Samacheer Kalvi 10th Maths Exercise 3.6 Solutions

10th Maths Exercise 3.6 solutions (Chapter 3 Algebra) with step-by-step answers for all 11 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance10 answers · full steps below

Q1 (i) $\dfrac{2x}{x-2}$
Q1 (ii) $\dfrac{2x^2+2x-7}{(x+3)(x-2)}=\dfrac{2x^2+2x-7}{x^2+x-6}$
Q1 (iii) $x^2+xy+y^2$
Q2 (i) $\dfrac{2(x-2)}{x-4}$
Q2 (ii) $\dfrac{1-x}{1+x}$
Q3 $\dfrac{2x^3+1}{(x^2+2)^2}$
Q4 $\dfrac{x+1}{x^2-2x+4}$
Q5 $\dfrac{4x^2-1}{2(4x^2+1)}$
Q7 $\dfrac{12}{5}$ hours $=2$ hours $24$ minutes
Q8Apples $=30$ kg and Bananas $=20$ kg

Question 1

(i)

Simplify (i) $\dfrac{x(x+1)}{x-2}+\dfrac{x(1-x)}{x-2}$

Denominators are the same, so add the numerators.
$=\dfrac{x(x+1)+x(1-x)}{x-2}$
$=\dfrac{x^2+x+x-x^2}{x-2}$
$=\dfrac{2x}{x-2}$

Answer

$\dfrac{2x}{x-2}$

(ii)

Simplify (ii) $\dfrac{x+2}{x+3}+\dfrac{x-1}{x-2}$

LCM of the denominators is $(x+3)(x-2)$ .
$=\dfrac{(x+2)(x-2)+(x-1)(x+3)}{(x+3)(x-2)}$
Numerator $=(x^2-4)+(x^2+2x-3)=2x^2+2x-7$
$=\dfrac{2x^2+2x-7}{(x+3)(x-2)}$

Answer

$\dfrac{2x^2+2x-7}{(x+3)(x-2)}=\dfrac{2x^2+2x-7}{x^2+x-6}$

(iii)

Simplify (iii) $\dfrac{x^3}{x-y}+\dfrac{y^3}{y-x}$

$y-x=-(x-y)$ , so $\dfrac{y^3}{y-x}=\dfrac{-y^3}{x-y}$ .
$=\dfrac{x^3-y^3}{x-y}$
Use $x^3-y^3=(x-y)(x^2+xy+y^2)$ .
$=\dfrac{(x-y)(x^2+xy+y^2)}{x-y}=x^2+xy+y^2$

Answer

$x^2+xy+y^2$

Question 2

(i)

Simplify (i) $\dfrac{(2x+1)(x-2)}{x-4}-\dfrac{(2x^2-5x+2)}{x-4}$

Same denominator, so subtract the numerators.
$=\dfrac{(2x+1)(x-2)-(2x^2-5x+2)}{x-4}$
$(2x+1)(x-2)=2x^2-3x-2$
Numerator $=2x^2-3x-2-2x^2+5x-2=2x-4$
$=\dfrac{2(x-2)}{x-4}$

Answer

$\dfrac{2(x-2)}{x-4}$

(ii)

Simplify (ii) $\dfrac{4x}{x^2-1}-\dfrac{x+1}{x-1}$

$x^2-1=(x+1)(x-1)$ , so LCM is $(x+1)(x-1)$ .
$=\dfrac{4x-(x+1)(x+1)}{(x+1)(x-1)}$
Numerator $=4x-(x^2+2x+1)=-(x^2-2x+1)=-(x-1)^2$
$=\dfrac{-(x-1)^2}{(x+1)(x-1)}=\dfrac{-(x-1)}{x+1}$
$=\dfrac{1-x}{1+x}$

