Linear Inequations

Solve the inequation $\;\;$ $\dfrac{-x}{3} \leq \dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}, \; x \in R$
Write the solution set and represent it on a number line.

Consider $\;$ $\dfrac{-x}{3} \leq \dfrac{x}{2} - \dfrac{4}{3}$

i.e. $\;$ $\dfrac{4}{3} \leq \dfrac{x}{2} + \dfrac{x}{3}$

i.e. $\;$ $\dfrac{4}{3} \leq \dfrac{5x}{6}$

i.e. $\;$ $8 \leq 5x$

i.e. $\;$ $\dfrac{8}{5} \leq x$

i.e. $\;$ $1.6 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{1}{6} + \dfrac{4}{3}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{9}{6}$

i.e. $\;$ $x < 3$ $\;\;\; \cdots \; (2)$

$\therefore \;$ From equations $(1)$ and $(2)$, the solution set of the given inequation is

$\left\{x \; \bigg | \; 1.6 \leq x < 3, \; \; x \in R \right\}$

Linear Inequations

Find the values of $x$ which satisfies the inequation $\;$ $-2 \leq \dfrac{1}{2} - \dfrac{2x}{3} \leq 1 \dfrac{5}{6}, \; x \in N$
Graph the solution set on a number line.

Consider $\;$ $-2 \leq \dfrac{1}{2} - \dfrac{2x}{3}$

i.e. $\;$ $-2 \leq \dfrac{3 - 4x}{6}$

i.e. $\;$ $-12 \leq 3 - 4x$

i.e. $\;$ $4x \leq 15$

i.e. $\;$ $x \leq \dfrac{15}{4}$

i.e. $\;$ $x \leq 3.75$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{1}{2} - \dfrac{2x}{3} \leq 1 \dfrac{5}{6}$

i.e. $\;$ $\dfrac{1}{2} - \dfrac{2x}{3} \leq \dfrac{11}{6}$

i.e. $\;$ $\dfrac{3 - 4x}{6} \leq \dfrac{11}{6}$

i.e. $\;$ $3 - 4x \leq 11$

i.e. $\;$ $-4x \leq 8$

i.e. $\;$ $-x \leq 2$

i.e. $\;$ $-2 \leq x$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, the values of $x$ which satisfy the given inequation as:

$\left\{x \; \big | -2 \leq x \leq 3.75 \right\}$

$\because \;$ $x \in N$, $\;$ the required values of $x$ are: $\;$ $\left\{1, \; 2, \; 3 \right\}$

Linear Inequations

Solve the inequation $\;$ $\dfrac{-1}{3} \leq \dfrac{x}{2} - 1\dfrac{1}{3} < \dfrac{1}{6}, \; x \in R$ $\;$ and represent the solution set on a number line.

Consider $\;$ $\dfrac{-1}{3} \leq \dfrac{x}{2} - 1\dfrac{1}{3}$

i.e. $\;$ $\dfrac{-1}{3} \leq \dfrac{x}{2} - \dfrac{4}{3}$

i.e. $\;$ $\dfrac{4}{3} - \dfrac{1}{3} \leq \dfrac{x}{2}$

i.e. $\;$ $1 \leq \dfrac{x}{2}$

i.e. $\;$ $2 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{x}{2} - 1\dfrac{1}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} - \dfrac{4}{3} < \dfrac{1}{6}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{1}{6} + \dfrac{4}{3}$

i.e. $\;$ $\dfrac{x}{2} < \dfrac{3}{2}$

i.e. $\;$ $x < 3$ $\;\;\; \cdots \; (2)$

$\therefore \;$ From equations $(1)$ and $(2)$, the solution set of the given inequation is

$\left\{x \; \big | \; 2 \leq x < 3, \; x \in R \right\}$

Linear Inequations

Given $P = \left\{x \; \Big| \; 9 < 2x -1 \leq 13, \; x \in R \right\}$, $\;$ $Q = \left\{x \; \Big| \; -5 \leq 3 + 4x < 15, \; x \in I \right\}$ $\;$ where $R$ is the set of real numbers and $I$ is the set of integers.

Represent $P$ and $Q$ on different number lines.

Write down the elements of $P \cap Q$.

