Question

III SYSTEMS OF LINEAR ELATIONS Solving a word problem using a system of linear equations of th... Two pools are being filled with water. To start the first pool contains 720 liters of water and the second pool is empty. Water is being added to the first pool at a rate of 20.25 liters per minute. Water is being added to the second pool at a rate of 42.75 liters per minute. After how many minutes will the two pools have the same amount of water? 0 minutes How much water will be in each pool when they have the same amount sters

Answer 1

Using a system of linear equations, it can be determined that the two pools will have the same amount of water after approximately 26.53 minutes, with each pool holding around 1144.99 liters.

Step-by-step explanation:

Let's define minutes as our variable, x. The problem can be tackled using a system of two linear equations. Our two equations are formed according to the information given about the two pools:

• The first pool already contains 720 liters, and the water is added at a rate of 20.25 liters per minute. This gives us the equation for the First Pool: 720 + 20.25x
• The second pool starts empty, but water is added at a faster rate of 42.75 liters per minute. This gives us the equation for the Second Pool: 0+42.75x

To find out when both pools will have the same amount of water, we create another equation where first pool's water is equal to the second pool's water: 720 + 20.25x = 42.75x. Solving this for x (which represents minutes), we get x = approximately 26.53 minutes. At this point, they will both have the same volume of water, approximately 1144.99 liters.

Learn more about Solving Systems of Linear Equations