Find the area and perimeter of the following rectangle. the lengths are 4x+y+2; and 3x+3y. the widths are 2x+y+1 and 12

Equating the lengths, we have:

Equating the widths, we have:

Solving the two equations simultaneously gives that x = 4 and y = 3.

Thus, the length is 4(4) + 3 + 2 = 21 and the width is 12.

Therefore area = length times width = 21 x 12 = 252 square units

perimeter = 2(length + width) = 2(21 + 12) = 2(33) = 66.

a_n = a_1 + (n - 1)d

the first term is 6.

the common difference d = ?

when you find d, replace n with 7 and d with the common difference to find your answer.