$P = \left\{x \; \Big| \; 9 < 2x -1 \leq 13, \; x \in R \right\}$

Consider $\;$ $9 < 2x - 1$

i.e. $\;$ $10 < 2x$ $\implies$ $5 < x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $2x - 1 \leq 13$

i.e. $\;$ $2x \leq 14$ $\implies$ $x \leq 7$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$ the solution set of $P$ is

$\left\{x \; \Big | \; 5 < x \leq 7, \; x \in R\right\}$

$Q = \left\{x \; \Big| \; -5 \leq 3 + 4x < 15, \; x \in I \right\}$ Consider $\;$ $-5 \leq 3 + 4x$

i.e. $\;$ $-8 \leq 4x$ $\implies$ $-2 \leq x$ $\;\;\; \cdots \; (3)$

Consider $\;$ $3 + 4x < 15$

i.e. $\;$ $4x < 12$ $\implies$ $x < 3$ $\;\;\; \cdots \; (4)$

$\therefore \;$ We have from equations $(3)$ and $(4)$ the solution set of $P$ is

$\left\{x \; \Big | \; -2 \leq x < 3, \; x \in I \right\}$

i.e. $\;$ $\left\{-2, -1, 0, 1, 2 \right\}$

Now, $\;$ $P \cap Q = \left\{x \; \Big | \; 5 < x \leq 7, \; x \in R\right\} \cap \left\{x \; \Big | \; -2 \leq x < 3, \; x \in I \right\}$

i.e. $\;$ $P \cap Q = \phi$ (null set)

Linear Inequations

Solve the inequation $\;$ $-3 + x \leq \dfrac{8x}{3} + 2 \leq \dfrac{14}{3} + 2x; \; x \in I$ $\;$ and represent the solution set on a number line.

Consider $\;$ $-3 + x \leq \dfrac{8x}{3} + 2$

i.e. $\;$ $-3 - 2\leq \dfrac{8x}{3} - x$

i.e. $\;$ $-5 \leq \dfrac{5x}{3}$

i.e. $\;$ $-1 \leq \dfrac{x}{3}$

i.e. $\;$ $-3 \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $\dfrac{8x}{3} + 2 \leq \dfrac{14}{3} + 2x$

i.e. $\;$ $\dfrac{8x}{3} - 2x \leq \dfrac{14}{3} - 2$

i.e. $\;$ $\dfrac{2x}{3} \leq \dfrac{8}{3}$

i.e. $\;$ $2x \leq 8$

i.e. $\;$ $x \leq 4$ $\;\;\; \cdots (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;$ $- 3 \leq x \leq 4$

$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -3 \leq x \leq 4, \; x \in I \right\}$

Linear Inequations

Solve the linear inequation and represent the solution set on a number line:
$-3 \left(x - 7\right) \geq 15 - 7x > \left(\dfrac{x + 1}{3}\right), \;\; x \in R$

Consider $\;$ $-3 \left(x - 7\right) \geq 15 - 7x$

i.e. $\;$ $-3x + 21 \geq 15 - 7x$

i.e. $\;$ $4x \geq -6$

i.e. $\;$ $x \geq -\dfrac{3}{2}$ $\implies$ $- \dfrac{3}{2} \leq x$ $\;\;\; \cdots \; (1)$

Consider $\;$ $15 - 7x > \dfrac{x + 1}{3}$

i.e. $\;$ $45 - 21x > x + 1$

i.e. $\;$ $44 > 22 x$

i.e. $\;$ $2 > x$ $\implies$ $x < 2$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;\;$ $- \dfrac{3}{2} \leq x < 2$

$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -\dfrac{3}{2} \leq x < 2, \; x \in R \right\}$

Linear Inequations

Solve the following inequation: $\;$ $\dfrac{1}{5} \leq \dfrac{3 x}{10} + 1 < 1 \dfrac{3}{5}, \; x \in R$
Write the solution set and represent the solution set on a number line.

Consider $\;$ $\dfrac{1}{5} \leq \dfrac{3 x}{10} + 1$

i.e. $\;$ $\dfrac{1}{5} \leq \dfrac{3x + 10}{10}$

i.e. $\;$ $1 \leq \dfrac{3x + 10}{2}$

i.e. $\;$ $2 \leq 3x + 10$

i.e. $\;$ $- 8 \leq 3x$

i.e. $- \dfrac{8}{3} \leq x$ $\;\;\; \cdots \; (1)$

$\implies$ $x \geq - \dfrac{8}{3}$

Consider $\;$ $\dfrac{3 x}{10} + 1 < 1 \dfrac{3}{5}$

i.e. $\;$ $\dfrac{3x + 10}{10} < \dfrac{8}{5}$

i.e. $\;$ $\dfrac{3x + 10}{2} < 8$

i.e. $\;$ $3x + 10 < 16$

i.e. $\;$ $3x < 6$ $\implies$ $x < 2$ $\;\;\; \cdots \; (2)$

$\therefore \;$ We have from equations $(1)$ and $(2)$, $\;$ $- \dfrac{8}{3} \leq x < 2$

$\therefore \;$ The solution set of the given inequation is: $\;$ $\left\{x \mid -\dfrac{8}{3} \leq x < 2, \; x \in R \right\}$