b. \( \sqrt{2 x-3}+x=3 \)

To solve the equation \( \sqrt{2x-3} + x = 3 \), start by isolating the square root. Subtract \( x \) from both sides to get \( \sqrt{2x-3} = 3 - x \). Next, square both sides, leading to \( 2x - 3 = (3 - x)^2 \). Expanding the right-hand side results in \( 2x - 3 = 9 - 6x + x^2 \). Rearranging gives the quadratic equation \( x^2 - 8x + 12 = 0 \). Factor this into \( (x-6)(x-2)=0\), yielding solutions \( x = 6 \) and \( x = 2 \). However, it's crucial to plug these values back into the original equation to check for any extraneous roots. When you substitute \( x = 6 \), you find that it satisfies the original equation. But when you check \( x = 2 \), you find it doesn't hold true. Hence, the only solution is \( x = 6 \).