10th Maths Chapter 3 Algebra 10th Maths Chapter 3 Algebra solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.
This chapter on Algebra moves from linear equations in three variables to GCD and LCM of algebraic expressions, rational expressions, square roots, and quadratic equations solved by factorisation, completing the square, and the formula. It also covers the nature of roots, relations between roots and coefficients, graphs, and matrices. The exercises give students practice in solving, forming equations, and working with matrix operations.
Exercises in this chapter Chapter 3 one-mark practice A system of three linear equations in three variables is inconsistent if their planes
(1) intersect only at a point (2) intersect in a line (3) coincides with each other (4) do not intersect Show answer Answer: (4) do not intersect
Inconsistent means the system has no solution.
No solution happens only when the three planes have no common point.
So the planes do not intersect.
Question 1 of 10 The solution of the system x + y − 3 z = − 6 x + y - 3z = -6 x + y − 3 z = − 6 , − 7 y + 7 z = 7 -7y + 7z = 7 − 7 y + 7 z = 7 , 3 z = 9 3z = 9 3 z = 9 is
(1) x = 1 , y = 2 , z = 3 x = 1,\ y = 2,\ z = 3 x = 1 , y = 2 , z = 3 (2) x = − 1 , y = 2 , z = 3 x = -1,\ y = 2,\ z = 3 x = − 1 , y = 2 , z = 3 (3) x = − 1 , y = − 2 , z = 3 x = -1,\ y = -2,\ z = 3 x = − 1 , y = − 2 , z = 3 (4) x = 1 , y = − 2 , z = 3 x = 1,\ y = -2,\ z = 3 x = 1 , y = − 2 , z = 3 Show answer Answer: (1) x = 1 , y = 2 , z = 3 x = 1,\ y = 2,\ z = 3 x = 1 , y = 2 , z = 3
From 3 z = 9 3z = 9 3 z = 9 , z = 3 z = 3 z = 3 .
Put z = 3 z=3 z = 3 in − 7 y + 7 z = 7 -7y + 7z = 7 − 7 y + 7 z = 7 : − 7 y + 21 = 7 ⇒ y = 2 -7y + 21 = 7 \Rightarrow y = 2 − 7 y + 21 = 7 ⇒ y = 2 .
Put y = 2 , z = 3 y=2,\ z=3 y = 2 , z = 3 in x + y − 3 z = − 6 x + y - 3z = -6 x + y − 3 z = − 6 : x + 2 − 9 = − 6 ⇒ x = 1 x + 2 - 9 = -6 \Rightarrow x = 1 x + 2 − 9 = − 6 ⇒ x = 1 .
Question 2 of 10 If ( x − 6 ) (x - 6) ( x − 6 ) is the HCF of x 2 − 2 x − 24 x^2 - 2x - 24 x 2 − 2 x − 24 and x 2 − k x − 6 x^2 - kx - 6 x 2 − k x − 6 then the value of k k k is
(1) 3 3 3 (2) 5 5 5 (3) 6 6 6 (4) 8 8 8 Show answer Answer: (2) 5 5 5
x 2 − 2 x − 24 = ( x − 6 ) ( x + 4 ) x^2 - 2x - 24 = (x-6)(x+4) x 2 − 2 x − 24 = ( x − 6 ) ( x + 4 ) , so ( x − 6 ) (x-6) ( x − 6 ) is a factor.
Since ( x − 6 ) (x-6) ( x − 6 ) is the HCF, it must divide x 2 − k x − 6 x^2 - kx - 6 x 2 − k x − 6 .
So x = 6 x = 6 x = 6 makes it zero: 36 − 6 k − 6 = 0 36 - 6k - 6 = 0 36 − 6 k − 6 = 0 .
6 k = 30 ⇒ k = 5 6k = 30 \Rightarrow k = 5 6 k = 30 ⇒ k = 5 .
Question 3 of 10 3 y − 3 y ÷ 7 y − 7 3 y 2 \dfrac{3y - 3}{y} \div \dfrac{7y - 7}{3y^2} y 3 y − 3 ÷ 3 y 2 7 y − 7 is
(1) 9 y 7 \dfrac{9y}{7} 7 9 y (2) 9 y 3 ( 21 y − 21 ) \dfrac{9y^3}{(21y - 21)} ( 21 y − 21 ) 9 y 3 (3) 21 y 2 − 42 y + 21 3 y 3 \dfrac{21y^2 - 42y + 21}{3y^3} 3 y 3 21 y 2 − 42 y + 21 (4) 7 ( y 2 − 2 y + 1 ) y 2 \dfrac{7(y^2 - 2y + 1)}{y^2} y 2 7 ( y 2 − 2 y + 1 ) Show answer Answer: (1) 9 y 7 \dfrac{9y}{7} 7 9 y
3 y − 3 y ÷ 7 y − 7 3 y 2 = 3 ( y − 1 ) y × 3 y 2 7 ( y − 1 ) \dfrac{3y-3}{y} \div \dfrac{7y-7}{3y^2} = \dfrac{3(y-1)}{y} \times \dfrac{3y^2}{7(y-1)} y 3 y − 3 ÷ 3 y 2 7 y − 7 = y 3 ( y − 1 ) × 7 ( y − 1 ) 3 y 2
Cancel ( y − 1 ) (y-1) ( y − 1 ) and one y y y .
= 9 y 7 = \dfrac{9y}{7} = 7 9 y
Question 4 of 10 y 2 + 1 y 2 y^2 + \dfrac{1}{y^2} y 2 + y 2 1 is not equal to
(1) y 4 + 1 y 2 \dfrac{y^4 + 1}{y^2} y 2 y 4 + 1 (2) ( y + 1 y ) 2 \left(y + \dfrac{1}{y}\right)^2 ( y + y 1 ) 2 (3) ( y − 1 y ) 2 + 2 \left(y - \dfrac{1}{y}\right)^2 + 2 ( y − y 1 ) 2 + 2 (4) ( y + 1 y ) 2 − 2 \left(y + \dfrac{1}{y}\right)^2 - 2 ( y + y 1 ) 2 − 2 Show answer Answer: (2) ( y + 1 y ) 2 \left(y + \dfrac{1}{y}\right)^2 ( y + y 1 ) 2
( y + 1 y ) 2 = y 2 + 2 + 1 y 2 \left(y + \dfrac{1}{y}\right)^2 = y^2 + 2 + \dfrac{1}{y^2} ( y + y 1 ) 2 = y 2 + 2 + y 2 1 , which is 2 2 2 more than y 2 + 1 y 2 y^2 + \dfrac{1}{y^2} y 2 + y 2 1 .
Options (A), (C), (D) all simplify to y 2 + 1 y 2 y^2 + \dfrac{1}{y^2} y 2 + y 2 1 .
So (B) is the one that is not equal.
Question 5 of 10 x x 2 − 25 − 8 x 2 + 6 x + 5 \dfrac{x}{x^2 - 25} - \dfrac{8}{x^2 + 6x + 5} x 2 − 25 x − x 2 + 6 x + 5 8 gives
(1) x 2 − 7 x + 40 ( x − 5 ) ( x + 5 ) \dfrac{x^2 - 7x + 40}{(x - 5)(x + 5)} ( x − 5 ) ( x + 5 ) x 2 − 7 x + 40 (2) x 2 + 7 x + 40 ( x − 5 ) ( x + 5 ) ( x + 1 ) \dfrac{x^2 + 7x + 40}{(x - 5)(x + 5)(x + 1)} ( x − 5 ) ( x + 5 ) ( x + 1 ) x 2 + 7 x + 40 (3) x 2 − 7 x + 40 ( x 2 − 25 ) ( x + 1 ) \dfrac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)} ( x 2 − 25 ) ( x + 1 ) x 2 − 7 x + 40 (4) x 2 + 10 ( x 2 − 25 ) ( x + 1 ) \dfrac{x^2 + 10}{(x^2 - 25)(x + 1)} ( x 2 − 25 ) ( x + 1 ) x 2 + 10 Show answer Answer: (3) x 2 − 7 x + 40 ( x 2 − 25 ) ( x + 1 ) \dfrac{x^2 - 7x + 40}{(x^2 - 25)(x + 1)} ( x 2 − 25 ) ( x + 1 ) x 2 − 7 x + 40
x 2 − 25 = ( x + 5 ) ( x − 5 ) x^2 - 25 = (x+5)(x-5) x 2 − 25 = ( x + 5 ) ( x − 5 ) and x 2 + 6 x + 5 = ( x + 5 ) ( x + 1 ) x^2 + 6x + 5 = (x+5)(x+1) x 2 + 6 x + 5 = ( x + 5 ) ( x + 1 ) .
LCM = ( x + 5 ) ( x − 5 ) ( x + 1 ) = (x+5)(x-5)(x+1) = ( x + 5 ) ( x − 5 ) ( x + 1 ) .
