Question

1. the quotient of a number and -5 increased by one is at most 7.
2. the product of -2 and a number minus six is greater than -18.

Explanation:

Problem 9:

Step1: Translate the problem into an inequality

Let the number be \( x \). The product of \(-2\) and \( x \) minus \( 6 \) is greater than \(-18\), so the inequality is \(-2x - 6 > -18\).

Step2: Add 6 to both sides

Adding \( 6 \) to both sides of the inequality \(-2x - 6 > -18\) gives \(-2x - 6 + 6 > -18 + 6\), which simplifies to \(-2x > -12\).

Step3: Divide both sides by -2 (and reverse the inequality sign)

Dividing both sides of \(-2x > -12\) by \(-2\) (and remembering to reverse the inequality sign because we are dividing by a negative number) gives \( x < \frac{-12}{-2}\), so \( x < 6 \).

Problem 7 (assuming the full problem is "The quotient of a number and -5 increased by one is at most 7"):

Step1: Translate the problem into an inequality

Let the number be \( x \). The quotient of \( x \) and \(-5\) is \( \frac{x}{-5}\), increased by one is \( \frac{x}{-5} + 1 \), and this is at most \( 7 \), so the inequality is \( \frac{x}{-5} + 1 \leq 7 \).

Step2: Subtract 1 from both sides

Subtracting \( 1 \) from both sides of \( \frac{x}{-5} + 1 \leq 7 \) gives \( \frac{x}{-5} + 1 - 1 \leq 7 - 1 \), which simplifies to \( \frac{x}{-5} \leq 6 \).

Step3: Multiply both sides by -5 (and reverse the inequality sign)

Multiplying both sides of \( \frac{x}{-5} \leq 6 \) by \(-5\) (and reversing the inequality sign because we are multiplying by a negative number) gives \( x \geq 6 \times (-5) \), so \( x \geq -30 \).

Answer:

\( x < 6 \)