Question

Решите систему уравнений {y-x^2=2 2x+3y=17 Смотреть ответ Mirage86 Mirage86 I hope this helps you

Answer 1

To solve the system of equations:

1. y - x^2 = 2
2. 2x + 3y = 17

There are several methods to solve this system, but I'll demonstrate the substitution method.

Step 1: Solve one of the equations for one variable in terms of the other variable.

Let's solve equation 1 for y: y = x^2 + 2

Step 2: Substitute the expression for y from Step 1 into the other equation (equation 2).

2x + 3(x^2 + 2) = 17

Step 3: Simplify and solve for x.

2x + 3x^2 + 6 = 17 3x^2 + 2x = 17 - 6 3x^2 + 2x = 11

Step 4: Set the equation equal to zero to use the quadratic formula.

3x^2 + 2x - 11 = 0

Step 5: Use the quadratic formula to find the values of x.

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = 3, b = 2, and c = -11.

x = (-(2) ± √(2^2 - 4 * 3 * -11)) / 2 * 3 x = (-2 ± √(4 + 132)) / 6 x = (-2 ± √136) / 6 x = (-2 ± 2√34) / 6 x = (-1 ± √34) / 3

So we have two potential values for x: (-1 + √34) / 3 and (-1 - √34) / 3.

Step 6: Find the corresponding y-values using the equation y = x^2 + 2.

For the first value of x, x = (-1 + √34) / 3: y = ((-1 + √34) / 3)^2 + 2 y = (1/9) * (1 - 2√34 + 34) + 2 y = (1/9) * (35 - 2√34) + 2 y = (35/9) - (2/9)√34 + 2

For the second value of x, x = (-1 - √34) / 3: y = ((-1 - √34) / 3)^2 + 2 y = (1/9) * (1 + 2√34 + 34) + 2 y = (1/9) * (35 + 2√34) + 2 y = (35/9) + (2/9)√34 + 2

So the solutions to the system of equations are:

1. x = (-1 + √34) / 3, y = (35/9) - (2/9)√34 + 2
2. x = (-1 - √34) / 3, y = (35/9) + (2/9)√34 + 2

I hope this helps! Let me know if you have any other questions.