Sure! Let's solve this problem step by step to determine the length of Lisa's rectangular painting.
### Step 1: Define Variables
Let [tex]\( w \)[/tex] be the width of the painting in inches.
Let [tex]\( l \)[/tex] be the length of the painting in inches.
### Step 2: Relate Length to Width
According to the problem, the length [tex]\( l \)[/tex] is three more than twice the width [tex]\( w \)[/tex]. This can be written as:
[tex]\[ l = 2w + 3 \][/tex]
### Step 3: Create the Equation for the Perimeter
The perimeter of a rectangle is the sum of all four sides, which is given as 30 inches. The formula for the perimeter [tex]\( P \)[/tex] of a rectangle is:
[tex]\[ P = 2l + 2w \][/tex]
We know that [tex]\( P = 30 \)[/tex]. Substituting [tex]\( l \)[/tex] from our earlier relationship into this equation gives us:
[tex]\[ 30 = 2(2w + 3) + 2w \][/tex]
### Step 4: Simplify and Solve the Equation
Let's simplify this equation step by step:
[tex]\[ 30 = 2(2w + 3) + 2w \][/tex]
[tex]\[ 30 = 4w + 6 + 2w \][/tex]
[tex]\[ 30 = 6w + 6 \][/tex]
Subtract 6 from both sides:
[tex]\[ 24 = 6w \][/tex]
Divide both sides by 6:
[tex]\[ w = 4 \][/tex]
### Step 5: Find the Length [tex]\( l \)[/tex]
Now that we have the width [tex]\( w = 4 \)[/tex] inches, we can calculate the length [tex]\( l \)[/tex]:
[tex]\[ l = 2w + 3 \][/tex]
[tex]\[ l = 2(4) + 3 \][/tex]
[tex]\[ l = 8 + 3 \][/tex]
[tex]\[ l = 11 \][/tex]
### Step 6: Verify the Options
We need to check the options provided in the problem to confirm which one matches our solution:
- [tex]\( 6w + 6 = 30 = 4 \)[/tex]
- [tex]\( 6w + 6 = 30 ; 11 \)[/tex]
- [tex]\( 3w + 3 = 30 ; 21 \)[/tex]
- [tex]\( 3w + 3 = 30 ; 9 \)[/tex]
From our calculations and the matching logical option, we can see the correct result is in the set:
[tex]\[ 6w + 6 = 30 ; 11 \][/tex]
Thus, the length of the painting is 11 inches.
