- Perimeter of a rectangle =
![2b+2h](https://img.qammunity.org/2019/formulas/mathematics/middle-school/cdnxsgxv7fure7rqi3nl9rqrbkta2r7vhm.png)
- Area of a rectangle =
![bh](https://img.qammunity.org/2019/formulas/mathematics/middle-school/v9a1xm22e2m51nsz7c7zqfcpatqfzqb1tx.png)
So for this, we will be doing a system of equations, one representing the sum of their areas and the other representing the sum of their perimeters:
![2x+9y=46\\\\4+2x+18+2y=40\\2x+2y+22=40](https://img.qammunity.org/2019/formulas/mathematics/middle-school/jm3iy1e106im1tqd7dokzgtb1m1jlftgmr.png)
So with this, I will be using the elimination method. So firstly, subtract 22 on both sides of the second equation:
![2x+9y=46\\2x+2y=18](https://img.qammunity.org/2019/formulas/mathematics/middle-school/ddm16sgezg7yy9yp8ywnoqfn0w28azcznf.png)
Next, subtract the second equation from the first equation and you should get
. From here we can solve for y.
For this, just divide both sides by 7 and your first answer will be y = 4.
Now that we have the value of y, substitute it into either equation to solve for x as such:
![2x+9*4=46\\2x+36=46\\2x=10\\x=5\\\\2x+2*4=18\\2x+8=18\\2x=10\\x=5](https://img.qammunity.org/2019/formulas/mathematics/middle-school/l9ufi6lcofisc7cg1dm8ro4wuo8spuej8c.png)
In short, x = 5 and y = 4.