Make a system of equations, the perimeter of the rectangle is 36 m.If its length is increased by 2 m

Make a system of equations, the perimeter of the rectangle is 36 m.If its length is increased by 2 m, then you get a rectangle whose area is more than the area of the original rectangle by 30 square meters. find the length and width of the original rectangle

Let the length of the rectangle be x meters, and the width of the rectangle equal to y meters, then the perimeter of the rectangle is 2 (x + y) meters or 36 meters. Let’s compose the first equation of the system. 2 (x + y) = 36.
If the length of the first rectangle is increased by 2 m, then the length of the rectangle becomes (x + 2) meters. The area of ​​the original rectangle is xy m ^ 2, the area of ​​the other rectangle is y (x + 2) m ^ 2. It is known that the area of ​​the original rectangle is less than the second rectangle by (y (x + 2) – xy) m ^ 2 or 30 m ^ 2. Let’s compose the second equation of the system. y (x + 2) – xy = 30.
Let’s combine the equations into a system and solve it.
2 (x + y) = 36; y (x + 2) – xy = 30;
x + y = 36: 2; xy + 2y – xy = 30;
x + y = 18; 2y = 30;
x = 18 – y; y = 30: 2;
x = 18 – y; y = 15;
x = 18 – 15; y = 15;
x = 3; y = 15.
Answer. 3 meters, 15 meters.