Question

3x1 + x2 + x₃ = -4- 3x1+5x2+6x3=36x1-4x2-2x3=–19​

Answer 1

Ответ:

3*1+x2+x³=4

3+x=4

x=4-3

x=1

____________

3*1+1=4

Answer 2

It looks like you've provided a system of linear equations with three variables: x₁, x₂, and x₃. The system can be represented as follows:

1. $3x₁ + x₂ + x₃ = -4$
2. $-3x₁ + 5x₂ + 6x₃ = 36$
3. $6x₁ - 4x₂ - 2x₃ = -19$

To solve this system of equations, you can use various methods such as substitution, elimination, or matrix operations. Let's solve it using the substitution method:

From equation (1), we can solve for $x₁$: $x₁ = \frac{-x₂ - x₃ - 4}{3}$

Substitute this value into equations (2) and (3):

Equation (2): $-3\left(\frac{-x₂ - x₃ - 4}{3}\right) + 5x₂ + 6x₃ = 36$ $x₂ + x₃ + 4 + 5x₂ + 6x₃ = 36$ $6x₂ + 7x₃ = 32$ $6x₂ = 32 - 7x₃$ $x₂ = \frac{32 - 7x₃}{6}$

Equation (3): $6\left(\frac{-x₂ - x₃ - 4}{3}\right) - 4x₂ - 2x₃ = -19$ $-2x₂ - 2x₃ - 8 - 4x₂ - 2x₃ = -19$ $-6x₂ - 4x₃ = 11$ $6x₂ + 4x₃ = -11$ $x₃ = \frac{-11 - 6x₂}{4}$

Now you have expressions for $x₂$ and $x₃$ in terms of $x₂$. Substitute these values back into the equation for $x₁$:

$x₁ = \frac{-x₂ - \frac{-11 - 6x₂}{4} - 4}{3}$ $x₁ = \frac{-x₂ + \frac{11 + 6x₂}{4} + 4}{3}$ $x₁ = \frac{-x₂ + 4 + \frac{11}{4} + \frac{6x₂}{4}}{3}$