Exercise 3.2 - Chapter 3 Algebra Term 1 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.2$
Question $1 .$
Fill in the blanks
(i) The addition of $-7 \mathrm{~b}$ and $2 \mathrm{~b}$ is
(ii) The subtraction of $5 \mathrm{~m}$ from $-3 \mathrm{~m}$ is
(iii) The additive inverse of $-37 x y z$ is
Solution:
(i) $-5 b$
(ii) $-8 \mathrm{~m}$
(iii) $37 \mathrm{xyz}$

Question $2 .$
Say True or False
(i) The expressions $8 x+3 y$ and $7 x+2 y$ cannot be added
(ii) If $x$ is a natural number, then $x+1$ is its predecessor.
Hint: $x-1$ is its predecessor.
(iii) Sum of $a-b+c$ and $-a+b-c$ is zero
Solution:
(i) False
(ii) False
(iii) True

Question $3 .$
Add: (i) $8 \mathrm{x}, 3 \mathrm{x}$
(ii) $7 \mathrm{mn}, 5 \mathrm{mn}$
(iii) $-9 \mathrm{y}, 11 \mathrm{y}, 2 \mathrm{y}$
Solution:
(i) $8 x+3 x=(8+3) x=11 x$
(ii) $7 \mathrm{mn}+5 \mathrm{mn}=(7+5) \mathrm{mn}=12 \mathrm{mn}$
(iii) $-9 y+11 y+2 y=(-9+11+2) y=(2+2) y=4 y$

Question $4 .$
Subtract:
(i) $4 \mathrm{k}$ from $12 \mathrm{k}$
(ii) $15 \mathrm{q}$ from $25 \mathrm{q}$
(iii) $7 x y z$ from $17 x y z$
Solution:
(i) $4 \mathrm{k}$ from $12 \mathrm{k}$
$12 \mathrm{k}-4 \mathrm{k}=(12-4) \mathrm{k}=8 \mathrm{k}$
(ii) $15 \mathrm{q}$ from $25 \mathrm{q}$
$25 \mathrm{q}-15 \mathrm{q}=(25-15) \mathrm{q}=10 \mathrm{q}$
(iii) $7 x y z$ from $17 x y z$
$17 x y z-7 x y z=(17-7) x y z=10 x y z$

Question $5 .$
Find the sum of the following expressions
(i) $7 \mathrm{p}+6 \mathrm{q}, 5 \mathrm{p}-\mathrm{q}, \mathrm{q}+16 \mathrm{p}$
Solution: $(7 p+6 q)+(5 p-q)+(q+16 p)=7 p+$ $=(7 p+5 p+16 p)+(6 q-q+q)$ $=(7+5+16) p+(6-1+1) q$ $=(12+16) p+6 q=28 p+6 q$
$\begin{aligned}
&(7 p+6 q)+(5 p-q)+(q+16 p)=7 p+6 q+5 p-q+q+16 p \\
&=(7 p+5 p+16 p)+(6 q-q+q) \\
&=(7+5+16) p+(6-1+1) q \\
&=(12+16) p+6 q=28 p+6 q
\end{aligned}$
(ii) $a+5 b+7 c, 2 a+106+9 c$
Solution:
$\begin{aligned}
&(a+5 b+7 c)+(2 a+10 b+9 c)=a+5 b+7 c+2 a+10 b+9 c \\
&=a+2 a+5 b+10 b+7 c+9 c \\
&=(1+2) a+(5+10) b+(7+9) c \\
&=3 a+15 b+16 c
\end{aligned}$

(iii) $m n+t, 2 m n-2 t,-3 t+3 m n$
Solution:
$\begin{aligned}
&(\mathrm{mn}+\mathrm{t})+(2 \mathrm{mn}-2 \mathrm{t})+(-3 \mathrm{t}+3 \mathrm{mn}) \\
&=\mathrm{mn}+\mathrm{t}+2 \mathrm{mn}-2 \mathrm{t}+(-3 \mathrm{t})+3 \mathrm{mn} \\
&=(\mathrm{mn}+2 \mathrm{mn}+3 \mathrm{mn})+(\mathrm{t}-2 \mathrm{t}-3 \mathrm{t}) \\
&=(1+2+3) \mathrm{mn}+(1-2-3) \mathrm{t} \\
&=6 \mathrm{mn}+(1-5) \mathrm{t} \\
&=6 \mathrm{mn}+(-4) \mathrm{t} \\
&=6 \mathrm{mn}-4 \mathrm{t}
\end{aligned}$
(iv) $u+v, u-v, 2 u+5 v, 2 u-5 v$
Solution:
$(u+v)+(u-v)+(2 u+5 v)+(2 u-5 v)$
$\begin{aligned}
&=u+v+u-v+2 u+5 v+2 u-5 v \\
&=u+u+2 u+2 u+v-v+5 v-5 v \\
&=(1+1+2+2) u+(1-1+5-5) v=6 u+0 v \\
&=6 u
\end{aligned}$
(v) $5 x y z-3 x y, 3 z x y-5 y x$
Solution:
$\begin{aligned}
&5 x y z-3 x y+3 z x y-5 y x=5 x y z+3 x y z-3 x y-5 x y \\
&=(5+3) x y z+[(-3)+(-5)] x y=8 x y z+(-8) x y \\
&=8 x y z-8 x y
\end{aligned}$

