(5) \[ \begin{array}{l} 3 x^{2}-3(x-1)=4(2 x+1)+3 \\ x^{2}=x \end{array} \]

Let's solve the given system of equations step by step: \[ \begin{cases} 3x^{2} - 3(x - 1) = 4(2x + 1) + 3 \\ x^{2} = x \end{cases} \] ### Step 1: Solve the Second Equation The second equation is: \[ x^{2} = x \] Subtract \( x \) from both sides to set the equation to zero: \[ x^{2} - x = 0 \] Factor out \( x \): \[ x(x - 1) = 0 \] So, the solutions are: \[ x = 0 \quad \text{or} \quad x = 1 \] ### Step 2: Substitute the Solutions into the First Equation Now, let's check if these solutions satisfy the first equation. **First Solution: \( x = 0 \)** Substitute \( x = 0 \) into the first equation: \[ 3(0)^2 - 3(0 - 1) = 4(2(0) + 1) + 3 \] \[ 0 - (-3) = 4(0 + 1) + 3 \] \[ 3 = 4 \cdot 1 + 3 \] \[ 3 = 7 \] This is **not true**. **Second Solution: \( x = 1 \)** Substitute \( x = 1 \) into the first equation: \[ 3(1)^2 - 3(1 - 1) = 4(2(1) + 1) + 3 \] \[ 3 - 0 = 4(2 + 1) + 3 \] \[ 3 = 12 + 3 \] \[ 3 = 15 \] This is also **not true**. ### Conclusion Neither of the solutions from the second equation (\( x = 0 \) or \( x = 1 \)) satisfies the first equation. Therefore, **the system has no solution**. **Answer:** There is no real solution to the system; no value of x satisfies both equations simultaneously.