If `vecaandvecb` are non-collinear vector, find the value of x such that the vectors `vecalpha=(x-2)veca+vecbandvecbeta=(3+2x)veca-2vecb` are collinear.

Answer

To find the value of \( x \) such that the vectors \( \vec{\alpha} \) and \( \vec{\beta} \) are collinear, we start with the definitions of the vectors:

\[ \vec{\alpha} = (x-2)\vec{a} + \vec{b} \]
\[ \vec{\beta} = (3 + 2x)\vec{a} - 2\vec{b} \]

### Step 1: Set up the collinearity condition
For the vectors \( \vec{\alpha} \) and \( \vec{\beta} \) to be collinear, there must exist a scalar \( \lambda \) such that:

\[ \vec{\alpha} = \lambda \vec{\beta} \]

### Step 2: Substitute the expressions for \( \vec{\alpha} \) and \( \vec{\beta} \)
Substituting the expressions we have:

\[ (x-2)\vec{a} + \vec{b} = \lambda \left( (3 + 2x)\vec{a} - 2\vec{b} \right) \]

### Step 3: Expand the right-hand side
Expanding the right-hand side gives:

\[ (x-2)\vec{a} + \vec{b} = \lambda(3 + 2x)\vec{a} - 2\lambda\vec{b} \]

### Step 4: Rearrange the equation
Rearranging the equation, we have:

\[ (x-2)\vec{a} + \vec{b} = (3\lambda + 2\lambda x)\vec{a} - 2\lambda\vec{b} \]

### Step 5: Compare coefficients of \( \vec{a} \) and \( \vec{b} \)
Now, we can compare the coefficients of \( \vec{a} \) and \( \vec{b} \) on both sides:

1. Coefficients of \( \vec{a} \):
\[ x - 2 = 3\lambda + 2\lambda x \]

2. Coefficients of \( \vec{b} \):
\[ 1 = -2\lambda \]

### Step 6: Solve for \( \lambda \)
From the second equation, we can solve for \( \lambda \):

\[ -2\lambda = 1 \implies \lambda = -\frac{1}{2} \]

### Step 7: Substitute \( \lambda \) into the first equation
Now substituting \( \lambda \) back into the first equation:

\[ x - 2 = 3\left(-\frac{1}{2}\right) + 2\left(-\frac{1}{2}\right)x \]

This simplifies to:

\[ x - 2 = -\frac{3}{2} - x \]

### Step 8: Solve for \( x \)
Now, we can solve for \( x \):

\[ x + x = -\frac{3}{2} + 2 \]
\[ 2x = -\frac{3}{2} + \frac{4}{2} \]
\[ 2x = \frac{1}{2} \]
\[ x = \frac{1}{4} \]

### Final Answer
Thus, the value of \( x \) is:

\[ \boxed{\frac{1}{4}} \]