Question

Решите систему уравнений y+x^3=4 ;3y+y^2+2x^3=20

Answer 1

Ответ:

Объяснение:

Сделайте x ^ 3 предметом формулы, а затем подставьте выражение в уравнение с того места, где вы его не формировали

Answer 2

Solution to the System of Equations

To solve the system of equations y + x^3 = 4 and 3y + y^2 + 2x^3 = 20, we can use various methods such as substitution, elimination, or graphing. Let's solve it using the substitution method.

1. From the first equation, we can express y in terms of x as follows: y = 4 - x^3.

2. Substitute this expression for y into the second equation: 3(4 - x^3) + (4 - x^3)^2 + 2x^3 = 20.

3. Simplify the equation and solve for x:

12 - 3x^3 + 16 - 8x^3 + x^6 + 8x^3 = 20

x^6 - 10x^3 + 8 = 0

4. This equation is a sixth-degree polynomial equation. Solving it analytically can be quite complex. However, we can use numerical methods or a graphing calculator to find the approximate solutions.

Using a graphing calculator, we find that the approximate solutions are x ≈ -1.346 and x ≈ 1.346.

5. Substitute these values of x back into the first equation to find the corresponding values of y:

For x ≈ -1.346, y ≈ 4 - (-1.346)^3 ≈ 4 - (-2.448) ≈ 6.448.

For x ≈ 1.346, y ≈ 4 - (1.346)^3 ≈ 4 - 2.448 ≈ 1.552.

Therefore, the approximate solutions to the system of equations are (x ≈ -1.346, y ≈ 6.448) and (x ≈ 1.346, y ≈ 1.552).

Please note that these solutions are approximate and rounded to three decimal places.