Question

Часть 2 Алгебра . только полные решение!!!

Answer 1

Algebra: Solving Equations

To provide detailed solutions, I will need more specific information about the equations you would like me to solve. However, I can give you a general overview of solving equations.

# Linear Equations

A linear equation is an equation of the form ax + b = 0, where a and b are constants and x is the variable. To solve a linear equation, you can follow these steps:

1. Move the constant term to the other side of the equation. 2. Divide both sides of the equation by the coefficient of x to isolate x.

For example, let's solve the equation 2x + 3 = 7:

1. Move the constant term to the other side: 2x = 7 - 3 2x = 4

2. Divide both sides by the coefficient of x: x = 4/2 x = 2

So the solution to the equation 2x + 3 = 7 is x = 2.

# Quadratic Equations

A quadratic equation is an equation of the form ax^2 + bx + c = 0, where a, b, and c are constants and x is the variable. To solve a quadratic equation, you can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For example, let's solve the equation x^2 - 3x + 2 = 0:

1. Identify the values of a, b, and c: a = 1, b = -3, c = 2

2. Substitute these values into the quadratic formula: x = (-(-3) ± √((-3)^2 - 4(1)(2))) / (2(1)) x = (3 ± √(9 - 8)) / 2 x = (3 ± √1) / 2

3. Simplify the square root: x = (3 ± 1) / 2

4. Solve for both possible values of x: x = (3 + 1) / 2 or x = (3 - 1) / 2 x = 4 / 2 or x = 2 / 2 x = 2 or x = 1

So the solutions to the equation x^2 - 3x + 2 = 0 are x = 2 and x = 1.

Please provide specific equations you would like me to solve, and I will be happy to assist you further.