The width of a rectangle is 12 cm less than the length the perimeter is 156 cm. find the with and length

We know that the perimeter of a rectangle is twice the length, plus twice the width.  

P = 2L + 2W  

We also know that the perimeter is 156.  

P = 156  

Finally, we know that the width is 12 less than the length.  

W = L - 12.  

The next thing that we do is substitute the information that we have into the original equation:  

P = 2L + 2W  

156 = 2L + 2(L - 12)  

From this point we start to solve  

156 = 2L + 2L - 24 <---we multiplied the '2' through the parenthesis  

156 + 24 = 2L + 2L - 24 + 24  

180 = 2L + 2L <--- getting like terms on same sides  

180 = 4L <---combining like terms  

180/4 = 4L/4 <--- getting like terms on same sides  

45 = L <---now we have a value for L  

Now we take the known value for L and substitute it in to our equation for W  

W = L - 12  

W = 45 - 12  

W = 33  

So now we have Length = 45 and Width = 33.

w = l - 12

156 = 2l + 2w

Since we have a value of w, we can plug that into the variable w to find the exact value of l.

156 = 2l + 2(l - 12)

Distributive property.

156 = 2l + 2l - 24

Combine like terms.

156 = 4l - 24

Add 24 to both sides.

180 = 4l

Divide both sides by 4.

l = 45

Now that we have the exact value of l, we can find the exact value of w.

w = l - 12

w = 45 - 12

w = 33

We now know the width is equal to 33 cm, and the length is equal to 45 cm. (This is your answer.)

We can verify by plugging these values into the second equation.

156 = 2l + 2w

156 = 2(45) + 2(33)

156 = 90 + 66

156 = 156 √ this is correct.