Given:
The perimeter of a rectangle is 32 meters and the length is 4 meters longer than the width
Let, x = the length of the rectangle
And, y = the width of the rectangle
So, we have the following system of equations:
![\begin{gathered} 2x+2y=32\rightarrow(1) \\ x-y=4\rightarrow(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/611mnp9hu88vnsx2aewssr7w52kaxm5mt0.png)
We will use the method of substitution to solve the system
So, from equation 2:
![x=y+4\rightarrow(3)](https://img.qammunity.org/2023/formulas/mathematics/college/2fbzb47hhdl4xxft45j4siw9883or03ibw.png)
substitute with (x) from equation (3) intp eqaution (1)
![2(y+4)+2y=32](https://img.qammunity.org/2023/formulas/mathematics/college/1vz3k75okhabb2owxclwfoo4uhgtsoo5jy.png)
solve the equation to find (y):
![\begin{gathered} 2y+8+2y=32 \\ 4y+8=32 \\ 4y=32-8 \\ 4y=24 \\ y=(24)/(4)=6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/iyype2bo06rca2ntmqgxrtr0t5evx0zsz1.png)
Now, substitute with (y) into equation (3) to find (x):
![x=y+4=6+4=10](https://img.qammunity.org/2023/formulas/mathematics/college/c45cjw0nb6heqrlit5qq7rpcyd1jwubhy8.png)
So, the answer will be:
The length of the rectangle = 10 m
The width of the rectangle = 6 m