Answer

$\dfrac{1-x}{1+x}$

Q3. Subtract $\dfrac{1}{x^2+2}$ from $\dfrac{2x^3+x^2+3}{(x^2+2)^2}$

$\dfrac{2x^3+x^2+3}{(x^2+2)^2}-\dfrac{1}{x^2+2}$
LCM is $(x^2+2)^2$ .
$=\dfrac{2x^3+x^2+3-(x^2+2)}{(x^2+2)^2}$
$=\dfrac{2x^3+1}{(x^2+2)^2}$

Answer

$\dfrac{2x^3+1}{(x^2+2)^2}$

Q4. Which rational expression should be subtracted from $\dfrac{x^2+6x+8}{x^3+8}$ to get $\dfrac{3}{x^2-2x+4}$

Show steps (6)

Let the required expression be $p(x)$ .
$\dfrac{x^2+6x+8}{x^3+8}-p(x)=\dfrac{3}{x^2-2x+4}$
Factorise: $x^2+6x+8=(x+2)(x+4)$ and $x^3+8=(x+2)(x^2-2x+4)$ .
$\dfrac{x^2+6x+8}{x^3+8}=\dfrac{x+4}{x^2-2x+4}$
$p(x)=\dfrac{x+4}{x^2-2x+4}-\dfrac{3}{x^2-2x+4}$
$p(x)=\dfrac{x+1}{x^2-2x+4}$

Answer

$\dfrac{x+1}{x^2-2x+4}$

Q5. If $A=\dfrac{2x+1}{2x-1}$ , $B=\dfrac{2x-1}{2x+1}$ find $\dfrac{1}{A-B}-\dfrac{2B}{A^2-B^2}$

$\dfrac{1}{A-B}-\dfrac{2B}{A^2-B^2}=\dfrac{1}{A-B}-\dfrac{2B}{(A+B)(A-B)}$
$=\dfrac{(A+B)-2B}{(A+B)(A-B)}=\dfrac{A-B}{(A+B)(A-B)}=\dfrac{1}{A+B}$
$A+B=\dfrac{2x+1}{2x-1}+\dfrac{2x-1}{2x+1}=\dfrac{(2x+1)^2+(2x-1)^2}{(2x-1)(2x+1)}$
$=\dfrac{8x^2+2}{4x^2-1}$
So the value $=\dfrac{4x^2-1}{8x^2+2}=\dfrac{4x^2-1}{2(4x^2+1)}$

Answer

$\dfrac{4x^2-1}{2(4x^2+1)}$

Q6. If $A=\dfrac{x}{x+1}$ , $B=\dfrac{1}{x+1}$ , prove that $\dfrac{(A+B)^2+(A-B)^2}{A\div B}=\dfrac{2(x^2+1)}{x(x+1)^2}$

Show steps (7)

$A+B=\dfrac{x}{x+1}+\dfrac{1}{x+1}=\dfrac{x+1}{x+1}=1$
$A-B=\dfrac{x}{x+1}-\dfrac{1}{x+1}=\dfrac{x-1}{x+1}$
$(A+B)^2+(A-B)^2=1+\dfrac{(x-1)^2}{(x+1)^2}=\dfrac{(x+1)^2+(x-1)^2}{(x+1)^2}$
$=\dfrac{2x^2+2}{(x+1)^2}=\dfrac{2(x^2+1)}{(x+1)^2}$
$A\div B=\dfrac{x}{x+1}\times\dfrac{x+1}{1}=x$
$\dfrac{(A+B)^2+(A-B)^2}{A\div B}=\dfrac{2(x^2+1)}{(x+1)^2}\times\dfrac{1}{x}=\dfrac{2(x^2+1)}{x(x+1)^2}$
Hence proved.

Answer

$\dfrac{(A+B)^2+(A-B)^2}{A\div B}=\dfrac{2(x^2+1)}{x(x+1)^2}$ is proved.

Q7. Pari needs $4$ hours to complete a work. His friend Yuvan needs …

Pari needs $4$ hours to complete a work. His friend Yuvan needs $6$ hours to complete the same work. How long will it take to complete if they work together?

