Shweta was teaching "method to solve a linear inequality in one v… - Vidyadip

Question

Shweta was teaching "method to solve a linear inequality in one variable" to her daughter.
Step I Collect all terms involving the variable (x) on one side and constant terms on other side with the help of above rules and then reduce it in the form $\mathbf{a x}<\mathbf{b}$ or $\mathbf{a x} \leq \mathbf{b}$ or $\mathbf{a x}>\mathbf{b}$ or $\mathbf{a x} \geq \mathbf{b}$.
Step II Divide this inequality by the coefficient of variable (x). This gives the solution set of given inequality.
Step III Write the solution set.

Based on above information, answer the following questions.

(i) The solution set of $\mathbf{2 4 x}<\mathbf{1 0 0}$, when $\mathrm{x}$ is a natural number is
(a) $\{1,2,3,4\}$ (b) $(1,4)$ (c) $[1,4]$ (d) None of these

(ii) The solution set of $24100 \mathrm{x}<$, when $\mathrm{x}$ is an integer is
(a) $\{\ldots \ldots-4,-3,-2,-1,0,1,2,3,4\}$ (b) $(-\infty, 4]$ (c) $[4, \infty]$ (d) None of the above

(iii) The solution set of $-\mathbf{5 x}+\mathbf{2 5}>0$ is
(a) $[5, \infty)$ (b) $(-\infty, 5]$ (c) $(5, \infty)$ (d) $(-\infty, 5)$

(iv) The solution set of $\mathbf{3 x}-\mathbf{5}<\mathbf{x + 7}$ is
(a) $(6, \infty)$ (b) $[6, \infty)$ (c) $(-\infty, 6)$ (d) $(-\infty, 6]$

(v) The solution set of $x+\frac{x}{2}+\frac{x}{3}<11$ is
(a) $(-\infty, 6]$ (b) $(-\infty, 6)$ (c) $[6, \infty)$ (d) None of these

✓

Answer

(i) (a) We have, $\mathbf{2 4 x}<\mathbf{1 0 0}$
On dividing both sides by 24 , we get
$
=\frac{24 x}{24}<\frac{100}{24} \Rightarrow x<\frac{25}{6}
$
When $\mathbf{x}$ is a natural number, then solutions of the inequality $x<\frac{25}{6}$, are all natural numbers, which are less than $\frac{25}{6}$. In this case, the following values of $x$ make the statement true.
$
\mathrm{x}=1,2,3,4
$
Hence, the solution set of inequality is $\{\mathbf{1 , 2 , 3 , 4}\}$.

(ii) (a) We have, $24 \mathrm{x}<\mathbf{1 0 0}$
On dividing both sides by 24 , we get
$
=\frac{24 x}{24}<\frac{100}{24} \Rightarrow x<\frac{25}{6}
$
When $\mathbf{x}$ is an integer.
In this case, solutions of given inequality are
$
\ldots,-4,-3,-2,-1,0,1,2,3,4
$
Hence, the solution set of inequality is
$
\{\ldots,-4,-3,-2,-1,0,1,2,3,4\}
$

(iii) (d) We have, $-5 x+25>0$
On adding $5 \mathbf{x}$ both sides, we get
$
-5 x+25+5 x>0+5 x \Rightarrow 25>5 x \Rightarrow 5 x<25
$
On dividing both sides by 5 , we get
$
\Rightarrow \frac{5 x}{5}<\frac{25}{5} \Rightarrow x<5
$
Hence, the required solution set is $(-\infty, 5)$.

(iv)
(c) We have, $3 \mathbf{x}-\mathbf{5} < \mathbf{x}+\mathbf{7}$
$\Rightarrow 3 x-5+5 < x+7+5$
$\Rightarrow 3 \mathrm{x}<\mathrm{x}+12$
$\Rightarrow 3 \mathrm{x}-\mathrm{x}<\mathrm{x}+12-\mathrm{x}$
[adding 5 both sides]
[subtracting $\mathbf{x}$ from both sides]
$
\begin{aligned}
& \Rightarrow \quad 2 x<12 \\
& \Rightarrow \quad \frac{2 x}{2}<\frac{12}{2} \\
& \Rightarrow \quad x<6
\end{aligned}
$
[dividing both sides by 2]
The solution set is $\{\mathbf{x}: \mathbf{x} \in \mathbf{R}$ and $\mathbf{x}<6\}$ i.e., any real number less than 6 . This can also be written as $(-\infty, 6)$.

(v) (b) We have, $x+\frac{x}{2}+\frac{x}{3}<11$
$
\Rightarrow \frac{6 x+3 x+2 x}{6}<11 \Rightarrow \frac{11 x}{6}<11
$

On multiplying both sides by $\frac{6}{11}$, we get
$
\begin{aligned}
& \frac{11 x}{6} \times \frac{6}{11}<11 \times \frac{6}{11} \Rightarrow x<6 \\
& \therefore \quad x \in(-\infty, 6) \\
&
\end{aligned}
$

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