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Samacheer Kalvi 10th Maths Exercise 3.1 Solutions

10th Maths Exercise 3.1 solutions (Chapter 3 Algebra) with step-by-step answers for all 9 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance9 answers · full steps below
  1. Q1 (i)x=2, y=−1, z=4x=2,\ y=-1,\ z=4
  2. Q1 (ii)x=12, y=13, z=14x=\dfrac{1}{2},\ y=\dfrac{1}{3},\ z=\dfrac{1}{4}
  3. Q1 (iii)x=35, y=30, z=25x=35,\ y=30,\ z=25
  4. Q2 (i)The system is consistent and has infinitely many solutions
  5. Q2 (ii)The system is inconsistent and has no solution
  6. Q2 (iii)The system is consistent with a unique solution: x=3, y=9, z=15x=3,\ y=9,\ z=15
  7. Q3Vani is 2424 years, her father is 5151 years and her grandfather is 8484 years old
  8. Q4The original number is 137137
  9. Q5₹5 currencies =7=7, ₹10 currencies =3=3, ₹20 currencies =2=2

Q1. Solve the following system of linear equations in three variables

(i)

x+y+z=5x + y + z = 5 ; 2x−y+z=92x - y + z = 9 ; x−2y+3z=16x - 2y + 3z = 16

Show steps (7)

  1. x+y+z=5x+y+z=5 …(1) 2x−y+z=92x-y+z=9 …(2) x−2y+3z=16x-2y+3z=16 …(3)

  2. Adding (1) and (2): 3x+2z=143x+2z=14 …(4)

  3. 2×(2)2\times(2): 4x−2y+2z=184x-2y+2z=18

  4. 2×(2)−(3)2\times(2)-(3): 4x−2y+2z−(x−2y+3z)=18−164x-2y+2z-(x-2y+3z)=18-16, so 3x−z=23x-z=2 …(5)

  5. (4)−(5)(4)-(5): 3z=123z=12, so z=4z=4

  6. From (5): 3x−4=2⇒x=23x-4=2\Rightarrow x=2

  7. From (1): 2+y+4=5⇒y=−12+y+4=5\Rightarrow y=-1

Answer

x=2, y=−1, z=4x=2,\ y=-1,\ z=4

(ii)

1x−2y+4=0\dfrac{1}{x} - \dfrac{2}{y} + 4 = 0; 1y−1z+1=0\dfrac{1}{y} - \dfrac{1}{z} + 1 = 0; 2z+3x=14\dfrac{2}{z} + \dfrac{3}{x} = 14

Show steps (7)

  1. Let 1x=p, 1y=q, 1z=r\dfrac1x=p,\ \dfrac1y=q,\ \dfrac1z=r

  2. The equations become p−2q=−4p-2q=-4 …(1), q−r=−1q-r=-1 …(2), 3p+2r=143p+2r=14 …(3)

  3. From (1): p=2q−4p=2q-4; from (2): r=q+1r=q+1

  4. Substituting in (3): 3(2q−4)+2(q+1)=143(2q-4)+2(q+1)=14

  5. 6q−12+2q+2=14⇒8q=24⇒q=36q-12+2q+2=14\Rightarrow 8q=24\Rightarrow q=3

  6. So p=2(3)−4=2p=2(3)-4=2 and r=3+1=4r=3+1=4

  7. x=1p=12,y=1q=13,z=1r=14x=\dfrac1p=\dfrac12,\quad y=\dfrac1q=\dfrac13,\quad z=\dfrac1r=\dfrac14

Answer

x=12, y=13, z=14x=\dfrac{1}{2},\ y=\dfrac{1}{3},\ z=\dfrac{1}{4}

(iii)

x+20=3y2+10=2z+5=110−(y+z)x + 20 = \dfrac{3y}{2} + 10 = 2z + 5 = 110 - (y + z)

Show steps (7)

  1. x+20=3y2+10⇒2x−3y=−20x+20=\dfrac{3y}{2}+10\Rightarrow 2x-3y=-20 …(1)

