Answer:

Certainly! Let's go through each expression one by one and explain the values of the variable that make them undefined.

1. \( \frac{3x}{x-5} \):

In this expression, the denominator is \( x-5 \). For the expression to be undefined, the denominator must be equal to zero. So, we solve the equation \( x-5 = 0 \) and find that \( x = 5 \). Therefore, when \( x = 5 \), the expression becomes undefined.

2. \( \frac{2x+5}{x^2+11x+10} \):

Here, the denominator is \( x^2+11x+10 \). To make the expression undefined, we need to find the values of \( x \) that make the denominator zero. By factoring the quadratic equation \( x^2+11x+10 = 0 \), we get \( (x+1)(x+10) = 0 \). Solving for \( x \), we find that \( x = -1 \) and \( x = -10 \). Therefore, when \( x = -1 \) or \( x = -10 \), the expression becomes undefined.

3. \( \frac{5r}{r^2-16} \):

In this expression, the denominator is \( r^2-16 \). To make the expression undefined, we set the denominator equal to zero and solve the equation \( r^2-16 = 0 \). By factoring, we get \( (r-4)(r+4) = 0 \). Solving for \( r \), we find that \( r = -4 \) and \( r = 4 \). Therefore, when \( r = -4 \) or \( r = 4 \), the expression becomes undefined.

4. \( \frac{4y}{y+8} \):

Here, the denominator is \( y+8 \). To make the expression undefined, we need to find the values of \( y \) that make the denominator zero. Solving the equation \( y+8 = 0 \), we find that \( y = -8 \). Therefore, when \( y = -8 \), the expression becomes undefined.

5. \( \frac{7}{x^2-2x-15} \):

In this expression, the denominator is \( x^2-2x-15 \). To make the expression undefined, we set the denominator equal to zero and solve the equation \( x^2-2x-15 = 0 \). By factoring, we get \( (x-5)(x+3) = 0 \). Solving for \( x \), we find that \( x = 5 \) and \( x = -3 \). Therefore, when \( x = 5 \) or \( x = -3 \), the expression becomes undefined.

6. \( \frac{3x}{x^2-16x+15} \):

Here, the denominator is \( x^2-16x+15 \). To make the expression undefined, we need to find the values of \( x \) that make the denominator zero. Solving the equation \( x^2-16x+15 = 0 \), we find that \( x = 1 \) and \( x = 15 \). Therefore, when \( x = 1 \) or \( x = 15 \), the expression becomes undefined.

In summary, the values of the variable that make each expression undefined are:

1. \( x = 5 \)

2. \( x = -1 \) and \( x = -10 \)

3. \( r = -4 \) and \( r = 4 \)

4. \( y = -8 \)

5. \( x = 5 \) and \( x = -3 \)

6. \( x = 1 \) and \( x = 15 \)