x ( x + 1 ) − 8 ( x − 5 ) ( x + 5 ) ( x − 5 ) ( x + 1 ) = x 2 + x − 8 x + 40 ( x 2 − 25 ) ( x + 1 ) \dfrac{x(x+1) - 8(x-5)}{(x+5)(x-5)(x+1)} = \dfrac{x^2 + x - 8x + 40}{(x^2-25)(x+1)} ( x + 5 ) ( x − 5 ) ( x + 1 ) x ( x + 1 ) − 8 ( x − 5 ) = ( x 2 − 25 ) ( x + 1 ) x 2 + x − 8 x + 40
= x 2 − 7 x + 40 ( x 2 − 25 ) ( x + 1 ) = \dfrac{x^2 - 7x + 40}{(x^2-25)(x+1)} = ( x 2 − 25 ) ( x + 1 ) x 2 − 7 x + 40
Question 6 of 10 The square root of 256 x 8 y 4 z 10 25 x 6 y 6 z 6 \dfrac{256 x^8 y^4 z^{10}}{25 x^6 y^6 z^6} 25 x 6 y 6 z 6 256 x 8 y 4 z 10 is equal to
(1) 16 5 ∣ x 2 z 4 y 2 ∣ \dfrac{16}{5}\left|\dfrac{x^2 z^4}{y^2}\right| 5 16 y 2 x 2 z 4 (2) 16 ∣ y 2 x 2 z 4 ∣ 16\left|\dfrac{y^2}{x^2 z^4}\right| 16 x 2 z 4 y 2 (3) 16 5 ∣ y x z 2 ∣ \dfrac{16}{5}\left|\dfrac{y}{xz^2}\right| 5 16 x z 2 y (4) 16 5 ∣ x z 2 y ∣ \dfrac{16}{5}\left|\dfrac{xz^2}{y}\right| 5 16 y x z 2 Show answer Answer: (4) 16 5 ∣ x z 2 y ∣ \dfrac{16}{5}\left|\dfrac{xz^2}{y}\right| 5 16 y x z 2
256 x 8 y 4 z 10 25 x 6 y 6 z 6 = 256 x 2 z 4 25 y 2 \dfrac{256x^8y^4z^{10}}{25x^6y^6z^6} = \dfrac{256\,x^2 z^4}{25\,y^2} 25 x 6 y 6 z 6 256 x 8 y 4 z 10 = 25 y 2 256 x 2 z 4
Taking square root: 16 5 ∣ x z 2 y ∣ \dfrac{16}{5}\left|\dfrac{x z^2}{y}\right| 5 16 y x z 2
Question 7 of 10 Which of the following should be added to make x 4 + 64 x^4 + 64 x 4 + 64 a perfect square
(1) 4 x 2 4x^2 4 x 2 (2) 16 x 2 16x^2 16 x 2 (3) 8 x 2 8x^2 8 x 2 (4) − 8 x 2 -8x^2 − 8 x 2 Show answer Answer: (2) 16 x 2 16x^2 16 x 2
For ( x 2 + 8 ) 2 = x 4 + 16 x 2 + 64 (x^2 + 8)^2 = x^4 + 16x^2 + 64 ( x 2 + 8 ) 2 = x 4 + 16 x 2 + 64 .
Given expression has x 4 + 64 x^4 + 64 x 4 + 64 , so the missing middle term is 16 x 2 16x^2 16 x 2 .
Question 8 of 10 The solution of ( 2 x − 1 ) 2 = 9 \left(2x - 1\right)^2 = 9 ( 2 x − 1 ) 2 = 9 is equal to
(1) − 1 -1 − 1 (2) 2 2 2 (3) − 1 , 2 -1,\ 2 − 1 , 2 (4) None of these Show answer Answer: (3) − 1 , 2 -1,\ 2 − 1 , 2
( 2 x − 1 ) 2 = 9 ⇒ 2 x − 1 = ± 3 (2x-1)^2 = 9 \Rightarrow 2x - 1 = \pm 3 ( 2 x − 1 ) 2 = 9 ⇒ 2 x − 1 = ± 3 .
2 x − 1 = 3 ⇒ x = 2 2x - 1 = 3 \Rightarrow x = 2 2 x − 1 = 3 ⇒ x = 2 .
2 x − 1 = − 3 ⇒ x = − 1 2x - 1 = -3 \Rightarrow x = -1 2 x − 1 = − 3 ⇒ x = − 1 .
Question 9 of 10 The values of a a a and b b b if 4 x 4 − 24 x 3 + 76 x 2 + a x + b 4x^4 - 24x^3 + 76x^2 + ax + b 4 x 4 − 24 x 3 + 76 x 2 + a x + b is a perfect square are
(1) 100 , 120 100,120 100 , 120 (2) 10 , 12 10,12 10 , 12 (3) − 120 , 100 -120,100 − 120 , 100 (4) 12 , 10 12,10 12 , 10 Show answer Answer: (3) − 120 , 100 -120,100 − 120 , 100
Let 4 x 4 − 24 x 3 + 76 x 2 + a x + b = ( 2 x 2 − 6 x + c ) 2 4x^4 - 24x^3 + 76x^2 + ax + b = (2x^2 - 6x + c)^2 4 x 4 − 24 x 3 + 76 x 2 + a x + b = ( 2 x 2 − 6 x + c ) 2 .
Expanding: 4 x 4 − 24 x 3 + ( 36 + 4 c ) x 2 − 12 c x + c 2 4x^4 - 24x^3 + (36 + 4c)x^2 - 12cx + c^2 4 x 4 − 24 x 3 + ( 36 + 4 c ) x 2 − 12 c x + c 2 .
Compare x 2 x^2 x 2 terms: 36 + 4 c = 76 ⇒ c = 10 36 + 4c = 76 \Rightarrow c = 10 36 + 4 c = 76 ⇒ c = 10 .
So a = − 12 ( 10 ) = − 120 a = -12(10) = -120 a = − 12 ( 10 ) = − 120 and b = 10 2 = 100 b = 10^2 = 100 b = 1 0 2 = 100 .
Question 10 of 10
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