Chapter 1.7, Problem 1E

Fill in the blank with an appropriate inequality sign.

1. (a) If x < 5, then x − 3 _____ 2.
2. (b) If x ≤ 5, then 3x _____ 15.
3. (c) If x ≥ 2, then −3x _____ −6.
4. (d) If x < −2, then −x _____ 2.

To determine

To fill: The blank in the statement “If $x<5$, then $x-3\_\_\_\_2$.” with appropriate inequality sign.

Answer to Problem 1E

The complete statement is “If $x<5$, then $\underset{\_}{x-3<2}$”.

Explanation of Solution

Rule used:

Subtracting the same quantity from each side of an inequality gives an equivalent inequality.

That is, if $A\le B$, then $A-C\le B-C$, where A, B and C are real numbers or algebraic expressions.

Calculation:

Consider the given inequality $x<5$.

The left-hand side of the resulting inequality is given as $x-3$.

Therefore, subtract the same number 3, from both sides of the given inequality $x<5$.

Then, by the rule stated above, the inequality becomes as follows.

$\begin{array}{l}x<5\\ x-3<5-3\\ x-3<2\end{array}$

Thus, the complete statement is “If $x<5$, then $\underset{\_}{x-3<2}$”.

To determine

To fill: The blank in the statement “If $x\le 5$, then $3x\_\_\_\_15$.” with appropriate inequality sign.

Answer to Problem 1E

The complete statement is “If $x\le 5$, then $\underset{\_}{3x\le 15}$”.

Explanation of Solution

Rule used:

Multiplying each side of an inequality by the same positive quantity gives an equivalent inequality.

That is, if $A\le B$, then $CA\le CB$, where A, B and C are real numbers or algebraic expressions and $C>0$.

Calculation:

Consider the given inequality $x\le 5$.

The left-hand side of the resulting inequality is given as $3x$.

Therefore, multiply both sides of the given inequality $x\le 5$ by 3.

Then, by the rule stated above, the inequality becomes as follows.

$\begin{array}{l}x\le 5\\ 3\cdot x\le 3\cdot 5\\ 3x\le 15\end{array}$

Thus, the complete statement is “If $x\le 5$, then $\underset{\_}{3x\le 15}$”.

To determine

To fill: The blank in the statement “If $x\ge 2$, then $-3x\_\_\_\_-6$.” with appropriate inequality sign.

Answer to Problem 1E

The complete statement is “If $x\ge 2$, then $\underset{\_}{-3x\le -6}$”.

Explanation of Solution

Rule used:

Multiplying each side of an inequality by the same negative quantity reverses the direction of the inequality.

That is, if $A\le B$, then $CA\ge CB$, where A, B and C are real numbers or algebraic expressions and $C<0$.

Calculation:

Consider the given inequality $x\ge 2$.

The left-hand side of the resulting inequality is given as $-3x$.

Therefore, multiply the same number $-3$, on both sides of the given inequality $x\ge 2$.

Then, by the rule stated above, the inequality becomes as follows.

$\begin{array}{l}x\ge 2\\ -3\cdot x\le -3\cdot 2\\ -3x\le -6\end{array}$

Thus, the complete statement is “If $x\ge 2$, then $\underset{\_}{-3x\le -6}$”.

To determine

To fill: The blank in the statement “If $x<-2$, then $-x\_\_\_\_2$.” with appropriate inequality sign.

Answer to Problem 1E

The complete statement is “If $x<-2$, then $\underset{\_}{-x>2}$”.

Explanation of Solution

Consider the given inequality $x<-2$.

The left-hand side of the resulting inequality is given as $-x$.

Therefore, multiply the same number $-1$, on the both sides of the given inequality $x<-2$.

Then, by the rule stated in part (c), the inequality becomes as follows.

$\begin{array}{l}x<-2\\ -1\cdot x>-1\cdot -2\\ -x>2\end{array}$

Thus, the complete statement is “If $x<-2$, then $\underset{\_}{-x>2}$”.