10x+yx-y^2

To evaluate the expression \(10x + yx - y^2\) with the values \(x = 2\) and \(y = -3\), we can substitute these values into the expression:

1. Substitute \(x\) and \(y\): \[ 10(2) + (-3)(2) - (-3)^2 \]

2. Calculate each term:

   • \(10(2) = 20\)
   • \((-3)(2) = -6\)
   • \((-3)^2 = 9\)

3. Now substitute the calculated values back into the expression: \[ 20 - 6 - 9 \]

4. Simplify: \[ 20 - 6 = 14 \] \[ 14 - 9 = 5 \]

Therefore, the value of the expression \(10x + yx - y^2\) when \(x = 2\) and \(y = -3\) is \(5\).

Let's clarify the evaluation step by step, especially focusing on the signs involved in the calculation:

1. We start with the expression: \[ 10x + yx - y^2 \]

2. Substitute \(x = 2\) and \(y = -3\): \[ 10(2) + (-3)(2) - (-3)^2 \]

3. Calculate each term:

   • For \(10(2)\): \[ 10 \times 2 = 20 \]
   • For \((-3)(2)\): \[ -3 \times 2 = -6 \]
   • For \((-3)^2\): \[ (-3) \times (-3) = 9 \]

4. Now substitute these calculations back into the expression: \[ 20 + (-6) - 9 \]

5. Simplify step by step:

   • First, \(20 + (-6)\): \[ 20 - 6 = 14 \]
   • Now, subtract \(9\): \[ 14 - 9 = 5 \]

So the correct evaluation yields \(5\).

In summary:

• You were correct about \(-3 \times 2 = -6\), and the squaring of \(-3\) correctly yields \(9\) (not \(-9\) because squaring a negative number yields a positive result).

The final answer remains \(5\). Thank you for your question!