Solutions of the Questions Based on Expressions and Equations - Future Study Point

Solutions of the Questions Based on Expressions and Equations

Expressions and equations

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.Expressions and equations

 

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Example of expressions : (x +2), (3x -3), (3x²/2 + 3x + 4) , etc.

Example of equations : x + 2 = 0, 3x -3 = 10, 3x²/2 + 3x + 4 = 0, 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.

Q5.Write 1.3×10-3 as a decimal.

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

,

,

4 log3x + log3 7

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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

Putting x =3 and y =5

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

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

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

Applying the same codition in given pair of equation

3x + 6y = -2

-5x + ay = 4

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?

xy
0-4
4-20
-412
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?

 

rectangle

 

 

 

 

 

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

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

x – y = 5…………(iii)

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

 

Ans.We are given the equation

Squaring it both sides

Ans.We have to find the value of the expression

Squaring it we get

 

Putting the value of x² + 1/x² = 23

Ans. We are given that

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

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

Therefore

Putting the value of x = √5 + √2

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

Ans. We are given

Cubing it in both sides

Putting the given value in the question

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

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Chapter 1- Chemical reactions and equationsChapter 9- Heredity and Evolution
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Chapter 4- Carbon and its CompoundsChapter 12- Electricity
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Chapter 7-Control and CoordinationChapter 15-Environment
Chapter 8- How do organisms reproduce?Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-SetsChapter 9-Sequences and Series
Chapter 2- Relations and functionsChapter 10- Straight Lines
Chapter 3- TrigonometryChapter 11-Conic Sections
Chapter 4-Principle of mathematical inductionChapter 12-Introduction to three Dimensional Geometry
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NCERT solutions for class 12 maths

Chapter 1-Relations and FunctionsChapter 9-Differential Equations
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