Search

Samacheer Kalvi 10th Maths Exercise 3.4 Solutions

10th Maths Exercise 3.4 solutions (Chapter 3 Algebra) with step-by-step answers for all 8 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below
  1. Q1 (i)x−1x\dfrac{x-1}{x}
  2. Q1 (ii)x−9x−2\dfrac{x-9}{x-2}
  3. Q1 (iii)9x−1\dfrac{9}{x-1}
  4. Q1 (iv)p+52p(p−4)\dfrac{p+5}{2p(p-4)}
  5. Q2 (i)Excluded values are y=5y=5 and y=−5y=-5
  6. Q2 (ii)Excluded values are t=2t=2 and t=3t=3
  7. Q2 (iii)Excluded value is x=1x=1
  8. Q2 (iv)Excluded values are x=0, x=−3, x=2x=0,\ x=-3,\ x=2

Q1. Reduce each of the following rational expressions to its lowest form.

(i)

x2−1x2+x\dfrac{x^2-1}{x^2+x}

  1. Factorise numerator and denominator.

  2. x2−1=(x+1)(x−1)x^2-1=(x+1)(x-1), x2+x=x(x+1)x^2+x=x(x+1)

  3. x2−1x2+x=(x+1)(x−1)x(x+1)\dfrac{x^2-1}{x^2+x}=\dfrac{(x+1)(x-1)}{x(x+1)}

  4. Cancel the common factor (x+1)(x+1).

Answer

x−1x\dfrac{x-1}{x}

(ii)

x2−11x+18x2−4x+4\dfrac{x^2-11x+18}{x^2-4x+4}

  1. x2−11x+18=(x−9)(x−2)x^2-11x+18=(x-9)(x-2)

  2. x2−4x+4=(x−2)2x^2-4x+4=(x-2)^2

  3. x2−11x+18x2−4x+4=(x−9)(x−2)(x−2)(x−2)\dfrac{x^2-11x+18}{x^2-4x+4}=\dfrac{(x-9)(x-2)}{(x-2)(x-2)}

  4. Cancel (x−2)(x-2).

Answer

x−9x−2\dfrac{x-9}{x-2}

(iii)

9x2+81xx3+8x2−9x\dfrac{9x^2+81x}{x^3+8x^2-9x}

  1. 9x2+81x=9x(x+9)9x^2+81x=9x(x+9)

  2. x3+8x2−9x=x(x2+8x−9)=x(x+9)(x−1)x^3+8x^2-9x=x(x^2+8x-9)=x(x+9)(x-1)

  3. 9x2+81xx3+8x2−9x=9x(x+9)x(x+9)(x−1)\dfrac{9x^2+81x}{x^3+8x^2-9x}=\dfrac{9x(x+9)}{x(x+9)(x-1)}

  4. Cancel x(x+9)x(x+9).

Answer

9x−1\dfrac{9}{x-1}

(iv)

p2−3p−402p3−24p2+64p\dfrac{p^2-3p-40}{2p^3-24p^2+64p}

  1. p2−3p−40=(p−8)(p+5)p^2-3p-40=(p-8)(p+5)

  2. 2p3−24p2+64p=2p(p2−12p+32)=2p(p−8)(p−4)2p^3-24p^2+64p=2p(p^2-12p+32)=2p(p-8)(p-4)

  3. p2−3p−402p3−24p2+64p=(p−8)(p+5)2p(p−8)(p−4)\dfrac{p^2-3p-40}{2p^3-24p^2+64p}=\dfrac{(p-8)(p+5)}{2p(p-8)(p-4)}

  4. Cancel (p−8)(p-8).

Answer

p+52p(p−4)\dfrac{p+5}{2p(p-4)}

Q2. Find the excluded values, if any of the following expressions.

