ \frac{180}{4.5} = 40 $     |
| 3                   | $ \frac{180}{3} = 60 $       |
| 2.25                | $ \frac{180}{2.25} = 80 $    |

#### (b) Equation to find rate of travel:
The equation to find the rate $ r $ (in miles per hour) for any given time $ t $ (in hours) is:
\[
r = \frac{180}{t}
\]
This makes it easy to calculate the rate for any given time.

---

Would you like any further details or have any questions on this?

Here are 5 follow-up questions:
1. How would you rearrange equation (b) to solve for $ c $?
2. If the chef bought 10 pounds of chicken, how many pounds of ribs did they buy?
3. Can you explain why isolating a variable in an equation can be helpful?
4. For the car problem, what would the rate be if the travel time was 6 hours?
5. How could the car problem equation be used to find the travel time given a rate?

**Tip:** Always isolate the variable of interest when solving word problems involving equations.

3. A chef bought $17.01 worth of ribs and chicken. Ribs cost 1.89 per pound and chicken costs 0.90 per pound. The equation 0.9c + 1.89r = 17.01 represents the relationship between the quantities in this situation. Show that each of the following equations is equivalent to 0.9c + 1.89r = 17.01. Then, explain when it might be helpful to write the equation in these forms. (a) c = 18.9 - 2.1r, (b) r = -10/21c + 9. 4. A car traveled 180 miles at a constant rate. Complete the table to show the rate at which the car was traveling if it completed the same distance in each number of hours. Write an equation that would make it easy to find the rate at which the car was traveling in miles per hour r, if it traveled for t hours.

This lesson helps students solve linear equations using real-life scenarios such as purchasing ribs and chicken, and determining constant travel rates. Learn how to manipulate equations to find values for different variables. Suitable for students in Grades 8-10.

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