Solutions of the Questions Based on Expressions and Equations

Here you can study solutions to the questions on Expressions and Equation. Solutions of the questions based on the Expressions and the Equations are important not only for the students of 9 and 10-grade students but also for the candidates who are pursuing entrance exams like SSC, Bank, Railways, Defence, and other similar exams. Expressions are the combination of the terms, variables and the numbers and Equations show the relationship between the variables and represented by placing a sign = between two expressions.You can learn here the simplification of algebraic expressions and solutions of different linear equations by studying the Solutions of the questions based on the Expressions and the Equations.

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Example of expressions : (x +2), (3x -3), (3x²/2 + 3x + 4) , $\left | x-2 \right |+2$ etc.

Example of equations : x + 2 = 0, 3x -3 = 10, 3x²/2 + 3x + 4 = 0, $\left | x-2 \right |+2= 10$ etc.

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Solutions of the questions based on Expressions and Equations

Q1.Which real nu

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mbers are inverse of each other?

Ans. -1 and 1

Q2.Which real numbers are equal to their cubes?

Ans.1, 0, -1

Q3.Which real number has no inverse.

Ans.0 has no inverse since 1/0 = undefined

Q4.Write 6×10^-2as a decimal.

$\boldsymbol{Ans.\: \: 6\times 10^{-2}=\frac{6}{10^{2}}=\frac{6}{100}=0.06}$

Q5.Write 1.3×10^-3 as a decimal.

$\boldsymbol{Ans.\: \: 1.3\times 10^{-3}=\frac{1.3}{10^{3}}=\frac{1.3}{1000}=0.0013}$

Q6.Write 4 log₃x + log₃ 7 as a single logarithmic expression.

Ans We are given the expression

4 log₃x + log₃ 7

Applying both of the formulas

$\boldsymbol{\because nlog_{a}m=log_{a}m^{n}}$ , $\boldsymbol{and\: log_{a}m+log_{a}n=log_{a}mn}$

$\boldsymbol{\therefore 4log_{3}x=log_{3}x^{4}}$ ,

$\boldsymbol{log_{3}x^{4}+log_{3}7=log_{3}7x^{4}}$

4 log₃x + log₃ 7

$\boldsymbol{=log_{3}7x^{4}}$

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Q7.Factor the algebraic expression 6x²– 21 x y + 8 x z – 28 y z.

Ans .6x²– 21 x y + 8 x z – 28 y z.

3x(2x -7y) + 4z(2x -7y)

(2x – 7y)( 3x + 4z)

Q8.Factor the algebraic expression (x – 1)⁴– (y – 2)⁴

Applying the idenity a² – b² = (a + b)(a – b)

(x – 1)⁴– (y – 2)⁴

=[(x – 1)²]² -[(y – 2)²]²

=( x – 1 + y – 2)(x – 1 -y +2)

=( x + y – 3)(x -y +1)

Q9.Factor the algebraic expression x⁴– z⁸.

Applying the idenity a² – b² = (a + b)(a – b)

x⁴– z⁸

=(x²)² -(z⁴)²

=(x² +z⁴)(x² – z⁴)

=(x² +z⁴)[x² – (z²)²]

=(x² +z⁴)(x + z²)(x–z²)

Q10.Evaluate the algebraic expression |-4x – y + 3| for x = 3 and y = 5

$\boldsymbol{\left | -4x-y+3 \right |}$

Putting x =3 and y =5

$\boldsymbol{=\left | -4\times 3-5+3 \right |}$

$\boldsymbol{=\left | -12-5+3 \right |}$

$\boldsymbol{=\left | -17+3 \right |}$

$\boldsymbol{=\left | -14\right |=14}$

Q11.Simplify the algebraic expression -2(y – 3) + 4(-3 y + 8)

Ans.-2(y – 3) + 4(-3 y + 8)

