SOLUTION: B. Solve the ff. problems. The perimeter of a rectangle is 152 meters. The width is 22 meters less than the length. Find the length and the width.

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Question 1171899: B. Solve the ff. problems.
The perimeter of a rectangle is 152 meters. The width is 22 meters less than the length. Find the length and the width.
Found 2 solutions by Theo, MathTherapy:
Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
p = 2 * (l + w)
p is the perimeter
l is the length
w is the width.

the width is 22 meters less than the length.
this gets you w = l - 22
replace w with that in the original equation to get:
p = 2 * (l + w) becomes p = 2 * (l + l - 22)
combine like terms to get p = 2 * (2l - 22)
simplify to get p = 4l - 44
since p = 152, this becomes 152 = 4l - 44
add 44 to both sides of this equation and simplify to get:
196 = 4l
solve for l to get l = 196/4 = 49
since w = l - 22, then w = 49 - 22 = 27

you have:
l = 49 and w = 27
confirm these values to be correct by replacing l and w in the original equations to get:
p = 2 * (l + w) = 2 * (49 + 27) = 2 * (76) = 152.
since w = l - 22, then w = 49 - 22 = 27.
this confirms the values of l and w are correct.

your solution is that the length is 49 and the width is 27.
Answer by MathTherapy(10556)   (Show Source): You can put this solution on YOUR website!

B. Solve the ff. problems.
The perimeter of a rectangle is 152 meters. The width is 22 meters less than the length. Find the length and the width.
With perimeter being 152 m, 

                                               

                                W = L - 22, or L - W = 22 ------ eq (ii), since width is also 22 m less than the length

                                                  2L = 98 ------ Adding eqs (i) & (ii)

                                        Length or 

                                              49 + W = 76 ------- Substituting 49 for L in eq (i)

                                        Width, or  

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