(i)

yy2−25\dfrac{y}{y^2-25}

  1. The expression is undefined when the denominator is zero.

  2. y2−25=0y^2-25=0

  3. (y+5)(y−5)=0(y+5)(y-5)=0

  4. y=−5y=-5 or y=5y=5

Answer

Excluded values are y=5y=5 and y=−5y=-5

(ii)

tt2−5t+6\dfrac{t}{t^2-5t+6}

  1. Put the denominator equal to zero.

  2. t2−5t+6=0t^2-5t+6=0

  3. (t−2)(t−3)=0(t-2)(t-3)=0

  4. t=2t=2 or t=3t=3

Answer

Excluded values are t=2t=2 and t=3t=3

(iii)

x2+6x+8x2+x−2\dfrac{x^2+6x+8}{x^2+x-2}

  1. First reduce to lowest form.

  2. x2+6x+8=(x+2)(x+4)x^2+6x+8=(x+2)(x+4), x2+x−2=(x+2)(x−1)x^2+x-2=(x+2)(x-1)

  3. (x+2)(x+4)(x+2)(x−1)=x+4x−1\dfrac{(x+2)(x+4)}{(x+2)(x-1)}=\dfrac{x+4}{x-1}

  4. Now x−1=0x-1=0 gives x=1x=1.

Answer

Excluded value is x=1x=1

(iv)

x3−27x3+x2−6x\dfrac{x^3-27}{x^3+x^2-6x}

  1. x3−27=(x−3)(x2+3x+9)x^3-27=(x-3)(x^2+3x+9)

  2. x3+x2−6x=x(x2+x−6)=x(x+3)(x−2)x^3+x^2-6x=x(x^2+x-6)=x(x+3)(x-2)

  3. x3−27x3+x2−6x=(x−3)(x2+3x+9)x(x+3)(x−2)\dfrac{x^3-27}{x^3+x^2-6x}=\dfrac{(x-3)(x^2+3x+9)}{x(x+3)(x-2)} (already in lowest form)

  4. Denominator =0=0 gives x=0x=0, x=−3x=-3, x=2x=2.

Answer

Excluded values are x=0, x=−3, x=2x=0,\ x=-3,\ x=2

Quick check

  1. 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

  2. 3y−3y÷7y−73y2\dfrac{3y - 3}{y} \div \dfrac{7y - 7}{3y^2} is

    1. (1) 9y7\dfrac{9y}{7}
    2. (2) 9y3(21y−21)\dfrac{9y^3}{(21y - 21)}
    3. (3) 21y2−42y+213y3\dfrac{21y^2 - 42y + 21}{3y^3}
    4. (4) 7(y2−2y+1)y2\dfrac{7(y^2 - 2y + 1)}{y^2}
    Show answer

    Answer: (1) 9y7\dfrac{9y}{7}

    1. 3y−3y÷7y−73y2=3(y−1)y×3y27(y−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)(y-1) and one yy.

    3. =9y7= \dfrac{9y}{7}

    Question 2 of 5

  3. 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×1m \times 1.

    2. Its transpose has order 1×m1 \times m, which is a row matrix.

    Question 3 of 5

  4. The solution of (2x−1)2=9\left(2x - 1\right)^2 = 9 is equal to

    1. (1) −1-1
    2. (2) 22
    3. (3) −1, 2-1,\ 2
    4. (4) None of these
    Show answer

    Answer: (3) −1, 2-1,\ 2

    1. (2x−1)2=9⇒2x−1=±3(2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3.

    2. 2x−1=3⇒x=22x - 1 = 3 \Rightarrow x = 2.

    3. 2x−1=−3⇒x=−12x - 1 = -3 \Rightarrow x = -1.

    Question 4 of 5

  5. The values of aa and bb if 4x4−24x3+76x2+ax+b4x^4 - 24x^3 + 76x^2 + ax + b is a perfect square are

    1. (1) 100,120100,120
    2. (2) 10,1210,12
    3. (3) −120,100-120,100
    4. (4) 12,1012,10
    Show answer

    Answer: (3) −120,100-120,100

    1. Let 4x4−24x3+76x2+ax+b=(2x2−6x+c)24x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2.

    2. Expanding: 4x4−24x3+(36+4c)x2−12cx+c24x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2.

    3. Compare x2x^2 terms: 36+4c=76⇒c=1036 + 4c = 76 \Rightarrow c = 10.

    4. So a=−12(10)=−120a = -12(10) = -120 and b=102=100b = 10^2 = 100.

    Question 5 of 5
Chapter 3 formulas
  • Roots of a quadratic equation

    ax2+bx+c=0,  (a≠0)  ⇒  x=−b±b2−4ac2aax^2 + bx + c = 0,\; (a \neq 0) \;\Rightarrow\; x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}

