Exercise 3.5 - Chapter 3 Algebra 10th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

$\operatorname{Ex} 3.5$
Question $1 .$
Simplify
(i) $\frac{4 x^{2} y}{2 z^{2}} \times \frac{6 x z^{3}}{20 y^{4}}$
(ii) $\frac{p^{2}-10 p+21}{p-7} \times \frac{p^{2}+p-12}{(p-3)^{2}}$
(iii) $\frac{5 t^{3}}{4 t-8} \times \frac{6 t-12}{10 t}$
Solution:
(i) $\frac{4 x^{2} y}{2 z^{2}} \times \frac{6 x z^{3}}{20 y^{4}}=\frac{3 x^{3} z}{5 y^{3}}$
(ii) $\frac{p^{2}-10 p+21}{p-7} \times \frac{p^{2}+p-12}{(p-3)^{2}}$
$=\frac{(p-7)(p-3)}{(p-7)} \times \frac{(p+4)(p-3)}{(p-3)(p-3)}=p+4$
(iii) $\frac{5 t^{3}}{4 t-8} \times \frac{6 t-12}{10 t}=\frac{\not t^{3} \times 6^{3}(t-2)}{4(t-2) \times \underset{2}{2}+\frac{3 t^{2}}{4}}$

Question 2.
Simplify
(i) $\frac{x+4}{3 x+4 y} \times \frac{9 x^{2}-16 y^{2}}{2 x^{2}+3 x-20}$
(ii) $\frac{x^{3}-y^{3}}{3 x^{2}+9 x y+6 y^{2}} \times \frac{x^{2}+2 x y+y^{2}}{x^{2}-y^{2}}$
Solution:
(i) $\frac{x+4}{3 x+4 y} \times \frac{9 x^{2}-16 y^{2}}{2 x^{2}+3 x-20}$
$\begin{aligned}
&=\frac{(x+4)\left((3 x)^{2}-(4 y)^{2}\right)}{(3 x+4 y)(x+4)(2 x-5)} \\
&=\frac{(3 x+4 y)(3 x-4 y)}{(3 x+4 y)(2 x-5)}=\frac{3 x-4 y}{(2 x-5)}
\end{aligned}$
(ii)
$\begin{aligned}
&\frac{x^{3}-y^{3}}{3 x^{2}+9 x y+6 y^{2}} \times \frac{x^{2}+2 x y+y^{2}}{x^{2}-y^{2}} \\
&=\frac{(x-y)\left(x^{2}+x y+y^{2}\right)\left(x^{2}+2 x y+y^{2}\right)}{3\left(x^{2}+3 x y+2 y^{2}\right)(x+y)(x-y)} \\
&=\frac{\left(x^{2}+x y+y^{2}\right)(x+y)^{2}}{3(x+2 y)(x+y)(x+y)} \\
&=\frac{\left(x^{2}+x y+y^{2}\right)}{3(x+2 y)}
\end{aligned}$

Question $3 .$
Simplify
(i) $\frac{2 a^{2}+5 a+3}{2 a^{2}+7 a+6} \div \frac{a^{2}+6 a+5}{-5 a^{2}-35 a-50}$
(ii) $\frac{b^{2}+3 b-28}{b^{2}+4 b+4} \div \frac{b^{2}-49}{b^{2}-5 b-14}$
(iii) $\frac{x+2}{4 y} \div \frac{x^{2}-x-6}{12 y^{2}}$
(iv) $\frac{12 t^{2}-22 t+8}{3 t} \div \frac{3 t^{2}+2 t-8}{2 t^{2}+4 t}$
Solution:

$\text { (i) } \begin{aligned}
& \frac{2 a^{2}+5 a+3}{2 a^{2}+7 a+6} \div \frac{a^{2}+6 a+5}{-5 a^{2}-35 a-50} \\
=& \frac{2 a^{2}+5 a+3}{2 a^{2}+7 a+6} \times \frac{-5 a^{2}-35 a-50}{a^{2}+6 a+5} \\
=& \frac{(2 a+3)(a+1)}{2 a^{2}+7 a+6} \times \frac{-5\left(a^{2}+7 a+10\right)}{(a+5)(a+1)} \\
=& \frac{(2 a+3)(-5)(a+5)(a+2)}{\left(2 a^{2}+7 a+6\right)(a+5)} \\
=& \frac{(2 a+3)(-5)(a+2)}{(2 a+3)(a+2)}=-5 \\
\text { (ii) } & \frac{b^{2}+3 b-28}{b^{2}+4 b+4} \div \frac{b^{2}-49}{b^{2}-5 b-14} \\
=& \frac{b^{2}+3 b-28}{b^{2}+4 b+4} \times \frac{b^{2}-5 b-14}{b^{2}-49} \\
=& \frac{(b+7)(b-4)}{(b+2)(b+2)} \times \frac{(b-7)(b+2)}{(b+7)(b-7)} \\
=& \frac{b-4}{b+2}
\end{aligned}$

(iii) $\frac{x+2}{4 y} \div \frac{x^{2}-x-6}{12 y^{2}}=\frac{x+2}{4 y} \times \frac{312 y^{2}}{x^{2}-x-6}$
$=\frac{(x+2)}{1} \times \frac{3 y}{(x-3)(x+2)}$
$=\frac{3 y}{x-3}$
(iv)
$\begin{aligned}
&\frac{12 t^{2}-22 t+8}{3 t} \div \frac{3 t^{2}+2 t-8}{2 t^{2}+4 t} \\
&=\frac{12 t^{2}-22 t+8}{3 t} \times \frac{2 t^{2}+4 t}{3 t^{2}+2 t-8} \times \frac{-8^{4}}{6_{3}} \frac{-\beta^{1}}{6_{2}} \\
&=\frac{2\left(6 t^{2}-11 t+4\right)}{3 t} \times \frac{2 t(t+2)}{3 t^{2}+2 t-8} \times \frac{6^{2}}{\not 3} \\
&=\frac{2(3 t-4)(2 t-1)}{3} \times \frac{2(t-2)}{3(t+2)(3 t-4)} \\
&=\frac{4(2 t-1)}{3}
\end{aligned}$

Question $4 .$
If $x=\frac{a^{2}+3 a-4}{3 a^{2}-3}$ and $y=\frac{a^{2}+2 a-8}{2 a^{2}-2 a-4}$ find the value of $x^{2} y^{2}$.
Solution:
(i)
$\begin{aligned}
x &=\frac{a^{2}+3 a-4}{3 a^{2}-3}, y=\frac{a^{2}+2 a-8}{2 a^{2}-2 a-4} \\
x^{2} y^{-2} &=\left(\frac{a^{2}+3 a-4}{3 a^{2}-3}\right)\left(\frac{a^{2}+3 a-4}{3 a^{2}-3}\right) \times\left(\frac{2 a^{2}-2 a-4}{a^{2}+2 a-8}\right)^{2} \\
=& \frac{(a+4)(a-1)}{3(a+1)} \times \frac{(a-1)}{3(a+1)(a-1)} \\
\times \frac{2(a-2)(a+1) 2(a-2)(a+1)}{(a+4)(a-2)(a+4)(a-2)}
\end{aligned}$

$=\frac{4}{9}$

Question $5 .$
If a polynomial $p(x)=x^{2}-5 x-14$ is divided by another polynomial $q(x)$ we get $\frac{x-7}{x+2}$ find $q(x)$.
Solution:
$p(x)=x^{2}-5 x-14$
$\left(x^{2}-5 x-14\right) \div q(x)=\frac{x-7}{x+2}$
$\left(x^{2}-5 x-14\right) \times \frac{1}{q(x)}=\frac{x-7}{x+2}$
$\frac{1}{q(x)}=\frac{x-7}{x+2} \times \frac{1}{x^{2}-5 x-14}$
$\frac{1}{q(x)}=\frac{x-7}{x+2} \times \frac{1}{(x-7)(x+2)}=\frac{1}{(x+2)^{2}}$
$\therefore q(x)=(x+2)^{2}=x^{2}+4 x+4$