A piece of wire is 42 inches long. The wire is bent into a rectangle where the length is 2 times the width. What are the dimensions of the rectangle?
The first step to this problem is realizing that the length of the wire will be the new perimeter of the rectangle once it's bent. What is the equation of a rectangle with respect to its length, width, and perimeter? P = 2W + 2L The perimeter is equal to two times its width plus two times its length. We know P is equal to 42 inches in this case. Equation 1 gives us: (1) 42 = 2W + 2L We have one equation and two unknowns. We need another equation to solve this problem. What else was given in the problem? We know that the length is 2 times the width for the rectangle. We can write the second equation: (2) L = 2W Now, we have two equations and two unknowns! Let's substitute Equation 2 into Equation 1, and solve. 42 = 2W + 2(2W) 42 = 2W + 4W 42 = 6W W = 7 We were able to solve for the width (W) by substituting equation 2 into equation 1, giving us only one unknown! Now that we have the width, we can find the length by substituting our known width into equation 2. L = 2(7) L=14 The dimensions of the rectangle are 7"X14"!
Fred's income is 800 dollars per week. How much does Fred make in one year?
This is a simple word problem. There's a few questions we have to answer before we come to the final answer. First, we were given the income per WEEK, but asked how much he makes in one YEAR. How many weeks are in one year? 52! There are 52 weeks per year. Next, we know that Fred makes 800 dollars per WEEK. So how do we figure out how much he makes in one year? We have to use multiplication! We multiply 800 dollars X 52 weeks per year. $800 X 52 = $41,600 Fred makes $41,600 per year.
Solve the equation for X. 4X + 2 = 22 – 3(X+2)
To solve the equation for X, our mission is to get X on one side of the equation, and everything else on the other side. This is done by Algebra. 1. The first step is to distribute anywhere you have parenthesis. In this case, we have one set of parenthesis on the right side of the equation (X+2). We need to distribute the negative 3 into the parenthesis (don’t forget to distribute the negative sign!!). 4X + 2 = 22 -3(X) – 3(2) 4X + 2 = 22 -3X – 6 2. Next, we have to get all “like terms” on the same side of the equation. Let’s move all the constants to one side of the equation! On the right side, we have a 22 and a -6. Subtract to simplify which gives us 16. 4X + 2 = 16 – 3X We have a +2 on the left side, so to undo this; we have to subtract 2 on BOTH sides of the equation. 4X + 2 -2 = 16 – 2 – 3X 4X = 14 – 3X 3. Next, we have to get all “X”s on the same side of the equation. Let’s move the 3X from the right side to the left side. To undo the -3X, we have to ADD 3X to both sides of the equation. 4X + 3X = 14 -3X + 3X The -3X and +3X cancel out on the right side of the equation. Next, let’s add the 3X and 4X. 7X = 14 4. Our last step is to undo the 7 on the left side of the equation to get X by itself. To undo multiplication, we have to divide 7 on both sides of the equations! 7X / 7 = 14/7 X = 2 Our final answer is X=2!