The length of rectangle is 5 less than twice the width The perimeter of the rectangle is 80. Find the dimensions of the rectangle?

L= Length W= Width P= Perimeter Equation 1: L= 2W-5 Equation 2: 2L+2W=P=80 Then, From Equation 2, Solve the second equation for 2W. 2L + 2W= 80 2W = 80 - 2L From Equation 1, Substitute 80-2L for 2W in the first equation. This gives the equation one variable, which earlier algebra work.

L=(80-2L)-5 L=80-2L-5 2L + L= 80 - 5 3L= 75 L=25 Now, substitute 25 for L in either equation and solve for w. From Equation 1 25 = 2W - 5 5 +25 = 2W

30 = 2W 30 / 2 =W 15=W The solution is Lengh = 25 Width = 15 L= Length W= Width P= Perimeter Equation 1: L= 2W-5 Equation 2: 2L+2W=P=80 Then, From Equation 2, Solve the second equation for 2W. 2L + 2W= 80 2W = 80 - 2L From Equation 1, Substitute 80-2L for 2W in the first equation. This gives the equation one variable, which earlier algebra work.

L=(80-2L)-5 L=80-2L-5 2L + L= 80 - 5 3L= 75 L=25 Now, substitute 25 for L in either equation and solve for w. From Equation 1 25 = 2W - 5 5 +25 = 2W

30 = 2W 30 / 2 =W 15=W The solution is Lengh = 25 Width = 15