8. $$ 3(2 k+2)=4 \quad $$ 9. $$ 3 imes k^{2}-21 $$

Question

8. $$ 3(2 k+2)=4 \quad $$ 9. $$ 3 	imes k^{2}-21 $$

UpStudy AI · Accepted Answer

Let's solve the equations step by step.

### Problem 8: Solve $$ 3(2k + 2) = 4 $$

1. **Distribute the 3**:
   $$
   3 \cdot 2k + 3 \cdot 2 = 4
   $$
   This simplifies to:
   $$
   6k + 6 = 4
   $$

2. **Subtract 6 from both sides**:
   $$
   6k = 4 - 6
   $$
   This simplifies to:
   $$
   6k = -2
   $$

3. **Divide both sides by 6**:
   $$
   k = \frac{-2}{6}
   $$
   This simplifies to:
   $$
   k = -\frac{1}{3}
   $$

### Problem 9: Simplify $$ 3 \times k^{2} - 21 $$

Since there is no equation to solve, we can simply express this in terms of $$ k $$:
$$
3k^2 - 21
$$

This expression can be factored:
1. **Factor out the common term**:
   $$
   3(k^2 - 7)
   $$

So, the simplified expression is:
$$
3(k^2 - 7)
$$

### Summary of Solutions
- For Problem 8, $$ k = -\frac{1}{3} $$.
- For Problem 9, the expression simplifies to $$ 3(k^2 - 7) $$.
For Problem 8, $$ k = -\frac{1}{3} $$. For Problem 9, the expression simplifies to $$ 3(k^2 - 7) $$.

8. \( 3(2 k+2)=4 \quad \) 9. \( 3 \times k^{2}-21 \)

Solución de tutoría real

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Algebra Questions