Algebra Practice Questions — JAMB UTME — Page 10

Question 181Question →

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

9

11

13

15

Show answer & explanation

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 = 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

A

and

B

.

\mathcal{U} = \{1, 2, 3, \dots, 20\}

with

n(\mathcal{U}) = 20

.

A = \{4, 8, 12, 16, 20\}

with

n(A) = 5

.

B = \{1, 2, 4, 5, 10, 20\}

with

n(B) = 6

.

Listing the elements clearly determines the cardinality of each individual set.

2

Find the intersection and union of sets

A

and

B

.

A \cap B = \{4, 20\}

, so

n(A \cap B) = 2

.

A \cup B = \{1, 2, 4, 5, 8, 10, 12, 16, 20\}

, so

n(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 \cup B)'

.

n((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 $a \oplus b = a + b + kab$ , where $k$ is a non-zero real constant. If the inverse of $3$ under this operation is $-\frac{1}{2}$ , what is the inverse of $4$ under the same operation?

-\frac{12}{23}

-\frac{3}{5}

\frac{12}{17}

-\frac{5}{12}

Show answer & explanation

Answer: $-\frac{12}{23}$

Answer

-\frac{12}{23}

To find the inverse of

4

, first determine the identity element

e

from

a \oplus e = a

, which gives

e = 0

. Next, set

x \oplus x^{-1} = 0

to get

x^{-1} = -\frac{x}{1 + kx}

. Using

3^{-1} = -\frac{1}{2}

, solve

-\frac{3}{1 + 3k} = -\frac{1}{2}

to find

k = \frac{5}{3}

. Finally, substituting

x = 4

yields

4^{-1} = -\frac{4}{1 + 4(\frac{5}{3})} = -\frac{12}{23}

.

Step-by-Step Solution

1

Determine the identity element

e

of the operation.

e = 0

By definition of identity element,

a \oplus e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0

for all valid

a

.

2

Derive the general formula for the inverse

x^{-1}

of an element

x

.

x^{-1} = -\frac{x}{1 + kx}

An element and its inverse yield the identity element under the operation:

x \oplus x^{-1} = 0 \implies x + x^{-1} + kxx^{-1} = 0 \implies x^{-1}(1 + kx) = -x

.

3

Use the given inverse of

3

to solve for the constant

k

.

k = \frac{5}{3}

Given

3^{-1} = -\frac{1}{2}

, substitute

x = 3

into the inverse formula:

-\frac{3}{1 + 3k} = -\frac{1}{2} \implies 1 + 3k = 6 \implies 3k = 5 \implies k = \frac{5}{3}

.

4

Calculate the inverse of

4

using

k = \frac{5}{3}

.

4^{-1} = -\frac{12}{23}

Substitute

x = 4

and

k = \frac{5}{3}

into the inverse formula:

4^{-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 $\mathbb{R} \setminus \{1\}$ is defined by $a \star b = a + b - ab$ . If $y^{-1}$ denotes the inverse of an element $y$ under $\star$ , find the value of $x$ such that $(x \star 3)^{-1} = 2$ .

Show answer & explanation

Answer: 0.5

Answer

The value of

x

is

0.5

.

To solve for

x

, first calculate the identity element

e = 0

using

a \star e = a

. Next, use

y \star y^{-1} = 0

to get the inverse formula

y^{-1} = \frac{y}{y - 1}

. Substituting

y = x \star 3 = 3 - 2x

into the inverse formula yields

\frac{3 - 2x}{2 - 2x} = 2

, which simplifies to

2x = 1

, giving

x = 0.5

.

Step-by-Step Solution

1

Find the identity element

e

of the operation

\star

.

e = 0

By definition of the identity element,

a \star e = a \implies a + e - ae = a \implies e(1 - a) = 0

, so

e = 0

for all

a \neq 1

.

2

Derive the general expression for the inverse element

y^{-1}

of

y

.

y^{-1} = \frac{y}{y - 1}

By definition of inverse element,

y \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 \star 3

.

x \star 3 = 3 - 2x

Using the operational rule

a \star b = a + b - ab

, we obtain

x \star 3 = x + 3 - 3x = 3 - 2x

.

