Additional Questions - Chapter 3 Algebra Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Questions and Answers
Exercise $3.1$
Question $1 .$
If $4 x^{2}+y^{2}=40$ and $x y=b$ find the value of $2 x+y$.
Solution:
We have $(a+b)^{2}=a^{2}+2 a b+b^{2}$
$(2 x+y)^{2}=(2 x)^{2}+(2 \times 2 x \times y)+y^{2}=\left(4 x^{2}+y^{2}\right)+4 x y=40+4 \times 6=40+24$
$(2 x+y)^{2}=64$
$(2 x+y)^{2}=82$
$2 x+y=8$
 

Question $2 .$
If $x^{2}+\frac{1}{z^{2}}=23$ find $x+\frac{1}{x}$
Solution:
We have $(a+b)^{2}=a^{2}+2 a b+b^{2}$
So $\left(x+\frac{1}{x}\right)^{2}=x^{2}+2 \times x \times \frac{1}{x}+\frac{1}{x^{2}}$
$=x^{2}+2+\frac{1}{z^{2}}=x^{2}+\frac{1}{z^{2}}+2=23+2$
$\because x^{2}+\frac{1}{x^{2}}=23$
$\left(x+\frac{1}{x}\right)^{2}=25$
$\left(x+\frac{1}{x}\right)^{2}=52$
$x+\frac{1}{x}=5$

Question $3 .$
Find the product of $\left(\frac{2}{3} x^{2}+5 y^{2}\right)\left(\frac{2}{3} x^{2}+5 y^{2}\right)$
Solution:
$\left(\frac{2}{3} x^{2}+5 y^{2}\right)\left(\frac{2}{3} x^{2}+5 y^{2}\right)=\left(\frac{2}{3} x^{2}+5 y^{2}\right)^{2}$
We have $(a+b)^{2}=a^{2}+2 a b+b^{2}$
Here $a=\frac{2}{3} x^{2} b=5 y^{2}$ $\left(\frac{2}{3} x^{2}+5 y^{2}\right)^{2}=\left(\frac{2}{3} x^{2}\right)^{2}+2 \times \frac{2}{3} x^{2} \times 5 y^{2}+(5 y 2$ $=\left(\frac{2}{3}\right)^{2}\left(x^{2}\right)^{2}+\frac{20 x^{2} y^{2}}{3}+52\left(y^{2}\right)^{2}$ $\left(\frac{2}{3} x^{2}+5 y^{2}\right)^{2}=\frac{4}{9} x^{4}+\frac{20 x^{2} y^{2}}{3}+25 y^{4}$
$\begin{aligned}
&\left(\frac{2}{3} x^{2}+5 y^{2}\right)^{2}=\left(\frac{2}{3} x^{2}\right)^{2}+2 \times \frac{2}{3} x^{2} \times 5 y^{2}+\left(5 y^{2}\right)^{2} \\
&=\left(\frac{2}{3}\right)^{2}\left(x^{2}\right)^{2}+\frac{20 x^{2} y^{2}}{3}+52\left(y^{2}\right)^{2} \\
&\left(\frac{2}{3} x^{2}+5 y^{2}\right)^{2}=\frac{4}{9} x^{4}+\frac{20 x^{2} y^{2}}{3}+25 y^{4}
\end{aligned}$
 

Exercise $3.2$
Question $1 .$
Solve $2 x+5<15$ where $x$ is a natural number and represent the solution in a number line.
Solution:
$2 x+5<15$
Subtracting 5 on both sides $2 x+5-5<15-5$
$2 \mathrm{x}<10$
Dividing by 2 on both the sides
$\frac{2 x}{2}<\frac{10^{\circ}}{2}$.
Since $x$ is a natural number and it is less than 5 , the solution is $4,3,2$ and 1. It is shown in the number line as below.

Question $2 .$
Solve $2 \mathrm{c}+4 \leq 14$, where $\mathrm{c}$ is a whole number.
Solution:
$2 \mathrm{c}+4 \leq 14$
Subtracting 4 on both sides $2 c+4-4 \leq 14-4$
$2 \mathrm{c} \leq 10$
Dividing by 2 on both the sides
$\frac{2 c}{2} \leq \frac{10}{2}$
$\mathrm{c} \leq 5$
Since the solutions are whole numbers which are less than $r$ equal to 5 , the solution set is $0,1,2,3$, 4 and $5 .$

Question 3 .
Solve $-8<-2 \mathrm{n}+4, \mathrm{n}$ is a natural number.
Solution:
$-8<-2 n+4$
Subtracting 4 on both sides
$-8-4<-2 n+4-4$
$-12<-2 \mathrm{n}$
$+$ by $-2$, we have
$\frac{-2 n}{-2}<\frac{-12}{-2}[\because$ Dividing by negative number, the inequation get reversed]
$\mathrm{n}<6$
Since the solutions are natural numbers which are less then 6 , we have the solution as $1,2,3,4$ and
$5 .$