Answer:
The dimensions of rectangle whose perimeter is 70 inches and width is five less than three times the length of worktable is length = 10 and width = 25
Solution:
Consider ‘P’ as perimeter and ‘L’ as length and ‘W’ as width
P = 70 inches
W = 5 less than thrice of length
Therefore, [tex]\mathrm{W}=3 L-5 \quad(\text {equation } 1)[/tex]
Using perimeter formula for rectangle,
[tex]\begin{array}{l}{P=2(L+W)} \\ {70=2(L+3 L-5)} \\ {70=2(4 L-5)} \\ {70=8 L-10} \\ {8 L=80} \\ {L=10}\end{array}[/tex]
Substitute L= 10 in equation 1
[tex]\begin{aligned} W &=3 \times 10-5 \\ W &=30-5 \\ W &=25 \end{aligned}[/tex]