= -2y + 6 – 12y + 32

= -2y – 12y + 6 + 32

= -14y + 38

Q12.Expand and simplify the algebraic expression (z + 3)(z- 3) – (-z – 7)

Ans. (z + 3)(z- 3) – (-z – 7)

Applying the idenity a² – b² = (a + b)(a – b)

Substituting (z + 3)(z- 3) =z² – 3²= z² -9 in the given expression

z² -9 – (-z – 7)

= z² -9 +z + 7

=z² +z + 7 – 9

= z² + z – 2

Q13.Which property is used to write a(x + y+z) = a x + a y+az

Ans. a(x + y+z) = a x + a y+az shows the distributive property

Q14.Simplify ( 8 z³) / ( 2 x^-3)

Ans. We are given the expression

$\frac{8x^{3}}{2x^{-3}}$

$=4x^{3+3}$

$=4x^{6}$

Q15.Simplify (-x²y³)²(z²)⁰

Ans.We are give the expression

= (-x²y³)²(z²)⁰

= (-x²)²(y³)²×1

= -x⁴y⁶

Q16.For what value of k is the point (-3, k) on the line with equation -5 x + 3 y = 4?

Ans. Therefore putting x = -3 ad y = K in the equation

We have,

-5x+3y = 4

-5×-3 + 3K = 4

15 + 3K =4

3K = 4 – 15

3K = -11

$k=\frac{-11}{3}$

Q17.For what value of a will the system given below have no solutions?

3x + 6 y = -2
-5 x + a y = 4

Ans.If the standard pair of equation is a₁x + b₁y +c₁=0 and a₂x + b₂y +c₂=0, has no solution then we have

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}\neq \frac{c_{1}}{c_{2}}$

Applying the same codition in given pair of equation

3x + 6y = -2

-5x + ay = 4

$\boldsymbol{\frac{3}{-5}=\frac{6}{a}}$

3a = -30

a = -10

Therefore for the value of a = -10,the given pair of equation have no solutions

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Q18.Which equation best describes the relationship between x and y in this table?

x	y
0	-4
4	-20
-4	12
8	-36

(A) y = – x/4 – 4
(B) y = – x/4 + 4
(C) y = – 4x – 4
(D) y = – 4x + 4

Ans. Since all the values of table satisfies the equation y = – 4x – 4,so the answer is (C) y = – 4x – 4 .

Q19.Which equation best represents the area of the rectangle below?

A) area = 2(x+1) + 2(x-1)
B) area = 4(x+1)(x-1)
C) area = 2x²
D) area = x²– 1

Ans. The length of the given triangle is = x + 1

The breadth of given triangle is = x – 1

Area of the rectangle is = Length × Breadth =(x + 1)(x – 1) = x² – 1

So,the correct answer is (D) = x²– 1

21.Which line given by its equation below contains the points (1, -1) and (3, 5)?
(A) -2y -6x = 0
(B) 2y = 6x – 8
(C) y = 3x + 4
(D) y = -3x + 4

Ans. The given points (1, -1) and (3, 5) satifies the equation 2y = 6x – 8 ,so the required answer is (B) 2y = 6x – 8.

Q22.Solve the equation 2|3x – 2| – 3 = 7.

Ans.The given equation is

2|3x – 2| – 3 = 7

2|3x – 2| = 10

|3x – 2| = 5

In the case (3x -2) < 0

3x – 2 = -5

x = -1

In the when (3x -2) > 0

3x – 2 = 5

x = 7/3

Q23.Solve for x the equation (1/2)x²+ mx – 2 = 0.

