Exercise 3.6 - Chapter 3 Algebra 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 3.6$
Question $1 .$
Fill in the blanks:
(i) The value of $x$ in the equation $x+5=12$ is
Answer:
7
Hint:
Given, $x+5=12$
$x=12-5=7$ (by transposition method)
Value of $x$ is 7
(ii) The value of $y$ in the equation $y-9=(-5)+7$ is
Answer:
11
Hint:
Given, $y-9=(-5)+7$
$y-9=7-5$ (re-arranging)
$y-9=2$
$\therefore y=2+9=11$ (by transposition method)
(iii) The value of $\mathrm{m}$ in the equation $8 \mathrm{~m}=56$ is
Answer:
7
Hint:
Given, $8 m=56$
Divided by 8 on both sides
$\begin{aligned}
&\frac{8 \times m}{\mathrm{~s}}=\frac{5 \mathrm{fi}}{\mathrm{s}} \\
&\therefore \mathrm{m}=7
\end{aligned}$
(iv) The value of $p$ in the equation $\frac{2 p}{3}=10$ is
Answer:
15
Hint:
Given, $\frac{2 p}{3}=10$
Multiplying by 3 on both sides

$\begin{aligned}
&\frac{2 p}{\3} \times \not \3=10 \\
&\text { Dividing by } 2 \text { on both } \\
&\frac{2 p}{2}=\frac{30}{2} \\
&\therefore p=15
\end{aligned}$
(v) The linear equation in one variable has solution.
Answer:
one
 

Question $2 .$
Say True or False.
(i) The shifting of a number from one side of an equation to other is called transposition.
Answer:
True
(ii) Linear equation in one variable has only one variable with power $2 .$
Answer:
False
[Linear equation in one variable has only one variable with power one-correct statement]
 

Question 3 .
Match the following

(A) (i),(ii), (iv), (iii),(v)
(B) (iii), (iv), (i), (ii), (v)
(C) (iii),(i), (iv), (v), (ii)
(D) (iii), (i) , (v), (iv), (ii)
Answer:
(C) (iii),(i), (iv), (v), (ii)
a. $\frac{x}{2}=10$, multiplying by 2 on both sides, we get $\frac{2}{2} \times 2=10 \times 2 \Rightarrow x=20$
b. $20=6 x-4$ by transposition $\Rightarrow 20+4=6 x$ $6 x=24$ dividing by 6 on both sides, $\frac{6 x}{6}=\frac{24}{6} \Rightarrow x=4$

c. $2 x-5=3-x$
By transposing the variable ' $x$ ', we get $2 x-5+x=3$
by transposing $-5$ to other side, $2 x+x=3+5$
$\therefore 3 x=8$
$\therefore \mathrm{x}=\frac{8}{3}$
d. $7 x-4-8 x=20$
by transposing $-4$ to other side,
$7 x-8 x=20+4$
$-\mathrm{x}=24$
$\therefore \mathrm{x}=-24$
e. $\frac{4}{11}-x=\frac{-7}{11}$
Transposing $\frac{4}{11}$ to other side,
$-\mathrm{x}=\frac{-7}{11} \frac{-4}{11}=\frac{-7-4}{11}=\frac{-11}{11}=-1$
$\therefore-x=-1 \Rightarrow x=1$

Question $4 .$
Find $x$ :
(i) $\frac{2 x}{3}-4=\frac{10}{3}$
Answer:

Transposing $-4$ to other side, it becomes $+4$
$\therefore \frac{2 x}{3}=\frac{10}{3}+4$ Taking LCM \& adding,
$\frac{2 x}{3}=\frac{10}{3}+\frac{4}{1}=\frac{10+12}{3}=\frac{22}{3}$
$\frac{2 x}{3}=\frac{22}{3}$ Multiplying by 3 on both sides
$\frac{2 x}{\not \beta} \times \not \beta=\frac{22}{\not 3} \times \not \beta \Rightarrow 2 x=22$, dividing by 2 on both sides,
We get $\frac{2 x}{2}=\frac{22}{2}$
$\therefore x=11$
(ii) $y+\frac{1}{6}-3 y=\frac{2}{3}$
Answer:
Transposing to the other side,
$y-3 y=\frac{2}{3}-\frac{1}{6}$
Taking LCM,
$\begin{aligned}
-2 y &=\frac{2}{3}-\frac{1}{6}=\frac{2 \times 2-1}{6}=\frac{3}{6}=\frac{1}{2} \\
\therefore-2 y &=\frac{1}{2} \Rightarrow 2 y=-\frac{1}{2}, \text { dividing by } 2 \text { or both sides. } \\
\frac{\not y y}{2} &=-\frac{1}{2} \times \frac{1}{2} \Rightarrow y=-\frac{1}{4}
\end{aligned}$

