What are the dimensions of a rectangle whose length is 4 more than twice the width as whose perimeter is 3 less than 7 times the width?

L= length= 2w+4
w= width
P= perimeter= 7w-3

SOLVE FOR WIDTH
Perimeter of a Rectangle Formula
P= 2(l + w)

7w - 3= 2((2w + 4) + w)
combine like terms in parentheses

7w - 3= 2(3w + 4)
multiply 2 by parentheses

7w - 3= (2*3w) + (2*4)
multiply in parentheses

7w - 3= 6w + 8
add 3 to both sides

7w= 6w + 11
subtract 6w from both sides

w= 11 width

LENGTH
l= 2w + 4= 2(11) + 4= 22 + 4= 26

PERIMETER
P= 7w-3= 7(11) - 3= 77 - 3= 74

CHECK
74= 2(26 + 11)
74= 2(37)
74= 74

ANSWER: The length is 26, the width is 11 and the perimeter is 74.

Hope this helps! :)

The answer to your question is b’(16,14)