Exercise 3.6 - Chapter 3 - Algebra - 10th Maths Guide Samacheer Kalvi Solutions - Tamil Medium

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Ex 3.6
Question 1.

Simplify
(i) $\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}$
(ii) $\frac{x+2}{x+3}+\frac{x-1}{x-2}$
(iii) $\frac{x^{3}}{x-y}+\frac{y^{3}}{y-x}$
Solution:
(i) $\frac{x(x+1)}{x-2}+\frac{x(1-x)}{x-2}=\frac{x(x+1)+x(1-x)}{(x-2)}=\frac{2 x}{x-2}$
(ii) $\frac{x+2}{x+3}+\frac{x-1}{x-2}=\frac{(x-2)(x+2)+(x+3)(x-1)}{(x+3)(x-2)}$
$=\frac{x^{2}-4+x^{2}+2 x-3}{(x+3)(x-2)}$
$=\frac{2 x^{2}+2 x-7}{(x+3)(x-2)}$
(iii)
$\begin{aligned}
\frac{x^{3}}{x-y}+\frac{y^{3}}{y-x} &=\frac{x^{3}}{x-y}-\frac{y^{3}}{x-y}=\frac{x^{3}-y^{3}}{(x-y)} \\
&=\frac{(x-y)\left(x^{2}+x y+y^{2}\right)}{x-y} \\
&=x^{2}+x y+y^{2}
\end{aligned}$

Question 2.
Simplify
(i) $\frac{(2 x+1)(x-2)}{x-4}-\frac{\left(2 x^{2}-5 x+2\right)}{x-4}$
(ii) $\frac{4 x}{x^{2}-1}-\frac{x+1}{x-1}$
Solution:
(i) $\frac{(2 x+1)(x-2)}{x-4}-\frac{\left(2 x^{2}-5 x+2\right)}{x-4}$
$=\frac{2 x^{2}-3 x-2-2 x^{2}+5 x-2}{x-4}=\frac{2 x-4}{x-4}$
$=\frac{2(x-2)}{x-4}$
(ii) $\frac{4 x}{x^{2}-1}-\frac{x+1}{x-1}=\frac{4 x}{(x+1)(x-1)}-\frac{x+1}{(x-1)}$
$\begin{aligned}
&=\frac{4 x-(x+1)(x+1)}{(x+1)(x-1)}-\frac{4 x-\left(x^{2}+2 x+1\right)}{(x+1)(x-1)} \\
&=\frac{4 x-x^{2}-2 x-1}{(x+1)(x-1)}
\end{aligned}$

$\begin{aligned}
&=\frac{-x^{2}+2 x-1}{(x+1)(x-1)}=\frac{-\left(x^{2}-2 x+1\right)}{(x+1)(x-1)} \\
&=\frac{-(x-1)(x-1)}{(x+1)(x-1)} \\
&=\frac{-(x-1)}{(x+1)}=\frac{1-x}{1+x}
\end{aligned}$

Question $3 .$
Subtract

$\frac{2 x^{3}+x^{2}+3}{\left(x^{2}+2\right)^{2}}-\frac{1}{x^{2}+2}$
Solution:
$\frac{2 x^{3}+x^{2}+3}{\left(x^{2}+2\right)^{2}}-\frac{1}{x^{2}+2}$

$\begin{aligned}
&=\frac{2 x^{3}+x^{2}+3-\left(x^{2}+2\right)}{\left(x^{2}+2\right)^{2}} \\
&=\frac{2 x^{3}+x^{2}+3-x^{2}-2}{\left(x^{2}+2\right)^{2}}=\frac{2 x^{3}+1}{\left(x^{2}+2\right)^{2}}
\end{aligned}$

Question $4 .$
Which rational expression should be subtracted from $\frac{x^{2}+6 x+8}{x^{5}+8}$ to get $\frac{3}{x^{2}-2 x+4}$ Solution:
$\begin{aligned}
&\frac{x^{2}+6 x+8}{x^{3}+8}-q(x)=\frac{3}{x^{2}-2 x+4} \\
&q(x)=\frac{x^{2}+6 x+8}{x^{3}+8}-\frac{3}{x^{2}-2 x+4}
\end{aligned}$
$\begin{aligned}
&=\frac{(x+4)(x+2)}{(x+2)\left(x^{2}-2 x+4\right)}-\frac{3}{x^{2}-2 x+4} \\
&=\frac{(x+4)(x+2)}{(x+2)\left(x^{2}-2 x+4\right)}-\frac{3}{x^{2}-2 x+4} \\
&=\frac{x+4-3}{x^{2}-2 x+4} \\
&=\frac{x+1}{x^{2}-2 x+4}
\end{aligned}$