Work done by Pari in $1$ hour $=\dfrac{1}{4}$ .
Work done by Yuvan in $1$ hour $=\dfrac{1}{6}$ .
Work done together in $1$ hour $=\dfrac{1}{4}+\dfrac{1}{6}=\dfrac{3+2}{12}=\dfrac{5}{12}$ .
Time taken together $=\dfrac{12}{5}=2\dfrac{2}{5}$ hours.
$=2$ hours $24$ minutes.

Answer

$\dfrac{12}{5}$ hours $=2$ hours $24$ minutes

Q8. Iniya bought $50$ kg of fruits consisting of apples and bananas. She …

Iniya bought $50$ kg of fruits consisting of apples and bananas. She paid twice as much per kg for the apple as she did for the banana. If Iniya bought ₹ $1800$ worth of apples and ₹ $600$ worth bananas, then how many kgs of each fruit did she buy?

Show steps (6)

Let the cost of $1$ kg banana be ₹ $x$ , then $1$ kg apple costs ₹ $2x$ .
Quantity of apples $=\dfrac{1800}{2x}=\dfrac{900}{x}$ kg.
Quantity of bananas $=\dfrac{600}{x}$ kg.
Total weight is $50$ kg: $\dfrac{900}{x}+\dfrac{600}{x}=50$
$\dfrac{1500}{x}=50\Rightarrow x=30$
Apples $=\dfrac{900}{30}=30$ kg, Bananas $=\dfrac{600}{30}=20$ kg.

Answer

Apples $=30$ kg and Bananas $=20$ kg

Quick check

Graph of a linear equation is a _______
1. (1) straight line
2. (2) circle
3. (3) parabola
4. (4) hyperbola
Show answer
Answer: (1) straight line
1. A linear equation in two variables has degree one, and its graph is a straight line.
Question 1 of 5
$\dfrac{3y - 3}{y} \div \dfrac{7y - 7}{3y^2}$ is
1. (1) $\dfrac{9y}{7}$
2. (2) $\dfrac{9y^3}{(21y - 21)}$
3. (3) $\dfrac{21y^2 - 42y + 21}{3y^3}$
4. (4) $\dfrac{7(y^2 - 2y + 1)}{y^2}$
Show answer
Answer: (1) $\dfrac{9y}{7}$
1. $\dfrac{3y-3}{y} \div \dfrac{7y-7}{3y^2} = \dfrac{3(y-1)}{y} \times \dfrac{3y^2}{7(y-1)}$
2. Cancel $(y-1)$ and one $y$ .
3. $= \dfrac{9y}{7}$
Question 2 of 5
Transpose of a column matrix is
1. (1) unit matrix
2. (2) diagonal matrix
3. (3) column matrix
4. (4) row matrix
Show answer
Answer: (4) row matrix
1. A column matrix has order $m \times 1$ .
2. Its transpose has order $1 \times m$ , which is a row matrix.
Question 3 of 5
The solution of $\left(2x - 1\right)^2 = 9$ is equal to
1. (1) $-1$
2. (2) $2$
3. (3) $-1,\ 2$
4. (4) None of these
Show answer
Answer: (3) $-1,\ 2$
1. $(2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3$ .
2. $2x - 1 = 3 \Rightarrow x = 2$ .
3. $2x - 1 = -3 \Rightarrow x = -1$ .
Question 4 of 5
The values of $a$ and $b$ if $4x^4 - 24x^3 + 76x^2 + ax + b$ is a perfect square are
1. (1) $100,120$
2. (2) $10,12$
3. (3) $-120,100$
4. (4) $12,10$
Show answer
Answer: (3) $-120,100$
1. Let $4x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2$ .
2. Expanding: $4x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2$ .
3. Compare $x^2$ terms: $36 + 4c = 76 \Rightarrow c = 10$ .
4. So $a = -12(10) = -120$ and $b = 10^2 = 100$ .
Question 5 of 5

