# Suppose you are given a rectangle where the length is equal to 2x+4 and its width is equal to 3x-2. Assuming that the perimeter is equal to 54 cm, what's the value of x?

__Step 1: __

For a rectangle, we know that opposite sides are equal

Hence we use this information to draw a rectangle and label the 4 sides.

__Step 2:__

We know that the perimeter of a rectangle is equal to the sum of the length of the 4 sides.

This is equal to 2 x ** length **+ 2 x

**width**

Hence we write the following equation:

2*(2x+4) + 2*(3x-2) = Perimeter

2*(2x+4) + 2*(3x-2) = 54

__Step 3: __

Now we have to solve the above equation in terms of x:

2*(2x+4) + 2*(3x-2) = 54

Now using BIDMAS, we notice that we can't perform any operation in the brackets first, as we have x which is an unknown term, hence we proceed by multiplying out each of the brackets.

We get,

4x+8+6x-4 = 54

Now we proceed by adding the common terms on the LHS

This means we add 4x to 6x and subtract 4 from 8.

10x+4= 54

The next step is subtract 4 from each side, as this means will be left only with are unknown term on the LHS

10x=50

The final step is to divide both sides by 10:

x=5

Hence the value of x we have been looking for is 5.

__Step 4:__

Now we have to check if the value of x we have found above is correct:

As mentioned above, the sum of the length of the 4 sides of the rectangle are equal to the perimeter:

Adding all the sides together we get:

4x+8+6x-4 = 54

which is equal to:

10x+4= 54

as seen from above.

Now substituting x by 5, we get:

10*(5) + 4 = 50 + 4 = 54 using BIDMAS

Hence we have showed that LHS=RHS and hence we have verified that our answer is correct.