Future Study Point

# 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.

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

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.

You can also study

Number System (all types of numbers used in maths)

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

Three ways of solving quadratic equations

Achieve hundred percentage marks in maths

Mean, Mode and Median

The difference between rational and irrational numbers

Solutions of the questions based on expression and equations

Tips of developing memory power and qualifying a government entrance exams

Tips to get success in competitive exams

Lines, angles and triangles for 6 th to 10 th class cbse geometry

Finding the roots of the polynomial by the complete square method

Addition, subtraction, multiplication and division of polynomials

## Solutions of the questions based on Expressions and Equations

Q1.Which real nu

Click for online shopping

Future Study Point.Deal: Cloths, Laptops, Computers, Mobiles, Shoes etc

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

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

Techniques of Achieving Hundred Percent Marks in Mathematics.

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

= -x4 y6

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

Class 10 maths NCERT solution of Unique & important questions of chapter 3 pair of linear equations

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.

What to do to pass in maths CBSE X th board exam ?
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

You can compensate us

Paytm number 9891436286

The money collected by us will be used for the education of poor students who leaves their study because of a lack of money.

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

### NCERT Solutions for class 10 maths

CBSE Class 10-Question paper of maths 2021 with solutions

CBSE Class 10-Half yearly question paper of maths 2020 with solutions

CBSE Class 10 -Question paper of maths 2020 with solutions

CBSE Class 10-Question paper of maths 2019 with solutions

### 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

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

### 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

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

Solutions of class 12  maths question paper 2021 preboard exam CBSE Solution