  2. 3y2+10=2z+5⇒3y−4z=−10\dfrac{3y}{2}+10=2z+5\Rightarrow 3y-4z=-10 …(2)

  3. 2z+5=110−(y+z)⇒y+3z=1052z+5=110-(y+z)\Rightarrow y+3z=105 …(3)

  4. From (3): y=105−3zy=105-3z; substitute in (2):

  5. 3(105−3z)−4z=−10⇒315−13z=−10⇒13z=325⇒z=253(105-3z)-4z=-10\Rightarrow 315-13z=-10\Rightarrow 13z=325\Rightarrow z=25

  6. y=105−3(25)=30y=105-3(25)=30

  7. From (1): 2x=3(30)−20=70⇒x=352x=3(30)-20=70\Rightarrow x=35

Answer

x=35, y=30, z=25x=35,\ y=30,\ z=25

Q2. Discuss the nature of solutions of the following system of equations

(i)

x+2y−z=6x + 2y - z = 6 ; −3x−2y+5z=−12-3x - 2y + 5z = -12 ; x−2z=3x - 2z = 3

  1. x+2y−z=6x+2y-z=6 …(1) −3x−2y+5z=−12-3x-2y+5z=-12 …(2) x−2z=3x-2z=3 …(3)

  2. Adding (1) and (2): −2x+4z=−6-2x+4z=-6

  3. Dividing by −2-2: x−2z=3x-2z=3

  4. This is exactly equation (3), so (3)−(3)(3)-(3) gives 0=00=0, an identity.

  5. Since we get an identity, the three planes are not independent.

Answer

The system is consistent and has infinitely many solutions

(ii)

2y+z=3(−x+1)2y + z = 3(-x + 1); −x+3y−z=−4-x + 3y - z = -4 ; 3x+2y+z=−123x + 2y + z = -\dfrac{1}{2}

  1. 2y+z=3(−x+1)⇒3x+2y+z=32y+z=3(-x+1)\Rightarrow 3x+2y+z=3 …(1)

  2. −x+3y−z=−4-x+3y-z=-4 …(2)

  3. 3x+2y+z=−123x+2y+z=-\dfrac12 …(3)

  4. Subtracting (3) from (1): 0=3+12=720=3+\dfrac12=\dfrac72

  5. We arrive at the contradiction 0=720=\dfrac72.

Answer

The system is inconsistent and has no solution

(iii)

y+z4=z+x3=x+y2\dfrac{y+z}{4} = \dfrac{z+x}{3} = \dfrac{x+y}{2}; x+y+z=27x + y + z = 27

Show steps (7)

  1. Let y+z4=z+x3=x+y2=k\dfrac{y+z}{4}=\dfrac{z+x}{3}=\dfrac{x+y}{2}=k

  2. Then y+z=4k, z+x=3k, x+y=2ky+z=4k,\ z+x=3k,\ x+y=2k

  3. Adding: 2(x+y+z)=9k2(x+y+z)=9k

  4. Given x+y+z=27x+y+z=27, so 54=9k⇒k=654=9k\Rightarrow k=6

  5. y+z=24⇒x=27−24=3y+z=24\Rightarrow x=27-24=3

  6. z+x=18⇒y=27−18=9z+x=18\Rightarrow y=27-18=9

  7. x+y=12⇒z=27−12=15x+y=12\Rightarrow z=27-12=15

Answer

The system is consistent with a unique solution: x=3, y=9, z=15x=3,\ y=9,\ z=15

Q3. Vani, her father and her grand father have an average age of …

Vani, her father and her grand father have an average age of 5353. One-half of her grand father's age plus one-third of her father's age plus one fourth of Vani's age is 6565. Four years ago if Vani's grandfather was four times as old as Vani then how old are they all now ?