Question $5 .$
If $A=\frac{2 x+1}{2 x-1}, B=\frac{2 x-1}{2 x+1}$ find $\frac{1}{A-B}-\frac{2 B}{A^{2}-B^{2}}$
Solution:
$\begin{aligned}
&\mathrm{A}=\frac{2 x+1}{2 x-1}, \mathrm{~B}=\frac{2 x-1}{2 x+1} \\
&=\frac{1}{\mathrm{~A}-\mathrm{B}}-\frac{2 \mathrm{~B}}{\mathrm{~A}^{2}-\mathrm{B}^{2}}=\frac{\mathrm{A}+\mathrm{B}-2 \mathrm{~B}}{(\mathrm{~A}+\mathrm{B})(\mathrm{A}-\mathrm{B})} \\
&=\frac{(\mathrm{A}-\mathrm{B})}{(\mathrm{A}+\mathrm{B})(\mathrm{A}-\mathrm{B})}=\frac{1}{\mathrm{~A}+\mathrm{B}} \\
&=\frac{\frac{1}{2 x+1}+\frac{2 x-1}{2 x+1}}{2 x-1}=\frac{(2 x+1)^{2}+(2 x-1)^{2}}{(2 x-1)(2 x+1)} \\
&=\frac{4 x^{2}+4 x+1+4 x^{2}-4 x+1}{(2 x-1)(2 x+1)}=\frac{(2 x-1)(2 x+1)}{8 x^{2}+2} \\
&=\frac{4 x^{2}-1}{2\left(4 x^{2}+1\right)}
\end{aligned}$

Question 6.

If $A=, B=$, prove that $\frac{(A+B)^{2}+(A-B)^{2}}{A \div B}=\frac{2\left(x^{2}+1\right)}{x(x+1)^{2}}$

Solution:
$\mathrm{A}=\frac{x}{x+1}, \mathrm{~B}=\frac{1}{x+1}$
$=\frac{(A+B)^{2}+(A-B)^{2}}{A \div B}$
$=\frac{\mathrm{A}^{2}+2 \mathrm{AB}+\mathrm{B}^{2}+\mathrm{A}^{2}-2 \mathrm{AB}+\mathrm{B}^{2}}{(\mathrm{~A} \div \mathrm{B})}$
$=\frac{2 \mathrm{~A}^{2}+2 \mathrm{~B}^{2}}{\mathrm{~A} \div \mathrm{B}}=\frac{2\left(\mathrm{~A}^{2}+\mathrm{B}^{2}\right)}{(\mathrm{A} \div \mathrm{B})}$
$=\frac{2\left(\left(\frac{x}{x+1}\right)^{2}+\left(\frac{1}{x+1}\right)^{2}\right)}{\frac{x}{x+1} \div \frac{1}{x+1}}$
$=\frac{2\left(\frac{x^{2}}{(x+1)^{2}}+\frac{1}{(x+1)^{2}}\right)}{\frac{x(x+1)}{(x+1)}}=2\left(\frac{\left(x^{2}+1\right)}{x(x+1)^{2}}\right)$
 

Question $7 .$
Pari needs 4 hours to complete a work. His friend Yuvan needs 6 hours to complete the same work. How long will be take to complete if they work together?
Answer:
Let the work done by Pari and Yuvan together be $x$
Work done by part $=\frac{1}{4}$
Work done by Yuvan $=\frac{1}{6}$

By the given condition
$\begin{aligned}
&\frac{1}{4}+\frac{1}{6}=\frac{1}{x} \Rightarrow \frac{3+2}{12}=\frac{1}{x} \\
&\frac{5}{12}=\frac{1}{x} \\
&5 x=12 \Rightarrow x=\frac{12}{5} \\
&x=2 \frac{2}{5} \text { hours (or) } 2 \text { hours } 24 \text { minutes }
\end{aligned}$
 

Question $8 .$
Iniya bought $50 \mathrm{~kg}$ of fruits consisting of apples and bananas. She paid twice as much per $\mathrm{kg}$ for the apple as she did for the banana. If Iniya bought Rs. 1800 worth of apples and Rs. 600 worth bananas, then how many kgs of each fruit did she buy?
Answer:
Let the quantity of apples and bananas purchased be ' $x$ ' and ' $y$ '
By the given condition
$x+y=50 \ldots \ldots . .(1)$
Cost of one kg of apple $=\frac{1800}{x}$
Cost of one kg of banana $=\frac{6000}{y}$
By the given condition
One kg of apple $=2 \frac{(600)}{y}$
Total cost of fruits purchased $=1800+600$
$x \times 2 \frac{(600)}{y}+y \frac{(600)}{y}=2400$
$\frac{1200 x}{y}=24000-600$
$\frac{1200 x}{y}=1800$
$1200 \mathrm{x}=1800 \times \mathrm{y}$
$\mathrm{x}=\frac{1800 x}{12000}=\frac{3 y}{2}$

Substitute the value of $x$ in (1)
$\begin{aligned}
&\frac{3 y}{2}+y=50 \\
&\frac{5 y}{2}=50 \\
&5 y=100 \Rightarrow y=\frac{100}{5}=20 \\
&x=\frac{3 y}{2}=\frac{3 \times 20}{2} \\
&=30
\end{aligned}$
The quantity of apples $=30 \mathrm{~kg}$
The quantity of bananas $=20 \mathrm{~kg}$