The solution set of the inequality $\frac{2x+3}{x-1} < 0$ is:

The fraction $\frac{2x+3}{x-1}$ is negative when the numerator and denominator have opposite signs.

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Finding where the numerator equals zero:

$2x + 3 = 0$

$2x = -3$

$x = -\frac{3}{2}$

Finding where the denominator equals zero:

$x - 1 = 0$

$x = 1$

The expression is undefined at $x = 1$.

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These critical points divide the number line into three regions. Testing each region:

For $x < -\frac{3}{2}$, test $x = -2$:

Numerator: $2(-2) + 3 = -1$ (negative)

Denominator: $-2 - 1 = -3$ (negative)

Fraction: $\frac{(-)}{(-)} = (+)$ (positive)

For $-\frac{3}{2} < x < 1$, test $x = 0$:

Numerator: $2(0) + 3 = 3$ (positive)

Denominator: $0 - 1 = -1$ (negative)

Fraction: $\frac{(+)}{(-)} = (-)$ (negative)

For $x > 1$, test $x = 2$:

Numerator: $2(2) + 3 = 7$ (positive)

Denominator: $2 - 1 = 1$ (positive)

Fraction: $\frac{(+)}{(+)} = (+)$ (positive)

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The fraction is negative only when $-\frac{3}{2} < x < 1$.

At $x = -\frac{3}{2}$, the fraction equals 0, not negative.

At $x = 1$, the expression is undefined.

Therefore, the solution set is $x \in \left(-\frac{3}{2}, 1\right)$.