Ans.The given equation is

(1/2)x²+ mx – 2 = 0

^{Multiplying the given equation by 2}

x² + 2mx – 4 = 0

Applying the formula of quadratic equation

$\boldsymbol{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

a = 1, b = 2m, c = 4

$\boldsymbol{x=\frac{-2m\pm \sqrt{4m^{2}-4\times 1\times -4}}{2\times 1}}$

$\boldsymbol{x=\frac{-2m\pm \sqrt{4m^{2}+16}}{2}}$

$\boldsymbol{=\frac{-2m\pm \sqrt{4\left ( m^{2} +4\right )}}{2}}$

$\boldsymbol{=\frac{2\left ( -m\pm \sqrt{m^{2}+4} \right )}{2}}$

$\boldsymbol{x=-m\pm \sqrt{m^{2}+4}}$

Q24.For what values of k the equation -x² + 2kx – 4 = 0 has one real solution?

Ans. The given equation is -x² + 2kx – 4 = 0

The condition for real solution of a quadratic equation (ax² +bx +c) is

b² – 4ac ≥ 0

Where a = -1, b = 2k, c = -4

(2k)² – 4 × -1 ×-4 ≥ 0

(2k)² -16 ≥ 0

4k² ≥ 16

k² ≥ 4

k ≥ 2

So, for all values of k ≥ 2, the given equation has one real solution.

Q25.For what values of both equation x² – 4x + 4b = 0 has two real solutions?

Ans. The condition for real solution of a quadratic equation (Ax² +Bx +C) is

B² – 4AC ≥ 0

Where A= 1, B = -4, C = 4b

(-4)² – 4 × 1 × 4b ≥ 0

16 – 16b ≥ 0

-16b ≥ -16

16b ≤ 16

b ≤1

Therefore either the value of b must be less than 1 or equal to 1 so that the given equation has two solutions.

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Q26.Function f is described by the equation f (x) = -x²+ 7. What is the set of values of f(x) corresponding to the set for the independent variable x given by {1, 5, 7, 12}?

Ans. The given function is

f (x) = -x²+ 7

The set of variable = {1, 5, 7, 12}

Corresponding values of f(x) are calculated as follows

f(1) = -1² +7 = 8, f(5) = -5² +7= 32, f(7) = -7² +7 =56, f(12) =-12² +7=151

Therefore required set of the values of f(x)

{8, 32, 56, 151}

Q27. Find the length and width of a rectangle whose perimeter is equal to 160 cm and its length is equal to triple its width.

Ans. Let the length of the rectangle = x and Wdth = y

According to first condition

The perimeter of the rectangle = 160 cm

2(x + y) = 160

x + y = 80…….(i)

According to second condition

Length = 3× Width

x = 3y……….(ii)

Putting x = 3y from equation (ii) in equation (ii)

3y + y =80

4y = 80

y = 20

Putting y = 20, in equation (i)

x + 20 = 80

x = 60

Therefore, the length of the rectangle is 60 m and widh is 20 m.

How to creat and solve algebraic equations like linear and quadratic equations

Q28.Simplify: |- x| + |3 x| – |- 2 x| + 3|x|29.

Ans. The given expression is

|- x| + |3 x| – |- 2 x| + 3|x|29

=x + 3x -2x +87x

= 91x -2x = 89 x
29.If (x²– y²) = 10 and (x + y) = 2, find x and y.

Ans. The given equations are

(x²– y²) = 10 …..(i), (x + y) = 2……(ii)

Factorizing LHS of equation (i) and dividing it by equation (ii), we get equation (iii)

$\boldsymbol{\frac{\left ( x+y \right )\left ( x-y \right )}{\left ( x+y \right )}=\frac{10}{2}}$

x – y = 5…………(iii)

