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) , 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^{-2}as a decimal.
Q5.Write 1.3×10^{-3} as a decimal.
Q6.Write 4 log_{3}x + log_{3} 7 as a single logarithmic expression.
Ans We are given the expression
4 log_{3}x + log_{3} 7
Applying both of the formulas
,
,
4 log_{3}x + log_{3} 7
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Q7.Factor the algebraic expression 6x^{2}– 21 x y + 8 x z – 28 y z.
Ans .6x^{2}– 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 x^{4}– z^{8}.
Applying the idenity a² – b² = (a + b)(a – b)
x^{4}– z^{8}
=(x²)² -(z^{4})^{2}
=(x^{2} +z^{4})(x^{2} – z^{4})
=(x^{2} +z^{4})[x² – (z²)²]
=(x^{2} +z^{4})(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 z^{3}) / ( 2 x^{-3})
Ans. We are given the expression
Q15.Simplify (-x^{2}y^{3})^{2}(z^{2})^{0}
Ans.We are give the expression
= (-x^{2}y^{3})^{2}(z^{2})^{0}
= (-x²)²(y³)²×1
= -x^{4 }y^{6 }
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 a_{1}x + b_{1}y +c_{1 }=0 and a_{2}x + b_{2}y +c_{2 }=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?
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 = 2x^{2}
D) area = x^{2}– 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) = x^{2}– 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)x^{2}+ mx – 2 = 0.
Ans.The given equation is
(1/2)x^{2}+ 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 -x^{2} + 2kx – 4 = 0 has one real solution?
Ans. The given equation is -x^{2} + 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 x^{2} – 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) = -x^{2}+ 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) = -x^{2}+ 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.
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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 (x^{2}– y^{2}) = 10 and (x + y) = 2, find x and y.
Ans. The given equations are
(x^{2}– y^{2}) = 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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