Show steps (8)

  1. Let Vani's age be xx, her father's yy and her grandfather's zz.

  2. Average age 5353: x+y+z3=53⇒x+y+z=159\dfrac{x+y+z}{3}=53\Rightarrow x+y+z=159 …(1)

  3. z2+y3+x4=65\dfrac{z}{2}+\dfrac{y}{3}+\dfrac{x}{4}=65; multiplying by 1212: 3x+4y+6z=7803x+4y+6z=780 …(2)

  4. Four years ago: z−4=4(x−4)⇒z=4x−12z-4=4(x-4)\Rightarrow z=4x-12 …(3)

  5. From (1): y=159−x−zy=159-x-z; put in (2):

  6. 3x+4(159−x−z)+6z=780⇒−x+2z=1443x+4(159-x-z)+6z=780\Rightarrow -x+2z=144 …(4)

  7. Using (3) in (4): −x+2(4x−12)=144⇒7x=168⇒x=24-x+2(4x-12)=144\Rightarrow 7x=168\Rightarrow x=24

  8. z=4(24)−12=84z=4(24)-12=84 and y=159−24−84=51y=159-24-84=51

Answer

Vani is 2424 years, her father is 5151 years and her grandfather is 8484 years old

Q4. The sum of the digits of a three-digit number is 1111 . …

The sum of the digits of a three-digit number is 1111. If the digits are reversed, the new number is 4646 more than five times the former number. If the hundreds digit plus twice the tens digit is equal to the units digit, then find the original three digit number ?

Show steps (11)

  1. Let the digits be xx (hundreds), yy (tens), zz (units); number =100x+10y+z=100x+10y+z.

  2. Sum of digits: x+y+z=11x+y+z=11 …(1)

  3. Reversed number condition: 100z+10y+x=5(100x+10y+z)+46100z+10y+x=5(100x+10y+z)+46

  4. ⇒499x+40y−95z=−46\Rightarrow 499x+40y-95z=-46 …(2)

  5. Third condition: x+2y=z⇒x+2y−z=0x+2y=z\Rightarrow x+2y-z=0 …(3)

  6. (1)+(3)(1)+(3): 2x+3y=112x+3y=11 …(4)

  7. From (1): z=11−x−yz=11-x-y; put in (2): 499x+40y−95(11−x−y)=−46499x+40y-95(11-x-y)=-46

  8. 594x+135y=999594x+135y=999; dividing by 2727: 22x+5y=3722x+5y=37 …(5)

  9. From (4): y=11−2x3y=\dfrac{11-2x}{3}; put in (5): 66x+55−10x=111⇒56x=56⇒x=166x+55-10x=111\Rightarrow 56x=56\Rightarrow x=1

  10. y=11−23=3y=\dfrac{11-2}{3}=3 and z=11−1−3=7z=11-1-3=7

  11. The number is 100(1)+10(3)+7=137100(1)+10(3)+7=137

Answer

The original number is 137137

Q5. There are 1212 pieces of five, ten and twenty rupee currencies whose …

There are 1212 pieces of five, ten and twenty rupee currencies whose total value is ₹105105. When first 22 sorts are interchanged in their numbers its value will be increased by ₹2020. Find the number of currencies in each sort.

Show steps (9)

  1. Let the number of ₹5, ₹10 and ₹20 currencies be xx, yy, zz.

  2. Number of pieces: x+y+z=12x+y+z=12 …(1)

  3. Total value: 5x+10y+20z=105⇒x+2y+4z=215x+10y+20z=105\Rightarrow x+2y+4z=21 …(2)

  4. Interchanging the first two sorts: 5y+10x+20z=125⇒2x+y+4z=255y+10x+20z=125\Rightarrow 2x+y+4z=25 …(3)

  5. (3)−(2)(3)-(2): x−y=4⇒x=y+4x-y=4\Rightarrow x=y+4 …(4)

  6. Put (4) in (1): 2y+z=8⇒z=8−2y2y+z=8\Rightarrow z=8-2y

  7. Put in (2): (y+4)+2y+4(8−2y)=21(y+4)+2y+4(8-2y)=21

  8. −5y+36=21⇒y=3-5y+36=21\Rightarrow y=3

  9. x=3+4=7x=3+4=7 and z=8−6=2z=8-6=2

Answer

₹5 currencies =7=7, ₹10 currencies =3=3, ₹20 currencies =2=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

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