[tex]\sqrt{4x-2} + \sqrt{3x-3} =\sqrt{x+1}[/tex]

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To solve this equation, we need to isolate the square roots on one side and square both sides to eliminate the square roots. Let's begin with the given equation:

√(4x - 2) + √(3x - 3) = √(x + 1)

Step 1: Move √(3x - 3) to the right side.

√(4x - 2) = √(x + 1) - √(3x - 3)

Step 2: Square both sides.

(√(4x - 2))^2 = (√(x + 1) - √(3x - 3))^2

4x - 2 = (x + 1) - 2√((x + 1)(3x - 3)) + (3x - 3)

Step 3: Simplify the equation.

4x - 2 = x + 1 - 2√(3x^2 - 2x - 3) + 3x - 3
4x - 2 = 4x - 2 - 2√(3x^2 - 2x - 3)

Step 4: Combine like terms.

0 = -2√(3x^2 - 2x - 3)

Step 5: Square both sides again.

0 = 4(3x^2 - 2x - 3)
0 = 12x^2 - 8x - 12

Step 6: Divide by 4.

0 = 3x^2 - 2x - 3

Step 7: Factor the quadratic equation.

0 = (3x + 1)(x - 3)

Step 8: Solve for x.

3x + 1 = 0
3x = -1
x = -1/3

x - 3 = 0
x = 3

Therefore, the solutions to the given equation are x = -1/3 and x = 3.