Answer:
19,4 or 4,19
Step-by-step explanation:
Let the width of the rectangle be 'w' in
According to the question length of the rectangle (l) is one inch less than five times it's width;
[tex] \longrightarrow [/tex] l = (5w - 1) in
Perimeter of rectangle = 46 inches
Formula of perimeter of rectangle = 2(Length + Width)
So,
[tex]\rm \implies 2(Length + Width) = 46 \\ \\ \rm \implies 2(5w - 1 + w) = 46 \\ \\ \rm \implies 2(6w - 1) = 46 \\ \\ \rm \implies 6w - 1 = \dfrac{46}{2} \rm \implies 6w - 1 = 23 \\ \\ \rm \implies 6w = 23 + 1 \\ \\ \rm \implies 6w = 24 \\ \\ \rm \implies w = \dfrac{24}{6} \\ \\ \rm \implies w = 4 \: in[/tex]
[tex] \therefore [/tex] Width of the rectangle (w) = 4 in
Length of rectangle (l) = 5 × 4 - 1 = 19 in
