Introduction

We can find which values satisfy an inequality by trial and error, which involves substituting a number for the variable and checking to see if we get a true statement.

For example, which of the following values of y satisfy the inequality y - 1 > 3\,?

  1. y=6
  2. y=2
  3. y=3

Let's go through the possibilities one at a time.

  • Substituting the value {\color{blue}y}={\color{blue}6} into the given equation, we get: \begin{align} {\color{blue}y}-1 &> 3 \\ {\color{blue}6}-1 &> 3 \\ 5 & > 3 \quad \color{green}\large{\checkmark} \end{align} This is a true statement, so y=6 satisfies this inequality.

  • Substituting the value {\color{blue}y}={\color{blue}2} into the given equation, we get: \begin{align} {\color{blue}y}-1 &> 3 \\ {\color{blue}2}-1 &> 3 \\ 1 & > 3 \quad \color{red}\large{\times} \end{align} This is a false statement, so y=2 does not satisfy this inequality.

  • Substituting the value {\color{blue}y}={\color{blue}3} into the given equation, we get: \begin{align} {\color{blue}y}-1 &> 3 \\ {\color{blue}3}-1 &> 3 \\ 2 & > 3 \quad \color{red}\large{\times} \end{align} This is a false statement, so y=3 does not satisfy this inequality.

Therefore, only I satisfies the inequality.

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Example: Identifying Values that Satisfy a Given Inequality

Which of the following values of x satisfy the inequality 2x < 10\,?

  1. x=0
  2. x=2
  3. x=5
EXPLANATION

We substitute each value into the given inequality and check to see if the resulting statement is true or false.

  • We substitute the value {\color{blue}x}={\color{blue}0} into the given inequality. We get: \begin{align} 2{\color{blue}x} & < 10 \\ 2\cdot{\color{blue}x} & < 10 \\ 2\cdot{\color{blue}0} &< 10 \\ 0 &< 10 \quad \color{green}\large{\checkmark} \end{align} So, substituting x=0 into the inequality leads to a true statement.

    Therefore, x=0 satisfies this inequality.

  • We substitute the value {\color{blue}x}={\color{blue}2} into the given inequality. We get: \begin{align} 2{\color{blue}x} & < 10 \\ 2\cdot{\color{blue}x} & < 10 \\ 2\cdot{\color{blue}2} &< 10 \\ 4 &< 10 \quad \color{green}\large{\checkmark} \end{align} So, substituting x=2 into the inequality leads to a true statement.

    Therefore, x=2 satisfies this inequality.

  • We substitute the value {\color{blue}x}={\color{blue}5} into the given inequality. We get: \begin{align} 2{\color{blue}x} & < 10 \\ 2\cdot{\color{blue}x} & < 10 \\ 2\cdot{\color{blue}5} &< 10 \\ 10 &< 10 \quad \color{red}\large{\times} \end{align} So, substituting x=5 into the inequality leads to a false statement.

    Therefore, x=5 does not satisfy this inequality.

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Practice: Identifying Values that Satisfy a Given Inequality

Which of the following values of $w$ satisfy the inequality $5w \leq 20\,?$

  1. $w=5$
  2. $w=4$
  3. $w=3$
a
 I and II only
b
 II and III only
c
 III only
d
 I only
e
 II only
Practice: Identifying Values that Satisfy a Given Inequality

Which of the following values of $a$ satisfy the inequality $a+6 > 9\,?$

  1. $a=9$
  2. $a=1$
  3. $a=3$
a
 II only
b
 I only
c
 I and III only
d
 None
e
 II and III only
Example: Identifying Values that Satisfy a Given Inequality with Fractions

Which of the following values of t satisfy the inequality \dfrac{2t}{3} < 8\,?

  1. t=9
  2. t=6
  3. t=3
EXPLANATION

We substitute each value into the given inequality and check to see if the resulting statement is true or false.

  • We substitute the value {\color{blue}t}={\color{blue}9} into the given inequality. We get: \begin{align} \dfrac{2{\color{blue}t}}{3} &<8 \\[5pt] \dfrac{2\cdot{\color{blue}t}}{3} &<8 \\[5pt] \dfrac{2\cdot{\color{blue}9}}{3} &< 8\\[5pt] \dfrac{18}{3} &< 8\\ 6 & <8 \quad \color{green}\large{\checkmark} \end{align} So, substituting t = 9 into the inequality leads to a true statement.

    Therefore, t=9 satisfies this inequality.

  • We substitute the value {\color{blue}t}={\color{blue}6} into the given inequality. We get: \begin{align} \dfrac{2{\color{blue}t}}{3} &<8 \\[5pt] \dfrac{2\cdot{\color{blue}t}}{3} &<8 \\[5pt] \dfrac{2\cdot{\color{blue}6}}{3} &< 8\\[5pt] \dfrac{12}{3} &< 8\\ 4 & <8 \quad \color{green}\large{\checkmark} \end{align} So, substituting t=6 into the inequality leads to a true statement.

    Therefore, t=6 satisfies this inequality.

  • We substitute the value {\color{blue}t}={\color{blue}3} into the given inequality. We get: \begin{align} \dfrac{2{\color{blue}t}}{3} &<8 \\[5pt] \dfrac{2\cdot{\color{blue}t}}{3} &<8 \\[5pt] \dfrac{2\cdot{\color{blue}3}}{3} &< 8\\[5pt] \dfrac{6}{3} &< 8\\ 2 & <8 \quad \color{green}\large{\checkmark} \end{align} So, substituting t=3 into the inequality leads to a true statement.

    Therefore, t=3 satisfies this inequality.

