Let's denote:

• $w$ as the width of the rectangle (in cm)
• $l$ as the length of the rectangle (in cm)

Given conditions:

1. The length is 3 more than twice the width: $l = 2w + 3$

2. The perimeter of the rectangle is 42 cm: The formula for the perimeter of a rectangle is: $P = 2l + 2w$ Substituting the perimeter: $2l + 2w = 42$ Simplifying by dividing both sides by 2: $l + w = 21$

System of Equations

We now have the following system:

l = 2w + 3 \\ l + w = 21 \end{cases}$$ ### Solving the System 1. Substitute $$l = 2w + 3$$ from the first equation into the second equation: $$(2w + 3) + w = 21$$ 2. Combine like terms: $$3w + 3 = 21$$ 3. Subtract 3 from both sides: $$3w = 18$$ 4. Divide by 3: $$w = 6$$ 5. Substitute $$w = 6$$ back into the first equation to find $$l$$: $$l = 2(6) + 3 = 12 + 3 = 15$$ ### Solution The dimensions of the rectangle are: - **Width**: $$w = 6$$ cm - **Length**: $$l = 15$$ cm Would you like more details or have any questions? Here are some related questions to deepen understanding: 1. What would the perimeter be if the dimensions were doubled? 2. How would the system of equations change if the length were 5 more than twice the width? 3. Can you calculate the area of the rectangle with these dimensions? 4. What if the length were three times the width? How would you set up the equations? 5. How can you verify that the perimeter and area match these dimensions? **Tip:** When solving systems of equations, substitution is often useful when one equation is already solved for a variable.