Find the width

let $W$ be the width, $L$ be the length and $P$ be the perimeter of the rectangle

Then, $L=10+W$

The perimeter of a rectangle is the sum of its sides.

$P=2L+2W$

However, $L=10+W$ , we substitute

$120=2(10+W) +2W$

$120=20+2W+2W$

$120=20+4W$

$100=4W$

$25=W$

$\color{#3D99F6}\text{answer:} \color{#D61F06}\text{25 meters}$

Let the length be x and width be y

Two equation are,

x = y +10

or, x - y = 10......................................... (1)

Again, 2 ( x + y ) = 120

or, x + y = $\frac{120}{2}$

or, x + y = 60.........................................(2)

now,

x - y = 10....................................................(1)

x + y = 60....................................................(2)

Sum of the two equations,

x - y + x + y = 10 + 60

or, 2x = 70

or, x = $\frac{70}{2}$

Therefore, x = 35

'x' value in equation (1),

35 - y = 10

or, - y = 10 - 35

or, - y = - 25

y = 25

Therefore, the width of the classroom is 25 meters.