Solving (ii) and (iii), we get x = 7/2 and y = -3/2

$\boldsymbol{Q29. If\: \sqrt{x}+\frac{1}{\sqrt{x}}=3}$

$\boldsymbol{Find\: x+\frac{1}{x}=?}$

Ans.We are given the equation

$\boldsymbol{\sqrt{x}+\frac{1}{\sqrt{x}}=3}$

Squaring it both sides

$\boldsymbol{x+\frac{1}{x}+2=9}$

$\boldsymbol{x+\frac{1}{x}=9-2=7}$

$\boldsymbol{Q30.If\: x^{2}+\frac{1}{x^{2}}=23}$

$\boldsymbol{Find\: x+\frac{1}{x}=?}$

Ans.We have to find the value of the expression

$\boldsymbol{ x+\frac{1}{x}}$

Squaring it we get

$\boldsymbol{\left ( x+\frac{1}{x} \right )^{2} = x^{2}+\frac{1}{x^{2}}+2}$

Putting the value of x² + 1/x² = 23

$\boldsymbol{\left ( x+\frac{1}{x} \right )^{2} = 23+2=25}$

$\boldsymbol{ x+\frac{1}{x} =\sqrt{25}=5}$

$\boldsymbol{Q31.If\, x=\sqrt{7+2\sqrt{10}}}$

$\boldsymbol{Find\: x+\frac{1}{x}=?}$

Ans. We are given that

$\boldsymbol{\, x=\sqrt{7+2\sqrt{10}}}$

We can make 7 + 2√10,a complete square

7 + 2√10 = (√5)² +(√2)² +2√10 =(√5 + √2)²

$\boldsymbol{\therefore x=\sqrt{\left ( \sqrt{5} +\sqrt{2}\right )^{2}}=\sqrt{5}+\sqrt{2}}$

Therefore

$\boldsymbol{x+\frac{1}{x}}$

Putting the value of x = √5 + √2

$\boldsymbol{=\sqrt{5}+\sqrt{2}+\frac{1}{\sqrt{5}+\sqrt{2}}}$

$\boldsymbol{=\sqrt{5}+\sqrt{2}+\frac{\left ( \sqrt{5}-\sqrt{2} \right )}{(\sqrt{5}+\sqrt{2})\left ( \sqrt{5}-\sqrt{2} \right )}}$

$\boldsymbol{=\sqrt{5}+\sqrt{2}+\frac{\left ( \sqrt{5}-\sqrt{2} \right )}{5-2}}$

$\boldsymbol{=\sqrt{5}+\sqrt{2}+\frac{\left ( \sqrt{5}-\sqrt{2} \right )}{3}}$

$\boldsymbol{=\frac{3\sqrt{5}+3\sqrt{2}+\sqrt{5}-\sqrt{2}}{3}}$

$\boldsymbol{x+\frac{1}{x}=\frac{4\sqrt{5}+2\sqrt{2}}{3}}$

Q32. If αβ + βγ + γα = 20, α² + β² + γ² = 60, then find the value of α + β + γ = ?

Ans. We are given the values

αβ + βγ + γα = 20, α² + β² + γ² = 60

Using the following identity in solving this question

(α + β + γ )² = α² + β² + γ² + 2αβ + 2βγ + 2γα

(α + β + γ )² = α² + β² + γ² + 2(αβ + βγ + γα)

(α + β + γ )² = 60 + 2× 20 = 100

α + β + γ = √(100) = 10

Hence, α + β + γ = 10

$\boldsymbol{Q33.If \: x+\frac{1}{x}=2,\: then\: find\: x^{3}+\frac{1}{x^{3}}=?}$

Ans. We are given

$\boldsymbol{ x+\frac{1}{x}=2}$

Cubing it in both sides

$\boldsymbol{\left ( x+\frac{1}{x} \right )^{3}=x^{3}+\frac{1}{x^{3}}+3x^{2}.\frac{1}{x}+3.\frac{1}{x^{2}}.x}$

$\boldsymbol{\left ( x+\frac{1}{x} \right )^{3}=x^{3}+\frac{1}{x^{3}}+3x.\frac{1}{x}\left ( x+\frac{1}{x} \right )}$

Putting the given value in the question

$\boldsymbol{2^{3}=x^{3}+\frac{1}{x^{3}}+3\times 2}$

$\boldsymbol{x^{3}+\frac{1}{x^{3}}=8-6 =2}$

Hence, the required value of the expression is =2

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NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
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Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
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Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

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Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

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NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

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NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals