The length of a rectangular picture frame is 1IN more than twice the width. The perimeter is 119IN. Find the length and the width.

To solve for the length and width of the rectangular picture frame, let's define the variables:

- Let the width of the frame be \( w \) inches.
- The length is 1 inch more than twice the width, so the length can be expressed as \( 2w + 1 \) inches.

The formula for the perimeter \( P \) of a rectangle is given by:

\[
P = 2(\text{Length} + \text{Width})
\]

Given that the perimeter is 119 inches, we can set up the following equation:

\[
119 = 2((2w + 1) + w)
\]

### Step 1: Simplify the equation
\[
119 = 2(3w + 1)
\]

### Step 2: Distribute the 2
\[
119 = 6w + 2
\]

### Step 3: Subtract 2 from both sides
\[
117 = 6w
\]

### Step 4: Solve for \( w \)
\[
w = \frac{117}{6} = 19.5 \, \text{inches}
\]

### Step 5: Find the length
Substitute \( w = 19.5 \) into the equation for the length:

\[
\text{Length} = 2w + 1 = 2(19.5) + 1 = 39 + 1 = 40 \, \text{inches}
\]

### Final Answer:
- **Width** = 19.5 inches
- **Length** = 40 inches

Would you like further details or have any questions?

Here are 5 related questions to expand on this problem:
1. What would be the area of the frame with these dimensions?
2. If the perimeter were increased to 150 inches, how would the dimensions change?
3. How would you express the width in terms of the length?
4. What is the ratio of the length to the width?
5. How would the dimensions change if the width were 3 inches longer?

**Tip**: When solving word problems, defining variables clearly at the start helps avoid confusion during the calculation process.

Learn how to find the length and width of a rectangular picture frame where the length is 1 inch more than twice the width and the perimeter is 119 inches. This problem involves algebraic expressions and geometry, suitable for Grades 7-9 students.

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