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Practice: Identifying Values that Satisfy a Given Inequality with Fractions

Which of the following values of $t$ satisfy the inequality $\dfrac{2t}{5} > 3\,?$

  1. $t=5$
  2. $t=10$
  3. $t=15$
a
 II and III only
b
 I and II only
c
 III only
d
 I, II and III
e
 I only
Practice: Identifying Values that Satisfy a Given Inequality with Fractions

Which of the following values of $u$ satisfy the inequality $\dfrac{u}{3} \leq 5\,?$

  1. $u=15$
  2. $u=6$
  3. $u=12$
a
 I and II only
b
 II only
c
 I, II and III
d
 II and III only
e
 III only
Example: Identifying Values that Satisfy a Given Compound Inequality

Which of the following values of x satisfy the inequality 1 \leq \dfrac{x}{2}-1 < 3\,?

  1. x=0
  2. x=4
  3. x=10
EXPLANATION

We substitute each value into the given inequality and check to see if the resulting statement is true or false.

  • We substitute the value {\color{blue}x}={\color{blue}0} into the given inequality. We get: \begin{align} 1 &\leq \dfrac{{\color{blue}x}}{2}-1 < 3 \\[5pt] 1 &\leq \dfrac{{\color{blue}0}}{2}-1 < 3 \\[5pt] 1 &\leq 0-1 < 3 \\[5pt] 1 &\leq -1 < 3 \quad \color{red}\large{\times} \end{align} So, substituting x=0 into the inequality leads to a false statement.

    Therefore, x=0 does not satisfy this inequality.

  • We substitute the value {\color{blue}x}={\color{blue}4} into the given inequality. We get: \begin{align} 1 &\leq \dfrac{{\color{blue}x}}{2}-1 < 3 \\[5pt] 1 &\leq \dfrac{{\color{blue}4}}{2}-1 < 3 \\[5pt] 1 &\leq 2-1 < 3 \\[5pt] 1 &\leq 1 < 3 \quad \color{green}\large{\checkmark} \end{align} So, substituting x=4 into the inequality leads to a true statement.

    Therefore, x=4 satisfies this inequality.

  • We substitute the value {\color{blue}x}={\color{blue}10} into the given inequality. We get: \begin{align} 1 &\leq \dfrac{{\color{blue}x}}{2}-1 < 3 \\[5pt] 1 &\leq \dfrac{{\color{blue}10}}{2}-1 < 3 \\[5pt] 1 &\leq 5-1 < 3 \\[5pt] 1 &\leq 4 < 3 \quad \color{red}\large{\times} \end{align} So, substituting x=10 into the inequality leads to a false statement.

    Therefore, x=10 does not satisfy this inequality.

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Practice: Identifying Values that Satisfy a Given Compound Inequality

Which of the following values of $a$ satisfy the inequality $-3 \leq 2a + 1 < 8\,?$

  1. $a=-2$
  2. $a=2$
  3. $a=5$
a
 II only
b
 I only
c
 III only
d
 I and II only
e
 II and III only
Practice: Identifying Values that Satisfy a Given Compound Inequality

Which of the following values of $w$ satisfy the inequality $2 < 3w < 10\,?$

  1. $w=1$
  2. $w=4$
  3. $w=8$
a
 II only
b
 II and III only
c
 III only
d
 I only
e
 None
Example: Identifying Inequalities that are Satisfied by a Given Value

The value a = 2 satisfies which of the following inequalities?

  1. \dfrac{a}{2} + 2 > 5
  2. 1 \leq 2a < 6
  3. 3a - 1 < 10
EXPLANATION

We substitute the given value into each of the inequalities and check to see if the resulting statement is true or false.

  • We substitute the value {\color{blue}a} = {\color{blue}2} into the first inequality. We get: \begin{align} \dfrac{\color{blue}a}{2} + 2 &> 5 \\[5pt] \dfrac{\color{blue}2}{2} + 2 & > 5 \\[5pt] {\color{blue}1} + 2 & > 5 \\[5pt] 3 & > 5 \quad \color{red}\large{\times} \end{align} So, substituting a = 2 leads to a false statement.

    Therefore, a = 2 does not satisfy this inequality.

  • We substitute the value {\color{blue}a}={\color{blue}2} into the second inequality. We get: \begin{align} 1 &\leq 2{\color{blue}a} < 6 \\[0pt] 1 &\leq 2 \cdot {\color{blue}a} < 6 \\[0pt] 1 &\leq 2 \cdot {\color{blue}2} < 6 \\[0pt] 1 &\leq 4 < 6 \quad \color{green}\large{\checkmark} \end{align} So, substituting a = 2 leads to a true statement.

    Therefore, a = 2 satisfies this inequality.

  • We substitute the value {\color{blue}a} = {\color{blue}2} into the third inequality. We get: \begin{align} 3{\color{blue}a}-1 & < 10 \\ 3\cdot{\color{blue}a} - 1 & < 10 \\ 3\cdot{\color{blue}2} - 1 & < 10 \\ 5 &< 10 \quad \color{green}\large{\checkmark} \end{align} So, substituting a = 2 leads to a true statement.

    Therefore, a = 2 satisfies this inequality.

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Practice: Identifying Inequalities that are Satisfied by a Given Value

The value $t=4$ satisfies which of the following inequalities?

  1. $2t < -5$
  2. $2 \leq \dfrac{t}{2} \leq 4$
  3. $t + 1 < 5$
a
 II and III only
b
 None
c
 I and II only
d
 II only
e
 I only
Practice: Identifying Inequalities that are Satisfied by a Given Value

The value $q = 2$ satisfies which of the following inequalities?

  1. $q + \dfrac{1}{2} \leq \dfrac{5}{2}$
  2. $0 < 2q -1 < 1$
  3. $4q > 8$
a
 I only
b
 II only
c
 I and II only
d
 II and III only
e
 III only
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