Can you send me in English please step by step

First, convert the $5\text{A}$ current source on the left with its parallel $6\Omega$ resistor into an equivalent voltage source.
The voltage source value will be $V_L = I \times R = 5\text{A} \times 6\Omega = 30\text{V}$.
Since the current source points downwards, the positive terminal of the voltage source will be at the bottom (towards ground), and the negative terminal will be towards node $N_L$. So, it contributes $-30\text{V}$ to node $N_L$. This $30\text{V}$ source will be in series with the $6\Omega$ resistor.

Similarly, convert the $3\text{A}$ current source on the right with its parallel $4\Omega$ resistor into an equivalent voltage source.
The voltage source value will be $V_R = I \times R = 3\text{A} \times 4\Omega = 12\text{V}$.
Since the current source points upwards, the positive terminal of the voltage source will be towards node $N_R$, and the negative terminal will be at the bottom (towards ground). So, it contributes $+12\text{V}$ to node $N_R$. This $12\text{V}$ source will be in series with the $4\Omega$ resistor.

The circuit now looks like this, with two main nodes $N_L$ and $N_R$ (assuming the bottom wire is ground $0\text{V}$):

• From node $N_L$ to ground: A $3\Omega$ resistor in parallel with a series combination of a $6\Omega$ resistor and a $-30\text{V}$ voltage source.
• From node $N_R$ to ground: A $3\Omega$ resistor in parallel with a series combination of a $4\Omega$ resistor and a $+12\text{V}$ voltage source.
• Between nodes $N_L$ and $N_R$: A $7\Omega$ resistor (through which $i_o$ flows), and a series combination of a $1\Omega$ resistor and a $5\text{V}$ voltage source. The $5\text{V}$ source has its positive terminal towards $N_L$ (through the $1\Omega$ resistor) and its negative terminal towards $N_R$.

KCL at node $N_L$ (sum of currents leaving the node is zero):

$$\frac{N_L - N_R}{7\Omega} + \frac{N_L - (N_R + 5\text{V})}{1\Omega} + \frac{N_L}{3\Omega} + \frac{N_L - (-30\text{V})}{6\Omega} = 0$$ $$\frac{N_L - N_R}{7} + (N_L - N_R - 5) + \frac{N_L}{3} + \frac{N_L + 30}{6} = 0$$

Multiply the equation by 42 (the least common multiple of $7, 1, 3, 6$) to eliminate fractions:

$$6(N_L - N_R) + 42(N_L - N_R - 5) + 14N_L + 7(N_L + 30) = 0$$ $$6N_L - 6N_R + 42N_L - 42N_R - 210 + 14N_L + 7N_L + 210 = 0$$ $$(6 + 42 + 14 + 7)N_L + (-6 - 42)N_R = 0$$ $$69N_L - 48N_R = 0 \quad (\text{Equation A})$$

KCL at node $N_R$ (sum of currents leaving the node is zero):

$$\frac{N_R - N_L}{7\Omega} + \frac{N_R - (N_L - 5\text{V})}{1\Omega} + \frac{N_R}{3\Omega} + \frac{N_R - 12\text{V}}{4\Omega} = 0$$ $$\frac{N_R - N_L}{7} + (N_R - N_L + 5) + \frac{N_R}{3} + \frac{N_R - 12}{4} = 0$$

Multiply the equation by 84 (the least common multiple of $7, 1, 3, 4$) to eliminate fractions:

$$12(N_R - N_L) + 84(N_R - N_L + 5) + 28N_R + 21(N_R - 12) = 0$$ $$12N_R - 12N_L + 84N_R - 84N_L + 420 + 28N_R + 21N_R - 252 = 0$$ $$(-12 - 84)N_L + (12 + 84 + 28 + 21)N_R + 420 - 252 = 0$$ $$-96N_L + 145N_R + 168 = 0$$ $$-96N_L + 145N_R = -168 \quad (\text{Equation B})$$

From Equation A: $69N_L = 48N_R \implies N_L = \frac{48}{69}N_R = \frac{16}{23}N_R$.

Substitute the expression for $N_L$ into Equation B:

$$-96\left(\frac{16}{23}N_R\right) + 145N_R = -168$$ $$-\frac{1536}{23}N_R + 145N_R = -168$$

To combine the terms with $N_R$, find a common denominator:

$$\frac{-1536N_R + 145 \times 23 N_R}{23} = -168$$ $$\frac{-1536N_R + 3335N_R}{23} = -168$$ $$\frac{1799N_R}{23} = -168$$ $$1799N_R = -168 \times 23$$ $$1799N_R = -3864$$ $$N_R = -\frac{3864}{1799}\text{V}$$

Now, calculate $N_L$:

$$N_L = \frac{16}{23}N_R = \frac{16}{23} \times \left(-\frac{3864}{1799}\right)$$ $$N_L = -\frac{16 \times 3864}{23 \times 1799} = -\frac{61824}{41377}\text{V}$$

We can simplify $N_R$ and $N_L$ by noting that $1799 = 7 \times 257$.

$$N_R = -\frac{3864}{7 \times 257} = -\frac{552}{257}\text{V}$$ $$N_L = \frac{16}{23} \times \left(-\frac{552}{257}\right) = -\frac{16 \times 24}{257} = -\frac{384}{257}\text{V}$$

The current $i_o$ flows from node $N_L$ to node $N_R$.

$$i_o = \frac{N_L - N_R}{7\Omega}$$ $$i_o = \frac{-\frac{384}{257}\text{V} - \left(-\frac{552}{257}\text{V}\right)}{7\Omega}$$ $$i_o = \frac{-\frac{384}{257} + \frac{552}{257}}{7}\text{A}$$ $$i_o = \frac{\frac{552 - 384}{257}}{7}\text{A}$$ $$i_o = \frac{\frac{168}{257}}{7}\text{A}$$ $$i_o = \frac{168}{257 \times 7}\text{A}$$ $$i_o = \frac{24}{257}\text{A}$$

The current $i_o$ is $\boxed{\text{24/257 A}}$.