Question $6 .$
Subtract
(i) $13 x+12 y-5$ from $27 x+5 y-43$
Solution:
$\begin{aligned}
&27 x+5 y-43-(13 x+12 y-5)=27 z+5 y-43+(-13 x-12 y+5) \\
&=27 x+5 y-43-13 x-12 y+5 \\
&=(27-13) x+(5-12) y+(-43)+5 \\
&=14 x+(-7) y+(-38)=14 x-7 y-38
\end{aligned}$
(ii) $3 \mathrm{p}+5$ from $\mathrm{p}-2 \mathrm{q}+7$
Solution:
$\begin{aligned}
&\mathrm{p}-2 \mathrm{q}+7-(3 \mathrm{p}+5)=\mathrm{p}-2 \mathrm{q}+7+(-3 \mathrm{p}-5) \\
&=\mathrm{p}-2 \mathrm{q}+7-3 \mathrm{p}-5=\mathrm{p}-3 \mathrm{p}-2 \mathrm{q}+7-5 \\
&=(1-3) \mathrm{p}-2 \mathrm{q}+2=-2 \mathrm{p}-2 \mathrm{q}+2
\end{aligned}$
(iii) $m+n$ from $3 m-7 n$
Solution:
$\begin{aligned}
&3 m-7 n-(m+n)=3 m-7 n+(-m-n) \\
&=3 m-7 n-m-n=(3 m-m)+(-7 n-n) \\
&=(3-1) m+(-7-1) n=2 m+(-8) n \\
&=2 m-8 n
\end{aligned}$
(iv) $2 \mathrm{y}+\mathrm{z}$ from $6 \mathrm{z}-5 \mathrm{y}$
Solution:
$\begin{aligned}
&6 z-5 y-(2 y+z)=6 z-5 y+(-2 y-z) \\
&=6 z-5 y-2 y-z=6 z-z-5 y-2 y \\
&=(6-1) z+(-5-2) y=5 z+(-7) y \\
&=5 z-7 y=-7 y+5 z
\end{aligned}$

Question $7 .$
Simplify
(i) $(x+y-z)+(3 x-5 y+7 z)-(14 x+7 y-6 z)$
Solution:

$\begin{aligned}
&(x+y-z)+(3 x-5 y+7 z)-(14 x-7 y-6 z) \\
&=(x+y-z)+(3 x-5 y+7 z)+(-14 x-7 y+6 z) \\
&=(x+3 x-14 x)+(y-5 y-7 y)+(-z+7 z+6 z) \\
&=(1+3-14) x+(1-5-7) y+(-1+7+6) z \\
&=-10 x-11 y+12 z
\end{aligned}$
(ii) $p+p+2+p+3+p-4-p-5+p+10$
Solution:
$\begin{aligned}
&\mathrm{p}+\mathrm{p}+2+3-\mathrm{p}-4-\mathrm{p}-5+\mathrm{p}+10=(\mathrm{p}+\mathrm{p}+\mathrm{p}-\mathrm{p}-\mathrm{p}+\mathrm{p})+(2+3-4-5+10) \\
&=(1+1+1-1-1+1) \mathrm{p}+6=2 \mathrm{p}+6
\end{aligned}$
(iii) $n+(m+1)+(n+2)+(m+3)+(n+4)+(m+5)$
Solution:
$\begin{aligned}
&\mathrm{n}+(\mathrm{m}+1)+(\mathrm{n}+2)+(\mathrm{m}+3)+(\mathrm{n}+4)+(\mathrm{m}+5) \\
&=\mathrm{n}+\mathrm{m}+1+\mathrm{n}+2+\mathrm{m}+3+\mathrm{n}+4+\mathrm{m}+5 \\
&=\mathrm{n}+\mathrm{n}+\mathrm{n}+\mathrm{m}+\mathrm{m}+\mathrm{m}+1+2+3+4+5 \\
&=(1+1+1) \mathrm{n}+(1+1+1) \mathrm{m}+15 \\
&=3 \mathrm{n}+3 \mathrm{~m}+15=3 \mathrm{~m}+3 \mathrm{n}+15
\end{aligned}$

Objective Type Questions
Question 8 .
The addition of $3 \mathrm{mn},-5 \mathrm{mn}, 8 \mathrm{mn}$ and $-4 \mathrm{mn}$ is
(i) $\mathrm{mn}$
(ii) $-\mathrm{mn}$
(iii) $2 \mathrm{mn}$
(iv) $3 \mathrm{mn}$
Solution:
(iii) $2 \mathrm{mn}$
Hint: $=3 \mathrm{mn}+8 \mathrm{mn}-5 \mathrm{mn}-4 \mathrm{mn}=11 \mathrm{mn}-9 \mathrm{mn}=2 \mathrm{mn}$

Question $9 .$
When we subtract ' $\mathrm{a}$ ' from '-a', we get
(i) a
(ii) $2 \mathrm{a}$
(iii) $-2 \mathrm{a}$
(iv) -a
Solution:
(iii) $-2 \mathrm{a}$
Hint: $-a-a=-2 a$

Question 10 .
In an expression, we can add or subtract only
(i) like terms
(ii) unlike terms
(iii) all terms
(iv) None of the above
Solution:
(i) like terms