4

Set up and solve the equation

(x \star 3)^{-1} = 2

.

x = 0.5

Letting

y = 3 - 2x

, its inverse is

y^{-1} = \frac{3 - 2x}{(3 - 2x) - 1} = \frac{3 - 2x}{2 - 2x}

. Equating this to

2

gives

\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 $\mathbb{R}$ by $a \oplus b = 2a + 3b - 5$ . What is the value of $(4 \oplus 1) \oplus 2$ ?

Show answer & explanation

Answer: 13

Answer

The value of

(4 \oplus 1) \oplus 2

is 13.

First, evaluate the inner expression

4 \oplus 1

using

a = 4

and

b = 1

, which gives

2(4) + 3(1) - 5 = 8 + 3 - 5 = 6

. Then, substitute this result into the outer expression to evaluate

6 \oplus 2

using

a = 6

and

b = 2

, yielding

2(6) + 3(2) - 5 = 12 + 6 - 5 = 13

.

Step-by-Step Solution

1

Evaluate the inner operation

4 \oplus 1

6

Substitute

a = 4

and

b = 1

into the operational rule

a \oplus b = 2a + 3b - 5

.

2

Evaluate the outer operation using the result from Step 1:

6 \oplus 2

13

Substitute

a = 6

and

b = 2

into the operational rule

a \oplus b = 2a + 3b - 5

.

Key Concept

Evaluation of Binary Operations

Question 185Question →

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

2

3

4

5

Show answer & explanation

Answer: $3$

Answer

The positive value of

x

is

3

.

Substituting

y = x + 1

from the linear equation into the quadratic equation gives

x + 1 = x^2 - 5

. Rearranging terms yields

x^2 - x - 6 = 0

, which factors as

(x - 3)(x + 2) = 0

. The roots are

x = 3

and

x = -2

. Selecting the positive solution gives

3

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the linear equation.

y = x + 1

Isolating

y

simplifies substitution into the quadratic equation.

2

Substitute

y = x + 1

into the quadratic equation

y = x^2 - 5

.

x + 1 = x^2 - 5

This produces a quadratic equation in one variable,

x

.

3

Rearrange into standard quadratic form

ax^2 + bx + c = 0

.

x^2 - x - 6 = 0

Standard quadratic form allows direct factorization.

4

Factorize the quadratic equation.

(x - 3)(x + 2) = 0

Factoring determines the roots of the equation.

5

Solve for

x

and identify the positive value.

x = 3

or

x = -2

; positive value is

3

.

The question specifies the positive real solution for

x

.

Key Concept

Solving simultaneous linear and quadratic equations using substitution

Question 186Question →

If $x$ and $y$ satisfy the simultaneous equations $x - 2y = 1$ and $x^2 - xy - y^2 = 5$ , what is the positive value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Substituting

x = 2y + 1

into the non-linear equation yields

(2y + 1)^2 - (2y + 1)y - y^2 = 5

, which simplifies to

y^2 + 3y - 4 = 0

. Solving this gives

y = 1

or

y = -4

. Substituting

y = 1

into

x = 2y + 1

gives

x = 3

, which is the positive value of

x

.

Step-by-Step Solution

1

Make

x

the subject of the linear equation

x = 2y + 1

Substitution is the standard method for solving linear-quadratic simultaneous systems.

2

Substitute

x = 2y + 1

into

x^2 - xy - y^2 = 5

(2y + 1)^2 - (2y + 1)y - y^2 = 5

Eliminates

x

to create a single quadratic equation in terms of

y

.

3

Expand and simplify to standard quadratic form

y^2 + 3y - 4 = 0

Simplifying algebraic expressions allows factorization.

4

Solve for

y

by factoring

y = 1

or

y = -4

Factors of

-4

that sum to

3

are

+4

and

-1

.

5

Calculate corresponding values for

x

x = 3

when

y = 1

, and

x = -7

when

y = -4

Substitute

y

back into the linear expression for

x

.

6

Select the positive value of

x

3

The question explicitly asks for the positive value of

x

.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 187Question →

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

-6

-4

6

-8

Show answer & explanation

Answer: $-6$

Answer

The sum of all possible values of

u

is

-6

.

From the linear equation

u - 2v = 1

, we express

u

as

u = 2v + 1

. Substituting this into the quadratic equation

u^2 - 3v^2 = 13

gives

(2v + 1)^2 - 3v^2 = 13

, which expands and simplifies to

v^2 + 4v - 12 = 0

. Factoring gives

(v + 6)(v - 2) = 0

, so

v = 2

or

v = -6

. Substituting these back into

u = 2v + 1

yields

u = 5

(for

v = 2

) and

u = -11

(for

v = -6

). The sum of all possible values of

u

is

5 + (-11) = -6

.

Step-by-Step Solution

1

Express

u

in terms of

v

using the linear equation.

u = 2v + 1

Isolating

u

allows for straightforward substitution into the quadratic equation.

2

Substitute

u = 2v + 1

into the quadratic equation

u^2 - 3v^2 = 13

and expand.

(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

v

.

3

Solve the quadratic equation for

v

by factorisation.

(v + 6)(v - 2) = 0 \implies v = 2 \text{ or } v = -6

Finding the roots of the quadratic gives the

v

-coordinates of the solution pairs.

4

Calculate the corresponding values of

u

using

u = 2v + 1

.

For

v = 2

:

u = 2(2) + 1 = 5

.
For

v = -6

:

u = 2(-6) + 1 = -11

.

Each

v

value corresponds to a specific

u

value in the solution pairs.

5

Compute the sum of all possible values of

u

.

5 + (-11) = -6

The question asks specifically for the sum of the

u

-values.

Key Concept

Solving simultaneous linear and quadratic equations using substitution

Question 188Question →

If $(x_1, y_1)$ and $(x_2, y_2)$ are the real solution pairs to the simultaneous equations $3x - y = 5$ and $x^2 + 2xy - y^2 = 7$ , with $x_1 < x_2$ , what is the value of $y_2 - y_1$ ?

18

6

20

14

Show answer & explanation

Answer: 18

Answer

18

Expressing

y

as

3x - 5

and substituting into the quadratic equation yields

x^2 - 10x + 16 = 0

. Solving gives

x_1 = 2

and

x_2 = 8

. Evaluating

y = 3x - 5

for both values gives

y_1 = 1

and

y_2 = 19

. The difference

y_2 - y_1 = 19 - 1 = 18

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation

y = 3x - 5

Isolation of

y

facilitates substitution into the non-linear equation.

2

Substitute

y = 3x - 5

into the quadratic equation

x^2 + 2xy - y^2 = 7

x^2 + 2x(3x - 5) - (3x - 5)^2 = 7

Reduces the system to a single quadratic equation in terms of

x

.

3

Expand and simplify the algebraic expression

x^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

ax^2 + bx + c = 0

.

4

Solve the quadratic equation

x^2 - 10x + 16 = 0

by factorization

(x - 2)(x - 8) = 0 \implies x_1 = 2

and

x_2 = 8

Identifies the two roots with

x_1 < x_2

as specified by the condition.

5

Determine the corresponding

y

-values using

y = 3x - 5

y_1 = 3(2) - 5 = 1

and

y_2 = 3(8) - 5 = 19

Obtains the complete coordinate solution pairs

(2, 1)

and

(8, 19)

.

6

Calculate the required difference

y_2 - y_1

y_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 $\mathbb{R}$ is defined by $a \circ b = a + b - \frac{1}{2}ab$ . What is the inverse of $3$ under this operation?

6

-6

-3

\frac{1}{3}

Show answer & explanation

Answer: $6$

Answer

The inverse of

3

under the defined binary operation is

6

.

To find the inverse of

3

, we first find the identity element

e

using

a \circ e = a \implies a + e - \frac{1}{2}ae = a

, which gives

e = 0

. Then, setting

3 \circ x = 0

gives

3 + x - \frac{3}{2}x = 0 \implies 3 = \frac{1}{2}x \implies x = 6

.

Step-by-Step Solution

1

Find the identity element

e

of the operation.

e = 0

By definition of identity element,

a \circ e = a

. Substituting into the operation formula gives

a + e - \frac{1}{2}ae = a \implies e\left(1 - \frac{1}{2}a\right) = 0 \implies e = 0

for all

a \neq 2

.

2

Set up the inverse equation for the element

3

.

3 \circ x = 0

Let

x

be the inverse of

3

. By definition of inverse element,

3 \circ x = e

, where

e = 0

.

3

Apply the binary operation definition to the left-hand side.

3 + x - \frac{1}{2}(3)(x) = 0

Substitute

a = 3

and

b = x

into

a \circ b = a + b - \frac{1}{2}ab

.

4

Solve the linear equation for

x

.

x = 6

Simplify the equation:

3 + 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 $P = \begin{pmatrix} x & 1 & 2 \\ 3 & 0 & x \\ 1 & 2 & 1 \end{pmatrix}$ , for what positive value of $x$ is the determinant of $P$ equal to $3$ ?

Show answer & explanation

Answer: 2

Answer

The positive value of

x

is

2

.

Expanding the determinant of

P

along the first row yields

\det(P) = -2x^2 + x + 9

. Setting

\det(P) = 3

leads to

-2x^2 + x + 9 = 3

, which rearranges to

2x^2 - x - 6 = 0

. Factoring gives

(2x + 3)(x - 2) = 0

, leading to roots

x = -1.5

and

x = 2

. The positive value is

2

.

Step-by-Step Solution

1

Calculate the determinant of matrix

P

in terms of

x

\det(P) = -2x^2 + x + 9

Expanding along the first row gives

x(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

2x^2 - x - 6 = 0

Subtracting

3

from both sides gives

-2x^2 + x + 6 = 0

, which simplifies to

2x^2 - x - 6 = 0

.

3

Solve the quadratic equation for

x

x = 2 \text{ or } x = -1.5

Factoring

(2x + 3)(x - 2) = 0

yields

x = -\frac{3}{2}

and

x = 2

.

4

Select the positive solution as required by the problem statement

x = 2

The problem specifically requests the positive real value of

x

.

Key Concept

3x3 Determinant Evaluation and Quadratic Solving

Question 191Question →

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

2

3

4

7

Show answer & explanation

Answer: $3$

Answer

The value of

Q

when

p = 10

is

3

.

The relationship for partial inverse variation is

Q = k_1 + \frac{k_2}{p}

. Substituting

(p=2, Q=11)

gives

2k_1 + k_2 = 22

, and substituting

(p=5, Q=5)

gives

5k_1 + k_2 = 25

. Solving these simultaneous equations gives

k_1 = 1

and

k_2 = 20

. Substituting

p = 10

into

Q = 1 + \frac{20}{p}

results in

Q = 1 + 2 = 3

.

Step-by-Step Solution

1

Set up the general formula for partial variation.

Q = k_1 + \frac{k_2}{p}

, where

k_1

and

k_2

are constants.

Partial variation consists of a constant part and a part that varies inversely with

p

.

2

Substitute the given values to form two simultaneous linear equations.

For

p = 2, Q = 11 \implies 11 = k_1 + \frac{k_2}{2} \implies 2k_1 + k_2 = 22

For

p = 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

k_1

and

k_2

.

Subtracting the first equation from the second gives

3k_1 = 3 \implies k_1 = 1

.
Substituting

k_1 = 1

into

2(1) + k_2 = 22

yields

k_2 = 20

.

Determining

k_1

and

k_2

establishes the explicit relationship between

Q

and

p

.

4

Calculate

Q

when

p = 10

.

Q = 1 + \frac{20}{10} = 1 + 2 = 3

.

Substitute

p = 10

into the established variation equation

Q = 1 + \frac{20}{p}

.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:1m 30s

Question 192Question →

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

\frac{7}{3}

\frac{3}{5}

-\frac{7}{9}

-\frac{4}{3}

Show answer & explanation

Answer: $\frac{7}{3}$

Answer

\frac{7}{3}

Evaluating

3 \Delta x

gives

3 - 3x

. Then operating

(3 - 3x) \Delta 1

yields

(3 - 3x) + 3 - 2(3 - 3x) = 3x

. Equating

3x = 7

gives

x = \frac{7}{3}

.

Step-by-Step Solution

1

Evaluate the inner expression

3 \Delta x

using the definition

a \Delta b = a + 3b - 2ab

.

3 \Delta x = 3 + 3x - 2(3)(x) = 3 + 3x - 6x = 3 - 3x

.

Substitute

a = 3

and

b = x

into the operation rule.

2

Substitute the result

(3 - 3x)

as the first operand in the outer expression

(3 - 3x) \Delta 1

.

(3 - 3x) \Delta 1 = (3 - 3x) + 3(1) - 2(3 - 3x)(1)

.

Apply the binary operation definition with

a = 3 - 3x

and

b = 1

.

3

Expand and simplify the algebraic expression.

(3 - 3x) \Delta 1 = 3 - 3x + 3 - 6 + 6x = 3x

.

Distribute

-2

across

(3 - 3x)

to get

-6 + 6x

, then collect like terms.

4

Set the simplified expression equal to

7

and solve for

x

.

3x = 7 \implies x = \frac{7}{3}

.

Divide both sides by

3

to isolate

x

.

Key Concept

Non-commutative binary operation composition and algebraic equation solving

Question 193Question →

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

Show answer & explanation

Answer: 4

Answer

The value of

x

is 4.

Applying the binary operation definition

a * b = a^2 + 2b - 5

to

3 * x

gives

3^2 + 2x - 5 = 2x + 4

. Equating

2x + 4 = 12

yields

2x = 8

, so

x = 4

.

Step-by-Step Solution

1

Substitute

a = 3

and

b = x

into the operational rule

a * b = a^2 + 2b - 5

.

3 * x = 3^2 + 2x - 5 = 9 + 2x - 5 = 2x + 4

To express the operation

3 * x

as an algebraic expression in terms of

x

.

2

Set the simplified algebraic expression equal to the given value of 12.

2x + 4 = 12

The question states that

3 * x = 12

.

3

Solve the linear equation for

x

.

2x = 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 $\mathbb{R} \setminus \{2\}$ is defined by $a \ast b = 2a + 2b - ab - 2$ . If $y^{-1}$ represents the inverse of an element $y$ under the operation $\ast$ , find the value of $x$ such that $(x \ast 3) \ast 4^{-1} = 5$ .

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 $2x - y = 1$ and $2x^2 - xy + y^2 + x - y = 17$ , let $(x_1, y_1)$ and $(x_2, y_2)$ be the real solution pairs such that $x_1 > x_2$ . What is the value of $2x_1 + y_2$ ?

1

3

9

-3

Show answer & explanation

Answer: 1

Answer

1

Substituting

y = 2x - 1

into

2x^2 - xy + y^2 + x - y = 17

simplifies correctly to

4x^2 - 4x - 15 = 0

. Factorizing gives roots

x = \frac{5}{2}

and

x = -\frac{3}{2}

. Given

x_1 > x_2

, we set

x_1 = \frac{5}{2}

and

x_2 = -\frac{3}{2}

. Substituting

x_2

back into the linear equation gives

y_2 = -4

. Calculating

2x_1 + y_2 = 2\left(\frac{5}{2}\right) + (-4) = 5 - 4 = 1

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation.

y = 2x - 1

Substitution method requires expressing one variable in terms of the other.

2

Substitute

y = 2x - 1

into the quadratic equation

2x^2 - xy + y^2 + x - y = 17

.

2x^2 - x(2x - 1) + (2x - 1)^2 + x - (2x - 1) = 17

This converts the system into a single quadratic equation in terms of

x

.

3

Expand and simplify the algebraic expression.

2x^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

ax^2 + bx + c = 0

.

4

Solve the quadratic equation

4x^2 - 4x - 15 = 0

by factorization.

(2x - 5)(2x + 3) = 0 \implies x = \frac{5}{2}

or

x = -\frac{3}{2}

Finding the two real values for

x

.

5

Identify

x_1

and

x_2

according to

x_1 > x_2

and compute corresponding

y

-values.

x_1 = \frac{5}{2} \implies y_1 = 2\left(\frac{5}{2}\right) - 1 = 4

;

x_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

2x_1 + y_2

.

2\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 $\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{3}} + \frac{4\sqrt{7}}{\sqrt{7} - \sqrt{3}} = p + q\sqrt{21}$ , where $p$ and $q$ are integers, what is the value of $p + q$ ?

Show answer & explanation

Answer: 6

Answer

The value of

p + q

is 6.

Rationalising each fraction yields

(\sqrt{21} - 3)

and

(7 + \sqrt{21})

. Summing these expressions gives

4 + 2\sqrt{21}

. Comparing this to

p + q\sqrt{21}

yields

p = 4

and

q = 2

, so

p + q = 6

.

Step-by-Step Solution

1

Rationalise the first term

\frac{4\sqrt{3}}{\sqrt{7} + \sqrt{3}}

by multiplying the numerator and denominator by the conjugate

(\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

\frac{4\sqrt{7}}{\sqrt{7} - \sqrt{3}}

by multiplying the numerator and denominator by the conjugate

(\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 + q\sqrt{21}

.

(\sqrt{21} - 3) + (7 + \sqrt{21}) = 4 + 2\sqrt{21}

Combining like surd terms yields the simplified form

p + q\sqrt{21}

.

4

Identify the values of

p

and

q

and evaluate

p + q

.

p = 4, q = 2 \implies p + q = 4 + 2 = 6

Equating the rational parts gives

p = 4

and the irrational coefficients gives

q = 2

.

Key Concept

Rationalisation of surds with binomial denominators

Alternative Method

Combine the two fractions directly over the common denominator

(\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 $\mathbb{R}$ is given by $a \odot b = a + b + kab$ , where $k$ is a non-zero constant. If the inverse of $2$ under $\odot$ is $-4$ , what is the value of $(3 \odot 1)^{-1}$ ?

-\frac{52}{3}

\frac{13}{4}

-\frac{28}{5}

-\frac{13}{3}

Show answer & explanation

Answer: $-\frac{52}{3}$

Answer

The value of

(3 \odot 1)^{-1}

is

-\frac{52}{3}

.

First, the identity element is determined by solving

a \odot e = a

, which gives

a + e + kae = a \implies e = 0

. Next, using the inverse property

x \odot x^{-1} = 0

, we get

x^{-1} = \frac{-x}{1 + kx}

. Given

2^{-1} = -4

, substituting gives

\frac{-2}{1 + 2k} = -4

, leading to

k = -\frac{1}{4}

. Evaluating

3 \odot 1

yields

3 + 1 - \frac{3}{4} = \frac{13}{4}

. Finally, applying the inverse formula to

\frac{13}{4}

gives

\frac{-\frac{13}{4}}{1 - \frac{13}{16}} = -\frac{52}{3}

.

Step-by-Step Solution

1

Find the identity element

e

under the operation

\odot

e = 0

By definition of identity element,

a \odot e = a \implies a + e + kae = a \implies e(1 + ka) = 0 \implies e = 0

for all real numbers

a

.

2

Derive the formula for the inverse

x^{-1}

of an element

x

x^{-1} = \frac{-x}{1 + kx}

By definition of inverse,

x \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} = -4

to find the constant

k

k = -\frac{1}{4}

Substituting

x = 2

into the inverse formula gives

\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 \odot 1

3 \odot 1 = \frac{13}{4}

Using the operation definition with

k = -\frac{1}{4}

:

3 \odot 1 = 3 + 1 + \left(-\frac{1}{4}\right)(3)(1) = 4 - \frac{3}{4} = \frac{13}{4}

.

5

Calculate the inverse of

\frac{13}{4}

under

\odot

-\frac{52}{3}

Using the inverse formula

y^{-1} = \frac{-y}{1 + ky}

for

y = \frac{13}{4}

:

y^{-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 $\mathbb{R}$ is given by $a \ast b = a + b + 7$ . What is the identity element under this operation?

Show answer & explanation

Answer: -7

Answer

The identity element under the operation is

-7

.

The identity element

e

satisfies

a \ast e = a

for any real number

a

. Substituting into the definition gives

a + e + 7 = a

, which simplifies to

e = -7

.

Step-by-Step Solution

1

Set up the identity element equation using the definition

a \ast e = a

.

a + e + 7 = a

By definition, operating any element

a

with the identity element

e

yields

a

.

2

Subtract

a

from both sides of the equation.

e + 7 = 0

Isolating terms involving

e

.

3

Subtract

7

from both sides to solve for

e

.

e = -7

Determining the numerical value of the identity element.

Key Concept

Identity Element in Binary Operations

Question 199Question →

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

2

-2

3

5

Show answer & explanation

Answer: $2$

Answer

The positive value of

x

is 2.

Rearranging the linear equation gives

y = 2x + 3

. Setting this equal to the second expression for

y

gives

2x + 3 = x^2 + 3

, which simplifies to

x^2 - 2x = 0

. Factoring out

x

yields

x(x - 2) = 0

. The solutions for

x

are

0

and

2

. The positive solution is

2

.

Step-by-Step Solution

1

Express

y

in terms of

x

using the linear equation

y = 2x + 3

Isolating

y

prepares for substitution into the quadratic equation.

2

Substitute

y = 2x + 3

into the second equation

y = x^2 + 3

2x + 3 = x^2 + 3

This creates a single quadratic equation in terms of

x

.

3

Rearrange and factor the quadratic equation

x^2 - 2x = 0 \implies x(x - 2) = 0

Subtracting

3

and

2x

from both sides simplifies the equation to factorable form.

4

Solve for the non-zero (positive) value of

x

x = 2

Setting

x - 2 = 0

yields

x = 2

.

Key Concept

Solving simultaneous linear and quadratic equations by substitution

Question 200Question →

Find the sum of all integer values of $x$ that satisfy both the linear inequality $2x - 1 \ge 3$ and the quadratic inequality $x^2 - 5x - 14 \le 0$ .

Show answer & explanation

Answer: 27

Answer

The sum of all integer values of

x

satisfying both inequalities is

27

.

Solving the linear inequality

2x - 1 \ge 3

yields

x \ge 2

. Solving the quadratic inequality

x^2 - 5x - 14 \le 0

by factoring gives

(x - 7)(x + 2) \le 0

, which defines the interval

-2 \le x \le 7

. Taking the intersection of

x \ge 2

and

-2 \le x \le 7

gives

2 \le x \le 7

. The integer values satisfying this range are

2, 3, 4, 5, 6,

and

7

, and their sum is

27

.

Step-by-Step Solution

1

Solve the linear inequality

2x \ge 4 \implies x \ge 2

Adding 1 to both sides and dividing by 2 isolates the variable

x

.

2

Factor and solve the quadratic inequality

(x - 7)(x + 2) \le 0 \implies -2 \le x \le 7

The roots of the quadratic equation are

x = 7

and

x = -2

. The parabola opens upward, so the expression is non-positive between the roots.

3

Determine the intersection of both solution sets

2 \le x \le 7

The values of

x

must simultaneously satisfy

x \ge 2

and

-2 \le x \le 7

.

4

List all integer solutions within the valid interval

x \in \{2, 3, 4, 5, 6, 7\}

These are all the whole numbers contained in the closed interval

[2, 7]

.

5

Sum the integer solutions

2 + 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

Algebra

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