  • Sum of the roots

    α+β=−ba=−Co-efficient of xCo-efficient of x2\alpha + \beta = \dfrac{-b}{a} = \dfrac{-\text{Co-efficient of } x}{\text{Co-efficient of } x^2}

  • Product of the roots

    αβ=ca=Constant termCo-efficient of x2\alpha\beta = \dfrac{c}{a} = \dfrac{\text{Constant term}}{\text{Co-efficient of } x^2}

  • Quadratic equation formed from its roots

    x2−(α+β)x+αβ=0x^2 - (\alpha + \beta)x + \alpha\beta = 0

  • Discriminant

    Δ=b2−4ac\Delta = b^2 - 4ac

  • Nature of roots: real and unequal

    Δ>0  ⇒  the roots are real and unequal\Delta > 0 \;\Rightarrow\; \text{the roots are real and unequal}

  • Nature of roots: real and equal

    Δ=0  ⇒  the roots are real and equal\Delta = 0 \;\Rightarrow\; \text{the roots are real and equal}

  • Nature of roots: no real roots

    Δ<0  ⇒  there are no real roots\Delta < 0 \;\Rightarrow\; \text{there are no real roots}

  • Cases for a system of linear equations in three variables

    (i) Unique solution(ii) Infinitely many solutions(iii) No solution\text{(i) Unique solution}\quad \text{(ii) Infinitely many solutions}\quad \text{(iii) No solution}

  • LCM of algebraic expressions

    LCM=expression of lowest degree that each given expression exactly divides\text{LCM} = \text{expression of lowest degree that each given expression exactly divides}

  • Quadratic polynomial and its zeroes

    p(x)=ax2+bx+c  (a≠0)  ⇒  p(x) has atmost two zeroes, which lie on the x-axisp(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

    Order of A=(Number of rows)×(Number of columns)=m×n\text{Order of } A = (\text{Number of rows}) \times (\text{Number of columns}) = m \times n

  • Row matrix (row vector)

    A=(a11a12⋯a1n)1×nA = \begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1n} \end{pmatrix}_{1 \times n}

  • Column matrix (column vector)

    A=(a11a21⋮am1)m×1A = \begin{pmatrix} a_{11} \\ a_{21} \\ \vdots \\ a_{m1} \end{pmatrix}_{m \times 1}

  • Square matrix

    Number of rows=Number of columns,  order n×n\text{Number of rows} = \text{Number of columns},\; \text{order } n \times n

  • Zero matrix (null matrix)

    aij=0 for all i, ja_{ij} = 0 \text{ for all } i,\, j

  • Transpose of a matrix

    (AT)ij=aji(rows and columns of A interchanged)\left(A^{T}\right)_{ij} = a_{ji} \quad (\text{rows and columns of } A \text{ interchanged})

  • Diagonal matrix

    aij=0 for i≠j(square matrix)a_{ij} = 0 \text{ for } i \neq j \quad (\text{square matrix})

  • Scalar matrix

    aij={k,i=j0,i≠ja_{ij} = \begin{cases} k, & i = j \\ 0, & i \neq j \end{cases}

  • Identity matrix (unit matrix)

    aij={1,i=j0,i≠ja_{ij} = \begin{cases} 1, & i = j \\ 0, & i \neq j \end{cases}

  • Lower triangular matrix

    aij=0 for i<j(all entries above the leading diagonal are zero)a_{ij} = 0 \text{ for } i < j \quad (\text{all entries above the leading diagonal are zero})

  • Upper triangular matrix

    aij=0 for i>j(all entries below the leading diagonal are zero)a_{ij} = 0 \text{ for } i > j \quad (\text{all entries below the leading diagonal are zero})

  • Equality of matrices

    A=B  ⟺  same order and aij=bij for all i, jA = B \iff \text{same order and } a_{ij} = b_{ij} \text{ for all } i,\, j

Answers checked against the Samacheer Kalvi textbook · Updated

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