Chapter 1.1, Problem 49E

(a)

To determine

To indicate the values of $a$ and $b$ for the below equation when $x$ is a positive integer choosing the values from $-2,5,9,10$ -

  $ax=b-5$

Answer to Problem 49E

When $a=5$ and $b=10$ , $x$ is a positive integer.

Explanation of Solution

Given:

Equation: $ax=b-5$ ; $x$ is a positive integer

Values - $-2,5,9,10$ .

Calculation:

Given:

Equation: $ax=b-5$ ; $x$ is a positive integer.

Values - $-2,5,9,10$ .

Solving the above equation, we have:

  $\begin{array}{l}ax=b-5\\ x=\frac{b-5}{a}\end{array}$

To have $x$ as a positive integer, we need to have:

  $b=10$

  $a=5$ ; so that:

  $\begin{array}{l}x=\frac{10-5}{5}\\ x=\frac{5}{5}\\ x=1\end{array}$

Thus, when $a=5$ and $b=10$ , $x$ is a positive integer.

Conclusion:

Hence, when $a=5$ and $b=10$ , $x$ is a positive integer.

(b)

To determine

To indicate the values of $a$ and $b$ for the below equation when $x$ is a negative integer choosing the values from $-2,5,9,10$ -

  $ax=b-5$

Answer to Problem 49E

When $a=-2$ and $b=9$ , $x$ is a negative integer.

Explanation of Solution

Given:

Equation: $ax=b-5$ ; $x$ is a negative integer

Values - $-2,5,9,10$ .

Calculation:

Given:

Equation: $ax=b-5$ ; $x$ is a negative integer.

Values - $-2,5,9,10$ .

Solving the above equation, we have:

  $\begin{array}{l}ax=b-5\\ x=\frac{b-5}{a}\end{array}$

To have $x$ as a negative integer, we need to have:

  $b=9$

  $a=-2$ ; so that:

  $\begin{array}{l}x=\frac{9-5}{-2}\\ x=-\frac{4}{2}\\ x=-2\end{array}$

Thus, when $a=-2$ and $b=9$ , $x$ is a negative integer.

Conclusion:

Hence, when $a=-2$ and $b=9$ , $x$ is a negative integer.