Coach mac finds a new practice court for the basketball team. if the width of the rectangular lot is 16 feet less than the length. the perimeter is 160 feet, what are the dimensions of the new practice court? perimeter of rectangle is = 2w + 2l 160=2w+2l; w=

System of equations

Solution to system of equations

The width is 16 feet less than the length, so the appropriate equation for that is ...

... W = L - 16

Using this expression for W, we can substitute into the equation for perimeter:

... 160 = 2(L - 16) + 2L

... 160 = 4L -32 . . . . . . . collect terms

... 192 = 4L . . . . . . . . . . add 32

... 48 = L . . . . . . . . . . . . divide by 4

... W = 48 -16 = 32 . . . . find W using the relation with L

Length is 48 ft; width is 32 ft.

2 is the final ! : )

at least two

step-by-step explanation: