Algebra

239 questions

Question 181Question →

Let the universal set be U={x∈Z:1≤x≤20}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 20\}. Consider two subsets of U\mathcal{U} given by A={x∈U:x is a multiple of 4}A = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\} and B={x∈U:x is a factor of 20}B = \{x \in \mathcal{U} : x \text{ is a factor of } 20\}. What is the number of elements in the complement of (A∪B)(A \cup B), denoted as n((A∪B)′)n((A \cup B)')?

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Answer: 11

Answer

11
The universal set contains 20 elements. The set of multiples of 4 within this range has 5 elements, and the set of factors of 20 has 6 elements. Their shared elements are 4 and 20 (2 elements). Thus, the union contains 5+6−2=95 + 6 - 2 = 9 elements. Subtracting this from 20 gives 11 elements in the complement.

Step-by-Step Solution

1
Identify the elements of the universal set and the subsets AA and BB.
U={1,2,3,…,20}\mathcal{U} = \{1, 2, 3, \dots, 20\} with n(U)=20n(\mathcal{U}) = 20.
A={4,8,12,16,20}A = \{4, 8, 12, 16, 20\} with n(A)=5n(A) = 5.
B={1,2,4,5,10,20}B = \{1, 2, 4, 5, 10, 20\} with n(B)=6n(B) = 6.
Listing the elements clearly determines the cardinality of each individual set.
2
Find the intersection and union of sets AA and BB.
A∩B={4,20}A \cap B = \{4, 20\}, so n(A∩B)=2n(A \cap B) = 2.
A∪B={1,2,4,5,8,10,12,16,20}A \cup B = \{1, 2, 4, 5, 8, 10, 12, 16, 20\}, so n(A∪B)=5+6−2=9n(A \cup B) = 5 + 6 - 2 = 9.
Applying the principle of inclusion-exclusion avoids double-counting common elements.
3
Calculate the cardinality of the complement (A∪B)′(A \cup B)'.
n((A∪B)′)=n(U)−n(A∪B)=20−9=11n((A \cup B)') = n(\mathcal{U}) - n(A \cup B) = 20 - 9 = 11.
The complement contains all elements in the universal set that are not in the union.

Key Concept

Set Complements and Inclusion-Exclusion Principle
Question 182Question →

A binary operation ⊕\oplus defined on the set of real numbers is given by a⊕b=a+b+kaba \oplus b = a + b + kab, where kk is a non-zero real constant. If the inverse of 33 under this operation is −12-\frac{1}{2}, what is the inverse of 44 under the same operation?

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Answer: −1223-\frac{12}{23}

Answer

−1223-\frac{12}{23}
To find the inverse of 44, first determine the identity element ee from a⊕e=aa \oplus e = a, which gives e=0e = 0. Next, set x⊕x−1=0x \oplus x^{-1} = 0 to get x−1=−x1+kxx^{-1} = -\frac{x}{1 + kx}. Using 3−1=−123^{-1} = -\frac{1}{2}, solve −31+3k=−12-\frac{3}{1 + 3k} = -\frac{1}{2} to find k=53k = \frac{5}{3}. Finally, substituting x=4x = 4 yields 4−1=−41+4(53)=−12234^{-1} = -\frac{4}{1 + 4(\frac{5}{3})} = -\frac{12}{23}.

Step-by-Step Solution

1
Determine the identity element ee of the operation.
e=0e = 0
By definition of identity element, a⊕e=a  ⟹  a+e+kae=a  ⟹  e(1+ka)=0  ⟹  e=0a \oplus e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0 for all valid aa.
2
Derive the general formula for the inverse x−1x^{-1} of an element xx.
x−1=−x1+kxx^{-1} = -\frac{x}{1 + kx}
An element and its inverse yield the identity element under the operation: x⊕x−1=0  ⟹  x+x−1+kxx−1=0  ⟹  x−1(1+kx)=−xx \oplus x^{-1} = 0 \implies x + x^{-1} + kxx^{-1} = 0 \implies x^{-1}(1 + kx) = -x.
3
Use the given inverse of 33 to solve for the constant kk.
k=53k = \frac{5}{3}
Given 3−1=−123^{-1} = -\frac{1}{2}, substitute x=3x = 3 into the inverse formula: −31+3k=−12  ⟹  1+3k=6  ⟹  3k=5  ⟹  k=53-\frac{3}{1 + 3k} = -\frac{1}{2} \implies 1 + 3k = 6 \implies 3k = 5 \implies k = \frac{5}{3}.
4
Calculate the inverse of 44 using k=53k = \frac{5}{3}.
4−1=−12234^{-1} = -\frac{12}{23}
Substitute x=4x = 4 and k=53k = \frac{5}{3} into the inverse formula: 4−1=−41+4(53)=−41+203=−4233=−12234^{-1} = -\frac{4}{1 + 4\left(\frac{5}{3}\right)} = -\frac{4}{1 + \frac{20}{3}} = -\frac{4}{\frac{23}{3}} = -\frac{12}{23}.

Key Concept

Binary Operations: Identity and Inverse Elements with Unknown Parameters
Estimated Time:2m 0s
Question 183Question →

A binary operation ⋆\star on the set of real numbers R∖{1}\mathbb{R} \setminus \{1\} is defined by a⋆b=a+b−aba \star b = a + b - ab. If y−1y^{-1} denotes the inverse of an element yy under ⋆\star, find the value of xx such that (x⋆3)−1=2(x \star 3)^{-1} = 2.

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Answer: 0.5

Answer

The value of xx is 0.50.5.
To solve for xx, first calculate the identity element e=0e = 0 using a⋆e=aa \star e = a. Next, use y⋆y−1=0y \star y^{-1} = 0 to get the inverse formula y−1=yy−1y^{-1} = \frac{y}{y - 1}. Substituting y=x⋆3=3−2xy = x \star 3 = 3 - 2x into the inverse formula yields 3−2x2−2x=2\frac{3 - 2x}{2 - 2x} = 2, which simplifies to 2x=12x = 1, giving x=0.5x = 0.5.

Step-by-Step Solution

1
Find the identity element ee of the operation ⋆\star.
e=0e = 0
By definition of the identity element, a⋆e=a  ⟹  a+e−ae=a  ⟹  e(1−a)=0a \star e = a \implies a + e - ae = a \implies e(1 - a) = 0, so e=0e = 0 for all a≠1a \neq 1.
2
Derive the general expression for the inverse element y−1y^{-1} of yy.
y−1=yy−1y^{-1} = \frac{y}{y - 1}
By definition of inverse element, y⋆y−1=e  ⟹  y+y−1−yy−1=0  ⟹  y−1(1−y)=−y  ⟹  y−1=yy−1y \star y^{-1} = e \implies y + y^{-1} - y y^{-1} = 0 \implies y^{-1}(1 - y) = -y \implies y^{-1} = \frac{y}{y - 1}.
3
Evaluate the inner operation x⋆3x \star 3.
x⋆3=3−2xx \star 3 = 3 - 2x
Using the operational rule a⋆b=a+b−aba \star b = a + b - ab, we obtain x⋆3=x+3−3x=3−2xx \star 3 = x + 3 - 3x = 3 - 2x.
4
Set up and solve the equation (x⋆3)−1=2(x \star 3)^{-1} = 2.
x=0.5x = 0.5
Letting y=3−2xy = 3 - 2x, its inverse is y−1=3−2x(3−2x)−1=3−2x2−2xy^{-1} = \frac{3 - 2x}{(3 - 2x) - 1} = \frac{3 - 2x}{2 - 2x}. Equating this to 22 gives 3−2x2−2x=2  ⟹  3−2x=4−4x  ⟹  2x=1  ⟹  x=0.5\frac{3 - 2x}{2 - 2x} = 2 \implies 3 - 2x = 4 - 4x \implies 2x = 1 \implies x = 0.5.

Key Concept

Identity and Inverse Elements in Binary Operations
Question 184Question →

A binary operation ⊕\oplus is defined on the set of real numbers R\mathbb{R} by a⊕b=2a+3b−5a \oplus b = 2a + 3b - 5. What is the value of (4⊕1)⊕2(4 \oplus 1) \oplus 2?

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Answer: 13

Answer

The value of (4⊕1)⊕2(4 \oplus 1) \oplus 2 is 13.
First, evaluate the inner expression 4⊕14 \oplus 1 using a=4a = 4 and b=1b = 1, which gives 2(4)+3(1)−5=8+3−5=62(4) + 3(1) - 5 = 8 + 3 - 5 = 6. Then, substitute this result into the outer expression to evaluate 6⊕26 \oplus 2 using a=6a = 6 and b=2b = 2, yielding 2(6)+3(2)−5=12+6−5=132(6) + 3(2) - 5 = 12 + 6 - 5 = 13.

Step-by-Step Solution

1
Evaluate the inner operation 4⊕14 \oplus 1
6
Substitute a=4a = 4 and b=1b = 1 into the operational rule a⊕b=2a+3b−5a \oplus b = 2a + 3b - 5.
2
Evaluate the outer operation using the result from Step 1: 6⊕26 \oplus 2
13
Substitute a=6a = 6 and b=2b = 2 into the operational rule a⊕b=2a+3b−5a \oplus b = 2a + 3b - 5.

Key Concept

Evaluation of Binary Operations
Question 185Question →

What is the positive value of xx that satisfies the simultaneous equations y−x=1y - x = 1 and y=x2−5y = x^2 - 5?

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Answer: 33

Answer

The positive value of xx is 33.
Substituting y=x+1y = x + 1 from the linear equation into the quadratic equation gives x+1=x2−5x + 1 = x^2 - 5. Rearranging terms yields x2−x−6=0x^2 - x - 6 = 0, which factors as (x−3)(x+2)=0(x - 3)(x + 2) = 0. The roots are x=3x = 3 and x=−2x = -2. Selecting the positive solution gives 33.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation.
y=x+1y = x + 1
Isolating yy simplifies substitution into the quadratic equation.
2
Substitute y=x+1y = x + 1 into the quadratic equation y=x2−5y = x^2 - 5.
x+1=x2−5x + 1 = x^2 - 5
This produces a quadratic equation in one variable, xx.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
x2−x−6=0x^2 - x - 6 = 0
Standard quadratic form allows direct factorization.
4
Factorize the quadratic equation.
(x−3)(x+2)=0(x - 3)(x + 2) = 0
Factoring determines the roots of the equation.
5
Solve for xx and identify the positive value.
x=3x = 3 or x=−2x = -2; positive value is 33.
The question specifies the positive real solution for xx.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Question 186Question →

If xx and yy satisfy the simultaneous equations x−2y=1x - 2y = 1 and x2−xy−y2=5x^2 - xy - y^2 = 5, what is the positive value of xx?

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Answer: 3

Answer

The positive value of xx is 3.
Substituting x=2y+1x = 2y + 1 into the non-linear equation yields (2y+1)2−(2y+1)y−y2=5(2y + 1)^2 - (2y + 1)y - y^2 = 5, which simplifies to y2+3y−4=0y^2 + 3y - 4 = 0. Solving this gives y=1y = 1 or y=−4y = -4. Substituting y=1y = 1 into x=2y+1x = 2y + 1 gives x=3x = 3, which is the positive value of xx.

Step-by-Step Solution

1
Make xx the subject of the linear equation
x=2y+1x = 2y + 1
Substitution is the standard method for solving linear-quadratic simultaneous systems.
2
Substitute x=2y+1x = 2y + 1 into x2−xy−y2=5x^2 - xy - y^2 = 5
(2y+1)2−(2y+1)y−y2=5(2y + 1)^2 - (2y + 1)y - y^2 = 5
Eliminates xx to create a single quadratic equation in terms of yy.
3
Expand and simplify to standard quadratic form
y2+3y−4=0y^2 + 3y - 4 = 0
Simplifying algebraic expressions allows factorization.
4
Solve for yy by factoring
y=1y = 1 or y=−4y = -4
Factors of −4-4 that sum to 33 are +4+4 and −1-1.
5
Calculate corresponding values for xx
x=3x = 3 when y=1y = 1, and x=−7x = -7 when y=−4y = -4
Substitute yy back into the linear expression for xx.
6
Select the positive value of xx
3
The question explicitly asks for the positive value of xx.

Key Concept

Simultaneous Linear and Quadratic Equations
Question 187Question →

If (u,v)(u, v) is a pair of real numbers satisfying the simultaneous equations u−2v=1u - 2v = 1 and u2−3v2=13u^2 - 3v^2 = 13, what is the sum of all possible values of uu?

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Answer: −6-6

Answer

The sum of all possible values of uu is −6-6.
From the linear equation u−2v=1u - 2v = 1, we express uu as u=2v+1u = 2v + 1. Substituting this into the quadratic equation u2−3v2=13u^2 - 3v^2 = 13 gives (2v+1)2−3v2=13(2v + 1)^2 - 3v^2 = 13, which expands and simplifies to v2+4v−12=0v^2 + 4v - 12 = 0. Factoring gives (v+6)(v−2)=0(v + 6)(v - 2) = 0, so v=2v = 2 or v=−6v = -6. Substituting these back into u=2v+1u = 2v + 1 yields u=5u = 5 (for v=2v = 2) and u=−11u = -11 (for v=−6v = -6). The sum of all possible values of uu is 5+(−11)=−65 + (-11) = -6.

Step-by-Step Solution

1
Express uu in terms of vv using the linear equation.
u=2v+1u = 2v + 1
Isolating uu allows for straightforward substitution into the quadratic equation.
2
Substitute u=2v+1u = 2v + 1 into the quadratic equation u2−3v2=13u^2 - 3v^2 = 13 and expand.
(2v+1)2−3v2=13  ⟹  (4v2+4v+1)−3v2=13  ⟹  v2+4v−12=0(2v + 1)^2 - 3v^2 = 13 \implies (4v^2 + 4v + 1) - 3v^2 = 13 \implies v^2 + 4v - 12 = 0
This reduces the system to a single quadratic equation in terms of vv.
3
Solve the quadratic equation for vv by factorisation.
(v+6)(v−2)=0  ⟹  v=2 or v=−6(v + 6)(v - 2) = 0 \implies v = 2 \text{ or } v = -6
Finding the roots of the quadratic gives the vv-coordinates of the solution pairs.
4
Calculate the corresponding values of uu using u=2v+1u = 2v + 1.
For v=2v = 2: u=2(2)+1=5u = 2(2) + 1 = 5.
For v=−6v = -6: u=2(−6)+1=−11u = 2(-6) + 1 = -11.
Each vv value corresponds to a specific uu value in the solution pairs.
5
Compute the sum of all possible values of uu.
5+(−11)=−65 + (-11) = -6
The question asks specifically for the sum of the uu-values.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Question 188Question →

If (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) are the real solution pairs to the simultaneous equations 3x−y=53x - y = 5 and x2+2xy−y2=7x^2 + 2xy - y^2 = 7, with x1<x2x_1 < x_2, what is the value of y2−y1y_2 - y_1?

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Answer: 18

Answer

18
Expressing yy as 3x−53x - 5 and substituting into the quadratic equation yields x2−10x+16=0x^2 - 10x + 16 = 0. Solving gives x1=2x_1 = 2 and x2=8x_2 = 8. Evaluating y=3x−5y = 3x - 5 for both values gives y1=1y_1 = 1 and y2=19y_2 = 19. The difference y2−y1=19−1=18y_2 - y_1 = 19 - 1 = 18.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation
y=3x−5y = 3x - 5
Isolation of yy facilitates substitution into the non-linear equation.
2
Substitute y=3x−5y = 3x - 5 into the quadratic equation x2+2xy−y2=7x^2 + 2xy - y^2 = 7
x2+2x(3x−5)−(3x−5)2=7x^2 + 2x(3x - 5) - (3x - 5)^2 = 7
Reduces the system to a single quadratic equation in terms of xx.
3
Expand and simplify the algebraic expression
x2+6x2−10x−(9x2−30x+25)=7  ⟹  −2x2+20x−32=0  ⟹  x2−10x+16=0x^2 + 6x^2 - 10x - (9x^2 - 30x + 25) = 7 \implies -2x^2 + 20x - 32 = 0 \implies x^2 - 10x + 16 = 0
Simplifies the equation to standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation x2−10x+16=0x^2 - 10x + 16 = 0 by factorization
(x−2)(x−8)=0  ⟹  x1=2(x - 2)(x - 8) = 0 \implies x_1 = 2 and x2=8x_2 = 8
Identifies the two roots with x1<x2x_1 < x_2 as specified by the condition.
5
Determine the corresponding yy-values using y=3x−5y = 3x - 5
y1=3(2)−5=1y_1 = 3(2) - 5 = 1 and y2=3(8)−5=19y_2 = 3(8) - 5 = 19
Obtains the complete coordinate solution pairs (2,1)(2, 1) and (8,19)(8, 19).
6
Calculate the required difference y2−y1y_2 - y_1
y2−y1=19−1=18y_2 - y_1 = 19 - 1 = 18
Computes the final required target value.

Key Concept

Solving Simultaneous Linear and Quadratic Equations by Substitution
Question 189Question →

A binary operation ∘\circ on the set of real numbers R\mathbb{R} is defined by a∘b=a+b−12aba \circ b = a + b - \frac{1}{2}ab. What is the inverse of 33 under this operation?

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Answer: 66

Answer

The inverse of 33 under the defined binary operation is 66.
To find the inverse of 33, we first find the identity element ee using a∘e=a  ⟹  a+e−12ae=aa \circ e = a \implies a + e - \frac{1}{2}ae = a, which gives e=0e = 0. Then, setting 3∘x=03 \circ x = 0 gives 3+x−32x=0  ⟹  3=12x  ⟹  x=63 + x - \frac{3}{2}x = 0 \implies 3 = \frac{1}{2}x \implies x = 6.

Step-by-Step Solution

1
Find the identity element ee of the operation.
e=0e = 0
By definition of identity element, a∘e=aa \circ e = a. Substituting into the operation formula gives a+e−12ae=a  ⟹  e(1−12a)=0  ⟹  e=0a + e - \frac{1}{2}ae = a \implies e\left(1 - \frac{1}{2}a\right) = 0 \implies e = 0 for all a≠2a \neq 2.
2
Set up the inverse equation for the element 33.
3∘x=03 \circ x = 0
Let xx be the inverse of 33. By definition of inverse element, 3∘x=e3 \circ x = e, where e=0e = 0.
3
Apply the binary operation definition to the left-hand side.
3+x−12(3)(x)=03 + x - \frac{1}{2}(3)(x) = 0
Substitute a=3a = 3 and b=xb = x into a∘b=a+b−12aba \circ b = a + b - \frac{1}{2}ab.
4
Solve the linear equation for xx.
x=6x = 6
Simplify the equation: 3+x−32x=0  ⟹  3−12x=0  ⟹  12x=3  ⟹  x=63 + x - \frac{3}{2}x = 0 \implies 3 - \frac{1}{2}x = 0 \implies \frac{1}{2}x = 3 \implies x = 6.

Key Concept

Identity and Inverse Elements of Binary Operations
Question 190Question →

Given the matrix P=(x1230x121)P = \begin{pmatrix} x & 1 & 2 \\ 3 & 0 & x \\ 1 & 2 & 1 \end{pmatrix}, for what positive value of xx is the determinant of PP equal to 33?

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Answer: 2

Answer

The positive value of xx is 22.
Expanding the determinant of PP along the first row yields det⁡(P)=−2x2+x+9\det(P) = -2x^2 + x + 9. Setting det⁡(P)=3\det(P) = 3 leads to −2x2+x+9=3-2x^2 + x + 9 = 3, which rearranges to 2x2−x−6=02x^2 - x - 6 = 0. Factoring gives (2x+3)(x−2)=0(2x + 3)(x - 2) = 0, leading to roots x=−1.5x = -1.5 and x=2x = 2. The positive value is 22.

Step-by-Step Solution

1
Calculate the determinant of matrix PP in terms of xx
\det(P) = -2x^2 + x + 9
Expanding along the first row gives x(0⋅1−2x)−1(3⋅1−x⋅1)+2(3⋅2−0⋅1)=−2x2+x+9x(0 \cdot 1 - 2x) - 1(3 \cdot 1 - x \cdot 1) + 2(3 \cdot 2 - 0 \cdot 1) = -2x^2 + x + 9.
2
Set up the quadratic equation using det⁡(P)=3\det(P) = 3
2x^2 - x - 6 = 0
Subtracting 33 from both sides gives −2x2+x+6=0-2x^2 + x + 6 = 0, which simplifies to 2x2−x−6=02x^2 - x - 6 = 0.
3
Solve the quadratic equation for xx
x = 2 \text{ or } x = -1.5
Factoring (2x+3)(x−2)=0(2x + 3)(x - 2) = 0 yields x=−32x = -\frac{3}{2} and x=2x = 2.
4
Select the positive solution as required by the problem statement
x = 2
The problem specifically requests the positive real value of xx.

Key Concept

3x3 Determinant Evaluation and Quadratic Solving
Question 191Question →

A quantity QQ is partly constant and partly varies inversely as pp. Given that Q=11Q = 11 when p=2p = 2, and Q=5Q = 5 when p=5p = 5, what is the value of QQ when p=10p = 10?

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Answer: 33

Answer

The value of QQ when p=10p = 10 is 33.
The relationship for partial inverse variation is Q=k1+k2pQ = k_1 + \frac{k_2}{p}. Substituting (p=2,Q=11)(p=2, Q=11) gives 2k1+k2=222k_1 + k_2 = 22, and substituting (p=5,Q=5)(p=5, Q=5) gives 5k1+k2=255k_1 + k_2 = 25. Solving these simultaneous equations gives k1=1k_1 = 1 and k2=20k_2 = 20. Substituting p=10p = 10 into Q=1+20pQ = 1 + \frac{20}{p} results in Q=1+2=3Q = 1 + 2 = 3.

Step-by-Step Solution

1
Set up the general formula for partial variation.
Q=k1+k2pQ = k_1 + \frac{k_2}{p}, where k1k_1 and k2k_2 are constants.
Partial variation consists of a constant part and a part that varies inversely with pp.
2
Substitute the given values to form two simultaneous linear equations.
For p=2,Q=11  ⟹  11=k1+k22  ⟹  2k1+k2=22p = 2, Q = 11 \implies 11 = k_1 + \frac{k_2}{2} \implies 2k_1 + k_2 = 22
For p=5,Q=5  ⟹  5=k1+k25  ⟹  5k1+k2=25p = 5, Q = 5 \implies 5 = k_1 + \frac{k_2}{5} \implies 5k_1 + k_2 = 25
Substituting known data points provides equations to solve for the variation constants.
3
Solve the simultaneous equations for k1k_1 and k2k_2.
Subtracting the first equation from the second gives 3k1=3  ⟹  k1=13k_1 = 3 \implies k_1 = 1.
Substituting k1=1k_1 = 1 into 2(1)+k2=222(1) + k_2 = 22 yields k2=20k_2 = 20.
Determining k1k_1 and k2k_2 establishes the explicit relationship between QQ and pp.
4
Calculate QQ when p=10p = 10.
Q=1+2010=1+2=3Q = 1 + \frac{20}{10} = 1 + 2 = 3.
Substitute p=10p = 10 into the established variation equation Q=1+20pQ = 1 + \frac{20}{p}.

Key Concept

Partial Variation and Simultaneous Equations
Estimated Time:1m 30s
Question 192Question →

A binary operation Δ\Delta is defined on the set of real numbers R\mathbb{R} by aΔb=a+3b−2aba \Delta b = a + 3b - 2ab. If (3Δx)Δ1=7(3 \Delta x) \Delta 1 = 7, what is the value of xx?

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Answer: 73\frac{7}{3}

Answer

73\frac{7}{3}
Evaluating 3Δx3 \Delta x gives 3−3x3 - 3x. Then operating (3−3x)Δ1(3 - 3x) \Delta 1 yields (3−3x)+3−2(3−3x)=3x(3 - 3x) + 3 - 2(3 - 3x) = 3x. Equating 3x=73x = 7 gives x=73x = \frac{7}{3}.

Step-by-Step Solution

1
Evaluate the inner expression 3Δx3 \Delta x using the definition aΔb=a+3b−2aba \Delta b = a + 3b - 2ab.
3Δx=3+3x−2(3)(x)=3+3x−6x=3−3x3 \Delta x = 3 + 3x - 2(3)(x) = 3 + 3x - 6x = 3 - 3x.
Substitute a=3a = 3 and b=xb = x into the operation rule.
2
Substitute the result (3−3x)(3 - 3x) as the first operand in the outer expression (3−3x)Δ1(3 - 3x) \Delta 1.
(3−3x)Δ1=(3−3x)+3(1)−2(3−3x)(1)(3 - 3x) \Delta 1 = (3 - 3x) + 3(1) - 2(3 - 3x)(1).
Apply the binary operation definition with a=3−3xa = 3 - 3x and b=1b = 1.
3
Expand and simplify the algebraic expression.
(3−3x)Δ1=3−3x+3−6+6x=3x(3 - 3x) \Delta 1 = 3 - 3x + 3 - 6 + 6x = 3x.
Distribute −2-2 across (3−3x)(3 - 3x) to get −6+6x-6 + 6x, then collect like terms.
4
Set the simplified expression equal to 77 and solve for xx.
3x=7  ⟹  x=733x = 7 \implies x = \frac{7}{3}.
Divide both sides by 33 to isolate xx.

Key Concept

Non-commutative binary operation composition and algebraic equation solving
Question 193Question →

A binary operation ∗* defined on the set of real numbers R\mathbb{R} is given by a∗b=a2+2b−5a * b = a^2 + 2b - 5. If 3∗x=123 * x = 12, what is the value of xx?

Show answer & explanation

Answer: 4

Answer

The value of xx is 4.
Applying the binary operation definition a∗b=a2+2b−5a * b = a^2 + 2b - 5 to 3∗x3 * x gives 32+2x−5=2x+43^2 + 2x - 5 = 2x + 4. Equating 2x+4=122x + 4 = 12 yields 2x=82x = 8, so x=4x = 4.

Step-by-Step Solution

1
Substitute a=3a = 3 and b=xb = x into the operational rule a∗b=a2+2b−5a * b = a^2 + 2b - 5.
3∗x=32+2x−5=9+2x−5=2x+43 * x = 3^2 + 2x - 5 = 9 + 2x - 5 = 2x + 4
To express the operation 3∗x3 * x as an algebraic expression in terms of xx.
2
Set the simplified algebraic expression equal to the given value of 12.
2x+4=122x + 4 = 12
The question states that 3∗x=123 * x = 12.
3
Solve the linear equation for xx.
2x=8  ⟹  x=42x = 8 \implies x = 4
Subtract 4 from both sides and divide by 2.

Key Concept

Evaluating binary operations and solving algebraic equations involving defined operational rules.
Question 194Question →

A binary operation ∗\ast on the set of real numbers R∖{2}\mathbb{R} \setminus \{2\} is defined by a∗b=2a+2b−ab−2a \ast b = 2a + 2b - ab - 2. If y−1y^{-1} represents the inverse of an element yy under the operation ∗\ast, find the value of xx such that (x∗3)∗4−1=5(x \ast 3) \ast 4^{-1} = 5.

Show answer & explanation

Answer: 8

Answer

The value of x is 8.
To solve for x, first find the identity element by solving a * e = a, yielding e = 1. Next, compute 4^{-1} from 4 * 4^{-1} = 1, which gives 4^{-1} = 2.5. Then simplify x * 3 to 4 - x. Finally, substitute into (4 - x) * 2.5 = 5 and solve for x to get 8.

Step-by-Step Solution

1
Find the identity element e of the operation
e = 1
By definition, a * e = a. Substituting into the operation gives 2a + 2e - ae - 2 = a, which simplifies to (a - 2)(1 - e) = 0. Since a != 2, e must equal 1.
2
Calculate the inverse element 4^{-1}
4^{-1} = 2.5
By definition of an inverse element, 4 * 4^{-1} = e = 1. Applying the operation formula yields 2(4) + 2(4^{-1}) - 4(4^{-1}) - 2 = 1, which simplifies to 6 - 2(4^{-1}) = 1, so 4^{-1} = 2.5.
3
Express x * 3 in terms of x
x * 3 = 4 - x
Evaluating x * 3 using the operational definition gives 2x + 2(3) - 3x - 2 = 4 - x.
4
Solve the main equation (x * 3) * 4^{-1} = 5 for x
x = 8
Substituting x * 3 = 4 - x and 4^{-1} = 2.5 into the equation yields (4 - x) * 2.5 = 5. Applying the operation gives 2(4 - x) + 2(2.5) - 2.5(4 - x) - 2 = 5, which simplifies to 1 + 0.5x = 5, giving x = 8.

Key Concept

Identity and Inverse Elements in Binary Operations
Question 195Question →

Given the simultaneous equations 2x−y=12x - y = 1 and 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17, let (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) be the real solution pairs such that x1>x2x_1 > x_2. What is the value of 2x1+y22x_1 + y_2?

Show answer & explanation

Answer: 1

Answer

1
Substituting y=2x−1y = 2x - 1 into 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17 simplifies correctly to 4x2−4x−15=04x^2 - 4x - 15 = 0. Factorizing gives roots x=52x = \frac{5}{2} and x=−32x = -\frac{3}{2}. Given x1>x2x_1 > x_2, we set x1=52x_1 = \frac{5}{2} and x2=−32x_2 = -\frac{3}{2}. Substituting x2x_2 back into the linear equation gives y2=−4y_2 = -4. Calculating 2x1+y2=2(52)+(−4)=5−4=12x_1 + y_2 = 2\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1.

Step-by-Step Solution

1
Express yy in terms of xx from the linear equation.
y=2x−1y = 2x - 1
Substitution method requires expressing one variable in terms of the other.
2
Substitute y=2x−1y = 2x - 1 into the quadratic equation 2x2−xy+y2+x−y=172x^2 - xy + y^2 + x - y = 17.
2x2−x(2x−1)+(2x−1)2+x−(2x−1)=172x^2 - x(2x - 1) + (2x - 1)^2 + x - (2x - 1) = 17
This converts the system into a single quadratic equation in terms of xx.
3
Expand and simplify the algebraic expression.
2x2−2x2+x+4x2−4x+1+x−2x+1=17  ⟹  4x2−4x+2=17  ⟹  4x2−4x−15=02x^2 - 2x^2 + x + 4x^2 - 4x + 1 + x - 2x + 1 = 17 \implies 4x^2 - 4x + 2 = 17 \implies 4x^2 - 4x - 15 = 0
Combining like terms reveals the standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation 4x2−4x−15=04x^2 - 4x - 15 = 0 by factorization.
(2x−5)(2x+3)=0  ⟹  x=52(2x - 5)(2x + 3) = 0 \implies x = \frac{5}{2} or x=−32x = -\frac{3}{2}
Finding the two real values for xx.
5
Identify x1x_1 and x2x_2 according to x1>x2x_1 > x_2 and compute corresponding yy-values.
x1=52  ⟹  y1=2(52)−1=4x_1 = \frac{5}{2} \implies y_1 = 2\left(\frac{5}{2}\right) - 1 = 4; x2=−32  ⟹  y2=2(−32)−1=−4x_2 = -\frac{3}{2} \implies y_2 = 2\left(-\frac{3}{2}\right) - 1 = -4
Determining the complete coordinate solution pairs.
6
Evaluate the expression 2x1+y22x_1 + y_2.
2(52)+(−4)=5−4=12\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1
Answering the specific value requested in the problem statement.

Key Concept

Solving simultaneous linear and quadratic equations using substitution
Estimated Time:2m 30s
Question 196Question →

If 437+3+477−3=p+q21\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{3}} + \frac{4\sqrt{7}}{\sqrt{7} - \sqrt{3}} = p + q\sqrt{21}, where pp and qq are integers, what is the value of p+qp + q?

Show answer & explanation

Answer: 6

Answer

The value of p+qp + q is 6.
Rationalising each fraction yields (21−3)(\sqrt{21} - 3) and (7+21)(7 + \sqrt{21}). Summing these expressions gives 4+2214 + 2\sqrt{21}. Comparing this to p+q21p + q\sqrt{21} yields p=4p = 4 and q=2q = 2, so p+q=6p + q = 6.

Step-by-Step Solution

1
Rationalise the first term 437+3\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{3}} by multiplying the numerator and denominator by the conjugate (7−3)(\sqrt{7} - \sqrt{3}).
\frac{4\sqrt{3}(\sqrt{7} - \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{4\sqrt{21} - 12}{7 - 3} = \frac{4\sqrt{21} - 12}{4} = \sqrt{21} - 3
Multiplying by the conjugate eliminates the surd from the denominator using the difference of two squares.
2
Rationalise the second term 477−3\frac{4\sqrt{7}}{\sqrt{7} - \sqrt{3}} by multiplying the numerator and denominator by the conjugate (7+3)(\sqrt{7} + \sqrt{3}).
\frac{4\sqrt{7}(\sqrt{7} + \sqrt{3})}{(\sqrt{7})^2 - (\sqrt{3})^2} = \frac{28 + 4\sqrt{21}}{7 - 3} = \frac{28 + 4\sqrt{21}}{4} = 7 + \sqrt{21}
Conjugate rationalisation simplifies the second fraction into linear surd terms.
3
Add the two simplified expressions together and equate to p+q21p + q\sqrt{21}.
(\sqrt{21} - 3) + (7 + \sqrt{21}) = 4 + 2\sqrt{21}
Combining like surd terms yields the simplified form p+q21p + q\sqrt{21}.
4
Identify the values of pp and qq and evaluate p+qp + q.
p = 4, q = 2 \implies p + q = 4 + 2 = 6
Equating the rational parts gives p=4p = 4 and the irrational coefficients gives q=2q = 2.

Key Concept

Rationalisation of surds with binomial denominators

Alternative Method

Combine the two fractions directly over the common denominator (7+3)(7−3)=4(\sqrt{7} + \sqrt{3})(\sqrt{7} - \sqrt{3}) = 4: \frac{4\sqrt{3}(\sqrt{7} - \sqrt{3}) + 4\sqrt{7}(\sqrt{7} + \sqrt{3})}{4} = \frac{4\sqrt{21} - 12 + 28 + 4\sqrt{21}}{4} = \frac{16 + 8\sqrt{21}}{4} = 4 + 2\sqrt{21}.
Estimated Time:1m 30s
Question 197Question →

A binary operation ⊙\odot defined on the set of real numbers R\mathbb{R} is given by a⊙b=a+b+kaba \odot b = a + b + kab, where kk is a non-zero constant. If the inverse of 22 under ⊙\odot is −4-4, what is the value of (3⊙1)−1(3 \odot 1)^{-1}?

Show answer & explanation

Answer: −523-\frac{52}{3}

Answer

The value of (3⊙1)−1(3 \odot 1)^{-1} is −523-\frac{52}{3}.
First, the identity element is determined by solving a⊙e=aa \odot e = a, which gives a+e+kae=a  ⟹  e=0a + e + kae = a \implies e = 0. Next, using the inverse property x⊙x−1=0x \odot x^{-1} = 0, we get x−1=−x1+kxx^{-1} = \frac{-x}{1 + kx}. Given 2−1=−42^{-1} = -4, substituting gives −21+2k=−4\frac{-2}{1 + 2k} = -4, leading to k=−14k = -\frac{1}{4}. Evaluating 3⊙13 \odot 1 yields 3+1−34=1343 + 1 - \frac{3}{4} = \frac{13}{4}. Finally, applying the inverse formula to 134\frac{13}{4} gives −1341−1316=−523\frac{-\frac{13}{4}}{1 - \frac{13}{16}} = -\frac{52}{3}.

Step-by-Step Solution

1
Find the identity element ee under the operation ⊙\odot
e=0e = 0
By definition of identity element, a⊙e=a  ⟹  a+e+kae=a  ⟹  e(1+ka)=0  ⟹  e=0a \odot e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0 for all real numbers aa.
2
Derive the formula for the inverse x−1x^{-1} of an element xx
x−1=−x1+kxx^{-1} = \frac{-x}{1 + kx}
By definition of inverse, x⊙x−1=e  ⟹  x+x−1+kxx−1=0  ⟹  x−1(1+kx)=−x  ⟹  x−1=−x1+kxx \odot x^{-1} = e \implies x + x^{-1} + kxx^{-1} = 0 \implies x^{-1}(1 + kx) = -x \implies x^{-1} = \frac{-x}{1 + kx}.
3
Use the given inverse condition 2−1=−42^{-1} = -4 to find the constant kk
k=−14k = -\frac{1}{4}
Substituting x=2x = 2 into the inverse formula gives −21+2k=−4  ⟹  −2=−4(1+2k)  ⟹  −2=−4−8k  ⟹  8k=−2  ⟹  k=−14\frac{-2}{1 + 2k} = -4 \implies -2 = -4(1 + 2k) \implies -2 = -4 - 8k \implies 8k = -2 \implies k = -\frac{1}{4}.
4
Evaluate the operation 3⊙13 \odot 1
3⊙1=1343 \odot 1 = \frac{13}{4}
Using the operation definition with k=−14k = -\frac{1}{4}: 3⊙1=3+1+(−14)(3)(1)=4−34=1343 \odot 1 = 3 + 1 + \left(-\frac{1}{4}\right)(3)(1) = 4 - \frac{3}{4} = \frac{13}{4}.
5
Calculate the inverse of 134\frac{13}{4} under ⊙\odot
−523-\frac{52}{3}
Using the inverse formula y−1=−y1+kyy^{-1} = \frac{-y}{1 + ky} for y=134y = \frac{13}{4}: y−1=−1341+(−14)(134)=−1341−1316=−134316=−134×163=−523y^{-1} = \frac{-\frac{13}{4}}{1 + \left(-\frac{1}{4}\right)\left(\frac{13}{4}\right)} = \frac{-\frac{13}{4}}{1 - \frac{13}{16}} = \frac{-\frac{13}{4}}{\frac{3}{16}} = -\frac{13}{4} \times \frac{16}{3} = -\frac{52}{3}.

Key Concept

Binary Operations: Finding Identity Elements, Unknown Parameters, and Inverse Elements
Estimated Time:2m 0s
Question 198Question →

A binary operation ∗\ast defined on the set of real numbers R\mathbb{R} is given by a∗b=a+b+7a \ast b = a + b + 7. What is the identity element under this operation?

Show answer & explanation

Answer: -7

Answer

The identity element under the operation is −7-7.
The identity element ee satisfies a∗e=aa \ast e = a for any real number aa. Substituting into the definition gives a+e+7=aa + e + 7 = a, which simplifies to e=−7e = -7.

Step-by-Step Solution

1
Set up the identity element equation using the definition a∗e=aa \ast e = a.
a+e+7=aa + e + 7 = a
By definition, operating any element aa with the identity element ee yields aa.
2
Subtract aa from both sides of the equation.
e+7=0e + 7 = 0
Isolating terms involving ee.
3
Subtract 77 from both sides to solve for ee.
e=−7e = -7
Determining the numerical value of the identity element.

Key Concept

Identity Element in Binary Operations
Question 199Question →

Find the positive value of xx that satisfies the simultaneous equations y−2x=3y - 2x = 3 and y=x2+3y = x^2 + 3.

Show answer & explanation

Answer: 22

Answer

The positive value of xx is 2.
Rearranging the linear equation gives y=2x+3y = 2x + 3. Setting this equal to the second expression for yy gives 2x+3=x2+32x + 3 = x^2 + 3, which simplifies to x2−2x=0x^2 - 2x = 0. Factoring out xx yields x(x−2)=0x(x - 2) = 0. The solutions for xx are 00 and 22. The positive solution is 22.

Step-by-Step Solution

1
Express yy in terms of xx using the linear equation
y=2x+3y = 2x + 3
Isolating yy prepares for substitution into the quadratic equation.
2
Substitute y=2x+3y = 2x + 3 into the second equation y=x2+3y = x^2 + 3
2x+3=x2+32x + 3 = x^2 + 3
This creates a single quadratic equation in terms of xx.
3
Rearrange and factor the quadratic equation
x2−2x=0  ⟹  x(x−2)=0x^2 - 2x = 0 \implies x(x - 2) = 0
Subtracting 33 and 2x2x from both sides simplifies the equation to factorable form.
4
Solve for the non-zero (positive) value of xx
x=2x = 2
Setting x−2=0x - 2 = 0 yields x=2x = 2.

Key Concept

Solving simultaneous linear and quadratic equations by substitution
Question 200Question →

Find the sum of all integer values of xx that satisfy both the linear inequality 2x−1≥32x - 1 \ge 3 and the quadratic inequality x2−5x−14≤0x^2 - 5x - 14 \le 0.

Show answer & explanation

Answer: 27

Answer

The sum of all integer values of xx satisfying both inequalities is 2727.
Solving the linear inequality 2x−1≥32x - 1 \ge 3 yields x≥2x \ge 2. Solving the quadratic inequality x2−5x−14≤0x^2 - 5x - 14 \le 0 by factoring gives (x−7)(x+2)≤0(x - 7)(x + 2) \le 0, which defines the interval −2≤x≤7-2 \le x \le 7. Taking the intersection of x≥2x \ge 2 and −2≤x≤7-2 \le x \le 7 gives 2≤x≤72 \le x \le 7. The integer values satisfying this range are 2,3,4,5,6,2, 3, 4, 5, 6, and 77, and their sum is 2727.

Step-by-Step Solution

1
Solve the linear inequality
2x≥4  ⟹  x≥22x \ge 4 \implies x \ge 2
Adding 1 to both sides and dividing by 2 isolates the variable xx.
2
Factor and solve the quadratic inequality
(x−7)(x+2)≤0  ⟹  −2≤x≤7(x - 7)(x + 2) \le 0 \implies -2 \le x \le 7
The roots of the quadratic equation are x=7x = 7 and x=−2x = -2. The parabola opens upward, so the expression is non-positive between the roots.
3
Determine the intersection of both solution sets
2≤x≤72 \le x \le 7
The values of xx must simultaneously satisfy x≥2x \ge 2 and −2≤x≤7-2 \le x \le 7.
4
List all integer solutions within the valid interval
x∈{2,3,4,5,6,7}x \in \{2, 3, 4, 5, 6, 7\}
These are all the whole numbers contained in the closed interval [2,7][2, 7].
5
Sum the integer solutions
2+3+4+5+6+7=272 + 3 + 4 + 5 + 6 + 7 = 27
Summing the identified integer values yields the final required numerical answer.

Key Concept

Linear and Quadratic Inequalities
Estimated Time:1m 30s
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