Answer: L=22, W=9
Explanation:
The length of a rectangular poster is 4 more inches than two times its width. The area of the poster is 198 square inches. Solve for the dimensions (length and width) of the poster.
L=4+2W [Length] (1)
W=W [Width] (2)
P=2L+2W [perimeter] (3)
A=L*W = 198[area](4)
Subing (1) and (2) into (3) and (4)
P=2(4+2W)+2W
A=W(4+2W)
Simplify
P=8+4W+2W=8+6W
A=4W+2W^2=198
A=2W^2+4W-198
Set equal to 0 and solve for W
W=(-b+√(b^2-4ac))/2a
W=(-b-√(b^2-4ac))/2a
W=(-4+√(4^2-4(2)(-198)))/(2(2)) = 9
W=(-4-√(4^2-4(2)(-198)))/(2(2)) =-11
Disregard W =-11 since distance can’t be negative and keep W=9. Plug W=9 into Length equation (1)
L=4+2(9)=22
Final answer L=22 and W=9