(iii) $\frac{1}{3}-\frac{x}{3}=\frac{7 x}{12}+\frac{5}{4}$
Answer:
Transposing $\frac{-x}{3}$ to the other side, it becomes $+\frac{x}{3}$
$\therefore \frac{1}{3}=\frac{7 x}{12}+\frac{5}{4}+\frac{x}{3}$
Transposing $\frac{5}{4}$ to the other side, it becomes $\frac{-5}{4}$
$\frac{1}{3}-\frac{5}{4}=\frac{7 x}{12}+\frac{z}{3}$
Multiply by 12 throughout [we look at the denominators $3,4,12,3$ and take the LCM, which is 12]

$\begin{aligned}
&4-15=7 x+x \times 4 \\
&-11=7 x+4 x \\
&11 x=-11 \\
&x=-1
\end{aligned}$

Question $5 .$
Find $x$
(i) $-3(4 x+9)=21$
Answer:
$-3(4 x+9)=21$
Expanding the bracket,
$\begin{aligned}
&-3 \times 4+(-3) \times 9=21 \\
&\therefore-12 \mathrm{x}+(-27)=21 \\
&-12 \mathrm{x}-27=21
\end{aligned}$
Transposing $-27$ to other side, it becomes $+27$
$\begin{aligned}
&-12 \mathrm{x}=21+27=48 \\
&\therefore-12 \mathrm{x}=48 \Rightarrow 12 \mathrm{x}=-48
\end{aligned}$
Dividing by 12 on both sides
$\frac{12 x}{12}=\frac{-48}{12} \Rightarrow x=-1$
(ii) $20-2(5-p)=8$
Answer:
$20-2(5-p)=8$
Expanding the bracket,
$\begin{aligned}
&20-2 \times 5-2 \times(-p)=8 \\
&20-10+2 p=8 \\
&(-2 \times-p=2 p) \\
&10+2 p=8 \text { transposing lo to other side } \\
&2 p=8-10=-2 \\
&\therefore 2 p=-2 \\
&\therefore \mathrm{p}=-1
\end{aligned}$

(iii) $(7 x-5)-4(2+5 x)=10(2-x)$
Answer:
$(7 x-5)-4(2+5 x)=10(2-x)$
Expanding the brackets,
$7 \mathrm{x}-5-4 \times 2-4 \times 5 \mathrm{x}=10 \times 2+10 \times(-\mathrm{x})$
$\begin{aligned}
&7 x-5-8-20 x=20-10 x \\
&7 x-13-20 x=20-10 x
\end{aligned}$
Transposing $10 \mathrm{x} \&-13$, we get
$\begin{aligned}
&7 x-13-20 x+10 x=20 \\
&7 x-20 x+10 x=20+13 \text {, Simplifying } \\
&-3 x=33 \\
&3 x=33 \\
&x=\frac{-33}{3}=-11 \\
&x=-11
\end{aligned}$
 

Question $6 .$
Find $x$ and $m$ :
(i) $\frac{3 x-2}{4}-\frac{(x-3)}{5}=-1$
Answer:
Taking LCM on LHS,
$\frac{(3 x-2) \times 5-(x-3) \times 4}{20}=-1$

$\frac{(3 x-2) \times 5-(x-3) \times 4}{20}=-1$
Expanding brackets,
$\begin{aligned}
\frac{3 x \times 5-2 \times 5-x \times 4-(-3) \times 4}{20} &=-1 \\
\frac{15 x-10-4 x-(-12)}{20} &=-1 \\
\frac{15 x-10-4 x+12}{20} &=-1 \Rightarrow \frac{11 x+2}{20}=-1
\end{aligned}$
Multiplying both sides by 20
$\begin{aligned}
\frac{11 x+2}{20} \times 20 &=-1 \times 20 \\
\therefore 11 x+2 &=-20 \\
\therefore 11 x &=-20-2=-22 \\
x &=\frac{-22}{11}=-2 \quad \therefore \quad \text { Ins: } x=-2
\end{aligned}$

(ii) $\frac{m+9}{3 m+15}=\frac{5}{3}$
Answer:
Cross multiplying, we get
$\begin{aligned}
&\frac{m+9}{3 m+15} \ \frac{5}{3} \frac{m+9}{3 m+15} \ \frac{5}{3} \\
&\therefore(\mathrm{m}+9) \times 3=5 \times(3 \mathrm{~m}+15) \\
&\mathrm{m} \times 3+9 \times 3=5 \times 3 \mathrm{~m}+5 \times 15
\end{aligned}$
$3 m+27=15 m+75 \quad$ Transposing $3 m$ \& 75, we get
$\begin{aligned}
&27-75=15 m-3 m \\
&-48=12 m \\
&\frac{12 m}{12}=\frac{-48}{12} \\
&\Rightarrow m=-4
\end{aligned}$