Question What would be the equation (using x) to solve “What are the dimensions of a rectangle whose length is 4 more than twice the width and whose perimeter is 3 less than 7 times the width?”

Let w = the width Let 2w + 4 = the length Let 7w-3 = the perimeter P = 2l + 2w 7w-3 = 2w + 2(2w+4) 7w-3 = 2w + 4w + 8 w = 11 length = 2w + 4 = 2(11) + 4 length = 26 P = 7(11) -3 = 77 – 3 Perimeter = 74 Reply

Answer:Let w = the width Let 2w + 4 = the length Let 7w-3 = the perimeter P = 2l + 2w 7w-3 = 2w + 2(2w+4) 7w-3 = 2w + 4w + 8 w = 11 length = 2w + 4 = 2(11) + 4 length = 26 P = 7(11) -3 = 77 – 3 Perimeter = 74 Step-by-step explanation: Reply

Answer:Let w = the widthLet 2w + 4 = the lengthLet 7w-3 = the perimeterP = 2l + 2w7w-3 = 2w + 2(2w+4)7w-3 = 2w + 4w + 8w = 11length = 2w + 4 = 2(11) + 4length = 26P = 7(11) -3 = 77 – 3Perimeter = 74Step-by-step explanation: