Question

В двух канистрах 85 л бензина.После того,как из первой канистры вылили 3\4 ее содержимого,а из второй 5\9 ее содержимого,выяснилось,что всего вылели 55 л бензина.Сколько литров бензина было в каждой канистре первоначально? (решите пожалуйста)

Answer 1

Пусть х литров налито в первую канистру, а у литров во вторую, из условия получаем систему:
$\cfrac{3}{4}x+\cfrac{5}{9}y=55\\x+y=85\\x=85-y\\\cfrac{3}{4}(85-y)+\cfrac{5}{9}y=55\\\cfrac{255}{4}-\cfrac{3}{4}y+\cfrac{5}{9}y=55\\\cfrac{255}{4}-\cfrac{7}{36}y=55\\255\cdot 9-7y=55\cdot 36\\7y=315\\y=45\\x=85-y=85-45=40$
$y=45, x=40$
Ответ: 40 и 45 литров

Answer 2

Problem Analysis

We have two containers with a total of 85 liters of gasoline. After pouring out 3/4 of the first container and 5/9 of the second container, a total of 55 liters of gasoline were poured out. We need to determine the initial amount of gasoline in each container.

Solution

Let's assume the initial amount of gasoline in the first container is x liters and the initial amount of gasoline in the second container is y liters.

According to the given information, we can set up the following equations:

Equation 1: After pouring out 3/4 of the first container, the remaining amount of gasoline in the first container is (1/4)x liters. Equation 2: After pouring out 5/9 of the second container, the remaining amount of gasoline in the second container is (4/9)y liters. Equation 3: The total amount of gasoline poured out is 55 liters, so (1/4)x + (4/9)y = 55.

To solve these equations, we can use substitution or elimination method. Let's use the elimination method.

Multiplying Equation 1 by 9 and Equation 2 by 4, we get: Equation 4: (9/4)x = 9/4 * (1/4)x Equation 5: (4/9)y = 4/9 * (4/9)y

Multiplying Equation 3 by 36 to eliminate the fractions, we get: Equation 6: 9(1/4)x + 16(4/9)y = 55 * 36

Simplifying Equations 4, 5, and 6, we have: Equation 7: 9x/16 = (9/16)(1/4)x Equation 8: 4y/9 = (4/9)(4/9)y Equation 9: (9/4)x + (16/9)y = 55 * 36 / 36

Simplifying Equations 7 and 8 further, we get: Equation 10: 9x/16 = (1/64)x Equation 11: 4y/9 = (16/81)y

Now, let's solve Equations 10 and 11 simultaneously.

Multiplying Equation 10 by 64 and Equation 11 by 81, we get: Equation 12: 9x = x Equation 13: 4y = 16y

Simplifying Equations 12 and 13, we have: Equation 14: 8x = 0 Equation 15: 12y = 0

From Equation 14, we can see that x = 0. However, this is not a valid solution since we know there was gasoline in the containers initially.

From Equation 15, we can see that y = 0. However, this is also not a valid solution.

Since both x and y equal 0 in the equations, we cannot determine the initial amount of gasoline in each container using the given information.