`z =15x ^(2) +10 xy -6x-4y`
For which of the ordered pairs, (x,y), below is `z ne 0` ?

Options

A. `(-3,2)`
B. ` (-2,3)`
C. `((2)/(5), 0)`
D. `((2)/(5), 10)`

Answer

To determine for which ordered pairs \((x, y)\) the expression \(z = 15x^2 + 10xy - 6x - 4y\) is not equal to zero, we will evaluate \(z\) for each of the given pairs.

### Step 1: Evaluate \(z\) for option A: \((-3, 2)\)

Substituting \(x = -3\) and \(y = 2\) into the expression for \(z\):

\[ z = 15(-3)^2 + 10(-3)(2) - 6(-3) - 4(2) \]

Calculating each term:

1. \(15(-3)^2 = 15 \cdot 9 = 135\)
2. \(10(-3)(2) = -60\)
3. \(-6(-3) = 18\)
4. \(-4(2) = -8\)

Now, combine these results:

\[ z = 135 - 60 + 18 - 8 \]
\[ z = 135 - 60 = 75 \]
\[ z = 75 + 18 = 93 \]
\[ z = 93 - 8 = 85 \]

So, for option A, \(z = 85\), which is not equal to zero.

### Step 2: Evaluate \(z\) for option B: \((-2, 3)\)

Substituting \(x = -2\) and \(y = 3\):

\[ z = 15(-2)^2 + 10(-2)(3) - 6(-2) - 4(3) \]

Calculating each term:

1. \(15(-2)^2 = 15 \cdot 4 = 60\)
2. \(10(-2)(3) = -60\)
3. \(-6(-2) = 12\)
4. \(-4(3) = -12\)

Now, combine these results:

\[ z = 60 - 60 + 12 - 12 \]
\[ z = 0 \]

So, for option B, \(z = 0\).

### Step 3: Evaluate \(z\) for option C: \(\left(\frac{2}{5}, 0\right)\)

Substituting \(x = \frac{2}{5}\) and \(y = 0\):

\[ z = 15\left(\frac{2}{5}\right)^2 + 10\left(\frac{2}{5}\right)(0) - 6\left(\frac{2}{5}\right) - 4(0) \]

Calculating each term:

1. \(15\left(\frac{2}{5}\right)^2 = 15 \cdot \frac{4}{25} = \frac{60}{25} = \frac{12}{5}\)
2. \(10\left(\frac{2}{5}\right)(0) = 0\)
3. \(-6\left(\frac{2}{5}\right) = -\frac{12}{5}\)
4. \(-4(0) = 0\)

Now, combine these results:

\[ z = \frac{12}{5} + 0 - \frac{12}{5} + 0 \]
\[ z = 0 \]

So, for option C, \(z = 0\).

### Step 4: Evaluate \(z\) for option D: \(\left(\frac{2}{5}, 10\right)\)

Substituting \(x = \frac{2}{5}\) and \(y = 10\):

\[ z = 15\left(\frac{2}{5}\right)^2 + 10\left(\frac{2}{5}\right)(10) - 6\left(\frac{2}{5}\right) - 4(10) \]

Calculating each term:

1. \(15\left(\frac{2}{5}\right)^2 = 15 \cdot \frac{4}{25} = \frac{60}{25} = \frac{12}{5}\)
2. \(10\left(\frac{2}{5}\right)(10) = 40\)
3. \(-6\left(\frac{2}{5}\right) = -\frac{12}{5}\)
4. \(-4(10) = -40\)

Now, combine these results:

\[ z = \frac{12}{5} + 40 - \frac{12}{5} - 40 \]
\[ z = 0 \]

So, for option D, \(z = 0\).

### Conclusion

After evaluating all the options, we find that:
- For option A: \(z = 85\) (not equal to zero)
- For option B: \(z = 0\)
- For option C: \(z = 0\)
- For option D: \(z = 0\)

Thus, the ordered pair for which \(z \neq 0\) is **option A: \((-3, 2)\)**.