Exercise 2.2 - Chapter 2 Basic Algebra 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

EX 2.2
Question 1.
Solve for $\mathrm{x}$.
(i) $|3-x|<7$
Solution:
$
\begin{aligned}
& \Rightarrow-7<3-x<73-x>-7 \\
& -x>-7-3(=-10) \\
& -x>-10 \Rightarrow x<10 \\
& 3-x<7 \\
& -x<7-3(=4) \\
& -x<4 x>-4 \ldots \text { (2) }
\end{aligned}
$
From (1) and (2) $\Rightarrow x>-4$ and $x<10$
$
\Rightarrow-4<\mathrm{x}<10
$
(ii) $|4 x-5| \geq-2$
Solution:
$
\begin{aligned}
4 x-5 \leq-2 \text { or } 4 x-5 \geq-2 & & \\
4 x \leq 2+5(=7) & & 4 x \geq-2+5(=3) \\
& \Rightarrow x \leq \frac{7}{4} \ldots(1) & \Rightarrow x \geq \frac{3}{4}
\end{aligned}
$
From (1) and (2) $\Rightarrow \frac{3}{4} \leq x \leq \frac{7}{4}$
(iii)
$
\left|3-\frac{3}{4} x\right| \leq \frac{1}{4}
$

Solution:

from (1) and (2) $\Rightarrow \frac{11}{3} \leq x \leq \frac{13}{3}$
(iv) $|x|-10<-3$
Solution:
$
\begin{aligned}
& |x|<-3+10(=7) \\
& |x|<7 \Rightarrow-7<x<7
\end{aligned}
$
Question 2.
Solve $\frac{1}{|2 x-1|}<6$ and express the solution using the interval notation.
Solution:
$
\begin{aligned}
& \frac{1}{|2 x-1|}<6 \\
& \Rightarrow|2 x-1|>\frac{1}{6}
\end{aligned}
$

Question 3 .
Solve $-3|x|+5 \leq-2$ and graph the solution set in a number line.
Solution:
$
\begin{aligned}
& -3|\mathrm{x}|+5 \leq-2 \\
& \Rightarrow-3|\mathrm{x}| \leq-2-5(=-7) \\
& -3|\mathrm{x}| \leq-7 \Rightarrow 3|\mathrm{x}| \geq 7 \\
& \Rightarrow|x| \geq \frac{7}{3} \\
& \Rightarrow \quad x \leq \frac{-7}{3} \text { and } x \geq \frac{7}{3} \\
& \text { (i.e.) } \frac{-7}{3} \leq x \leq \frac{7}{3} \quad \text { (i.e.) } x \in\left(-\infty, \frac{-7}{3}\right] \cup\left[\frac{7}{3}, \infty\right] \\
&
\end{aligned}
$
Question 4.
Solve $2|x+1|-6 \leq 7$ and graph the solution set in a number line.
Solution:
$
\begin{aligned}
& 2|x+1|-6 \leq 7 \\
\Rightarrow & 2|x+1| \leq 7+6(=13) \\
\Rightarrow & |x+1| \leq \frac{13}{2} \\
\Rightarrow & x+1>\frac{-13}{2} \quad \text { (or) } x+1<\frac{13}{2}
\end{aligned}
$

Question 5.
Solve $\frac{1}{5}|10 x-2|<1$.
Solution:
$
\frac{1}{5}|10 x-2|<1 \Rightarrow|10 x-2|<1 \times 5(=5)
$

Question 6.
Solve $|5 \mathrm{x}-12|<-2$
Solution:
$
5 x-12>-2 \text { (or) } 5 x-12<2
$