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3w- plus 2 equals 2w plus 3?
3w - 2 = 2w + 3 Add 2 to both sides: 3w = 2w + 5 Subtract 2w from both sides: w = 5
What is 5-32 plus 2w equals -7?
Using basic algebra we can solve this problem. First we need to write out the problem:5-32+2w = -7-27+2w = -72w = 20w = 10
The lenght of a rectangle is 3 inches less than twice the width if the perimeter of the rectangel is 18 inches find the width of the rectangel?
In this rectangle we'll call the length L and the width W We know that L + 3 = 2W So, if we subtract 3 from each side we get L = 2W - 3 We also know that 2L + 2W = 18 Now we can substitute the L terms 2(2W - 3) + 2W = 18 4W - 6 + 2W = 18 4W + 2W = 18 + 6 6W = 24 W = 4 So, if L + 3 = 2W then L + 3 = 8.... therefore L = 5 So now, we'll check that the measurements we've ended up with work out... 2L + 2W = 18 5 + 5 + 4 + 4 does indeed = 18
If you know both the perimeter an the length of a rectangle how can you determine the width of the rectangle?
The perimeter is the total of all 4 sides of the rectangle, and a rectangle, by definition, has 2 sets of parallel sides. Perimeter = 2L + 2W (where L = length of ONE side, and W = width of ONE side). So, for example, if the perimeter is 40 cm, and the length is 15 cm, you figure out the width by subtraction and division. P = 2L + 2W 40 = 2(15) + 2W 40 = 30 + 2W 40-30 =30 + 2W - 30 10 = 2W 10/2 = 2W/2 5 = W So the width of the rectangle is 5 cm.
What are the dimensions of a rectangle with the area of 27 yds sq. and the length is 3 yds less than twice the width?
Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.Suppose the width is W yards.Then twice the width is 2W yards3 yards less that twice the width is (2W - 3) yards.That is, the length is (2W - 3) yardsSo, the area = Length *Width = (2W - 3)*W which is known to be 27Therefore 2W2 - 3W = 27or 2W2 - 3W - 27 = 0so (W + 3)*(2W - 9) = 0so W = -3 or W = 4.5Since W cannot be negative, W = 4.5 yards and then L = 6 yards.