Chapter 3 formulas

Roots of a quadratic equation
$ax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Sum of the roots
$\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}$
Product of the roots
$\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}$
Quadratic equation formed from its roots
$x^2 - (\alpha + \beta)x + \alpha\beta = 0$
Discriminant
$\Delta = b^2 - 4ac$
Nature of roots: real and unequal
$\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}$
Nature of roots: real and equal
$\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}$
Nature of roots: no real roots
$\Delta < 0 \;\Rightarrow\; \text{there are no real roots}$
Cases for a system of linear equations in three variables
$\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}$
LCM of algebraic expressions
$\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}$
Quadratic polynomial and its zeroes
$p(x) = ax^2 + bx + c \;(a \neq 0) \;\Rightarrow\; p(x) \text{ has atmost two zeroes, which lie on the } x\text{-axis}$
Order of a matrix
$\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n$
Row matrix (row vector)
$A = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}$
Column matrix (column vector)
$A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}$
Square matrix
$\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n$
Zero matrix (null matrix)
$a_{ij} = 0 \text{ for all } i,\, j$
Transpose of a matrix
$\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})$
Diagonal matrix
$a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})$
Scalar matrix
$a_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}$
Identity matrix (unit matrix)
$a_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}$
Lower triangular matrix
$a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})$
Upper triangular matrix
$a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})$
Equality of matrices
$A = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j$

Answers checked against the Samacheer Kalvi textbook · Updated 28 September 2026

Spotted a mistake? Email us at [email protected] and we will fix it.

Question 1

(i)

(ii)

(iii)

Question 2

(i)

(ii)

Q3. Subtract 1x2+2\dfrac{1}{x^2+2}x2+21​ from 2x3+x2+3(x2+2)2\dfrac{2x^3+x^2+3}{(x^2+2)^2}(x2+2)22x3+x2+3​

Q4. Which rational expression should be subtracted from x2+6x+8x3+8\dfrac{x^2+6x+8}{x^3+8}x3+8x2+6x+8​ to get 3x2−2x+4\dfrac{3}{x^2-2x+4}x2−2x+43​

Q5. If A=2x+12x−1A=\dfrac{2x+1}{2x-1}A=2x−12x+1​ , B=2x−12x+1B=\dfrac{2x-1}{2x+1}B=2x+12x−1​ find 1A−B−2BA2−B2\dfrac{1}{A-B}-\dfrac{2B}{A^2-B^2}A−B1​−A2−B22B​

Q6. If A=xx+1A=\dfrac{x}{x+1}A=x+1x​ , B=1x+1B=\dfrac{1}{x+1}B=x+11​ , prove that (A+B)2+(A−B)2A÷B=2(x2+1)x(x+1)2\dfrac{(A+B)^2+(A-B)^2}{A\div B}=\dfrac{2(x^2+1)}{x(x+1)^2}A÷B(A+B)2+(A−B)2​=x(x+1)22(x2+1)​

Q7. Pari needs 444 hours to complete a work. His friend Yuvan needs …

Q8. Iniya bought 505050 kg of fruits consisting of apples and bananas. She …

Quick check

Q3. Subtract $\dfrac{1}{x^2+2}$ from $\dfrac{2x^3+x^2+3}{(x^2+2)^2}$

Q4. Which rational expression should be subtracted from $\dfrac{x^2+6x+8}{x^3+8}$ to get $\dfrac{3}{x^2-2x+4}$

Q5. If $A=\dfrac{2x+1}{2x-1}$ , $B=\dfrac{2x-1}{2x+1}$ find $\dfrac{1}{A-B}-\dfrac{2B}{A^2-B^2}$

Q6. If $A=\dfrac{x}{x+1}$ , $B=\dfrac{1}{x+1}$ , prove that $\dfrac{(A+B)^2+(A-B)^2}{A\div B}=\dfrac{2(x^2+1)}{x(x+1)^2}$

Q7. Pari needs $4$ hours to complete a work. His friend Yuvan needs …

Q8. Iniya bought $50$ kg of fruits consisting of apples and bananas. She …