Additional Questions - Chapter 3 - Algebra - Term 1 - 7th Maths Guide Samacheer Kalvi Solutions

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Additional Questions and Answers
Exercise 3.1
Question $1 .$
Write any three expressions each having 4 terms:
Solution:
(i) $2 x^{3}-3 x^{2}+3 x y+8$
(ii) $7 \mathrm{x}^{3}+9 \mathrm{y}^{2}-2 \mathrm{xy}^{2}-6$
(iii) $9 x^{2}-2 x+3 x y-1$

Question 2 .
Identify the co-efficients of the terms of the following expressions
(i) $2 x-2 y$
(ii) $x+y+3$
Solution:
(i) $2 \mathrm{x}-2 \mathrm{y}$
The co-efficient of $x$ in $2 x$ is 2
The co-efficient of $y$ in $-2 y$ is $-2$
(ii) $x+y+3$
The co-efficient of $x$ is 1
The co-efficient ofy is 1
The constant term is 3
 

Question $3 .$
Group the like terms together from the following: $6 \mathrm{x}, 6,-5 \mathrm{x},-5,1, \mathrm{x}, 6 \mathrm{y}, \mathrm{y}, 7 \mathrm{y}, 16 \mathrm{x}, 3$

Solution:
We have $6 x,-5 x, x, 16 x$ are like terms
$6 y, y, 7 y$, are like terms
$6,-5,1,3$ are like terms

Question $4 .$
Give the algebraic expressions for the following cases:
(i) One half of the sum of a and b.
(ii) Numbers $p$ and $q$ both squared and added
Solution:
(i) $\frac{1}{2}(a+b)$
(ii) $\mathrm{p}^{2}+\mathrm{q}^{2}$
 

Exercise $3.2$
Question $1 .$
If $\mathrm{A}=2 \mathrm{a}^{2}-4 \mathrm{~b}-1 ; \mathrm{B}=5 \mathrm{a}^{2}+3 \mathrm{~b}-8$ and $\mathrm{C}=2 \mathrm{a}^{2}-9 \mathrm{~b}+3$ then find the value of $\mathrm{A}-\mathrm{B}+\mathrm{C}$.
Solution:
Given $A=2 a^{2}-4 b-1 ; B=5 a^{2}+3 b-8 ; C=2 a^{2}-9 b+3$
$A-B+C=\left(2 a^{2}-4 b-1\right)-\left(5 a^{2}+3 b-8\right)+\left(2 a^{2}-9 b+3\right)$
$=2 \mathrm{a}^{2}-4 \mathrm{~b}-1+\left(-5 \mathrm{a}^{2}-3 \mathrm{~b}+8\right)+2 \mathrm{a}^{2}-9 \mathrm{~b}+3$
$=2 \mathrm{a}^{2}-4 \mathrm{~b}-1-5 \mathrm{a}^{2}-3 \mathrm{~b}+8+2 \mathrm{a}^{2}-9 \mathrm{~b}+3$
$=2 \mathrm{a}^{2}-5 \mathrm{a}^{2}+2 \mathrm{a}^{2}-4 \mathrm{~b}-3 \mathrm{~b}-9 \mathrm{~b}-1+8+3$
$=(2-5+2) a^{2}+(-4-3-9) 6+(-1+8+3)$
$=-\mathrm{a}^{2}-16 \mathrm{~b}+10$

Question $2 .$
How much $2 x^{3}-2 x^{2}+3 x+5$ is greater than $2 x^{3}+7 x^{2}-2 x+7 ?$
Solution:
The required expression can be obtained as follows.
$\begin{aligned}
&=2 x^{3}-2 x^{2}+3 x+5-\left(2 x^{3}+7 x^{2}-2 x+7\right) \\
&=2 x^{3}-2 x^{2}+3 x+5+\left(-2 x^{3}-7 x^{2}+2 x-7\right) \\
&=2 x^{3}-2 x^{2}+3 x+5-2 x^{3}-7 x^{2}+2 x-7 \\
&=(2-2) x^{3}+(-2-7) x^{2}+(3+2) x+(5-7) \\
&=0 x^{3}+\left(-9 x^{2}\right)+5 x-2=-9 x^{2}+5 x-2 \\
&\therefore 2 x^{3}-2 x^{2}+3 x+5 \text { is greater than } 2 x^{3}+7 x^{2}-2 x+7 \text { by }-9 x^{2}+5 x-2
\end{aligned}$

Question $3 .$
What should be added to $2 b^{2}-a^{2}$ to get $b^{2}-2 a^{2}$
Solution:
The required expression is obtained by subtracting $2 b^{2}-a^{2}$ from $b^{2}-2 a^{2}$
$\begin{aligned}
&b^{2}-2 a^{2}-\left(2 b^{2}-a^{2}\right)=b^{2}-2 a^{2}+\left(-2 b^{2}+a^{2}\right) \\
&=b^{2}-2 a^{2}-2 b^{2}+a^{2} \\
&=(1-2) b^{2}+(-2+1) a^{2}=-b^{2}-a^{2}
\end{aligned}$

Exercise $3.3$
Question 1 .
Length of one side of an equilateral triangle is $3 x-4$ units. Find the perimeter.
Solution:
Equilateral triangle has three sides equal.
Perimeter $=$ Sum of three sides
$=(3 x-4)+(3 x-4)+(3 x-4)=3 x-4+3 x-4+3 x-4$
$=(3+3+3) x+[(-4)+(-4)+(-4)]=9 x+(-12)=9 x-12$
$\therefore$ Perimeter $=9 \mathrm{x}-12$ units.

Question $2 .$
Find the perimeter of a square whose side is $y-2$ units.
Solution:
Perimeter $=(y-2)+(y-2)+(y-2)+(y-2)$
$=\mathrm{y}-2+\mathrm{y}-2+\mathrm{y}-2+\mathrm{y}-2=4 \mathrm{y}-8$
Perimeter of the square $=4 \mathrm{y}-8$ units.

Question $3 .$
Simplify $3 x-5-x+9$ if $x=3$
Solution:
$\begin{aligned}
&3 x-5-x+9=3(3)-5-3+9 \\
&=9-5-3+9=18-8=10
\end{aligned}$