Exploring Algebraic Identities Class 9 — Competency Based Questions

20 Questions | 40 Marks

A rectangular garden has dimensions

(3 x + 5)

metres and

(2 x + 3)

metres. If the area of the garden is 60 square metres, find the value of

x

. If

x = 3

, what is the perimeter of the garden? 3marks

Assertion (A): The expression

(x - y) (x + y) = x^{2} - y^{2}

is an identity.Reason (R): An identity is an equation that is true for all possible values of the variables involved.(a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true. 1mark

The volume of a cube is given by

a^{3} + 6 a^{2} b + 12 a b^{2} + 8 b^{3}

. Find the side length of the cube. 2marks

A father wants to create a rectangular patio with an area of

x^{2} + 5 x + 6

square metres. He knows the width of the patio is

(x + 2)

metres. Find the length of the patio. If

x = 4

, what are the dimensions of the patio? 3marks

Consider the identity

(a + b)^{2} = a^{2} + 2 a b + b^{2}

. If

a = 5

and

b = - 3

, verify this identity. 2marks

Assertion (A):

(x + y)^{2} = x^{2} + y^{2}

.Reason (R):

(x + y)^{2}

expands to

x^{2} + 2 x y + y^{2}

, which is not equal to

x^{2} + y^{2}

unless

x y = 0

.(a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true. 1mark

Two friends, Rohan and Priya, are calculating

(105)^{2}

. Rohan uses the identity

(a + b)^{2}

and calculates it as

(100 + 5)^{2} = 100^{2} + 2 (100) (5) + 5^{2} = 10000 + 1000 + 25 = 11025

. Priya uses the identity

(a - b)^{2}

and calculates it as

(100 + 5)^{2}

. She mistakenly writes it as

(110 - 5)^{2} = 110^{2} - 2 (110) (5) + 5^{2} = 12100 - 1100 + 25 = 11025

.(a) Rohan's method is correct, while Priya's calculation, despite reaching the correct answer, used an incorrect initial step.(b) Both Rohan and Priya used incorrect methods.(c) Rohan used a correct method, but Priya's method is fundamentally flawed because

(110 - 5)

is not equal to

(105)^{2}

.(d) Both Rohan and Priya used correct methods. 2marks

Expand

(3 a - 4 b - 2 c)^{2}

using an appropriate algebraic identity. 3marks

Factorise the expression

16 x^{2} + 24 x y + 9 y^{2}

. 2marks

A farmer is planning to build a rectangular fence. He has

2 x^{2} + 11 x + 5

square metres of land available for the enclosure. If the length of the enclosure is

(2 x + 1)

metres, find its breadth. 3marks

Simplify the rational expression

\frac{x^{2} - 5 x + 6}{x^{2} - 4}

. 3marks

Assertion (A): The expression

(x - y)^{3} = x^{3} - y^{3}

.Reason (R): The correct expansion of

(x - y)^{3}

x^{3} - 3 x^{2} y + 3 x y^{2} - y^{3}

.(a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true. 1mark

The area of a square is given by

9 p^{2} - 12 p + 4

. Find the side length of the square. 2marks

Use the identity

a^{2} - b^{2} = (a + b) (a - b)

to find the value of

57^{2} - 43^{2}

. 2marks

A teacher asked her students to factorise

x^{2} + 13 x + 30

.Ravi wrote:

x^{2} + 10 x + 3 x + 30 = x (x + 10) + 3 (x + 10) = (x + 10) (x + 3)

.Meena wrote:

x^{2} + 15 x + 2 x + 30 = x (x + 15) + 2 (x + 15) = (x + 15) (x + 2)

.Who is correct and why? 2marks

Expand

(a + b + c)^{2}

and verify with

a = 1, b = 2, c = 3

. 3marks

Factorise the expression

x^{3} - y^{3}

. 1mark

A rectangular plot of land has an area of

4 x^{2} + 12 x y + 9 y^{2}

square metres. If the length of the plot is

(2 x + 3 y)

metres, find its breadth. 2marks

A student was asked to calculate the value of

(30 - 1)^{2}

. He wrote:

(30 - 1)^{2} = 30^{2} - 1^{2} = 900 - 1 = 899

.Identify the mistake in his calculation and provide the correct value using the appropriate identity. 3marks

Simplify the rational expression

\frac{2 p^{2} - 3 p q - 18 q^{2}}{p^{2} + 3 p q - 10 q^{2}}

. 3marks

Answers

The area of the rectangular garden is

(3 x + 5) (2 x + 3)

(3 x + 5) (2 x + 3) = 6 x^{2} + 9 x + 10 x + 15 = 6 x^{2} + 19 x + 15

.Given area = 60 sq. metres.

6 x^{2} + 19 x + 15 = 60

6 x^{2} + 19 x - 45 = 0

We need to find factors of $6 \times (- 45) = - 270$ that add up to 19. These are 27 and -10.

6 x^{2} + 27 x - 10 x - 45 = 0

3 x (2 x + 9) - 5 (2 x + 9) = 0

(3 x - 5) (2 x + 9) = 0

So, $3 x - 5 = 0$ or $2 x + 9 = 0$ .
$x = \frac{5}{3}$ or $x = - \frac{9}{2}$ . Since length cannot be negative, $x = \frac{5}{3}$ .
If $x = 3$ , dimensions are $(3 (3) + 5) = 14$ metres and $(2 (3) + 3) = 9$ metres.
Perimeter = $2 \times (length + breadth) = 2 \times (14 + 9) = 2 \times 23 = 46$ metres.

(a) Both A and R are true and R is the correct explanation of A.

We need to find

a

and

b

such that

(a + b)^{3} = a^{3} + 6 a^{2} b + 12 a b^{2} + 8 b^{3}

.Comparing with the identity

(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}

, we can rewrite the given expression as:

(a)^{3} + 3 (a)^{2} (2 b) + 3 (a) (2 b)^{2} + (2 b)^{3}

This matches the form $(a + b)^{3}$ with $a = a$ and $b = 2 b$ .
Therefore, the side length of the cube is $(a + 2 b)$ .

The area of the rectangular patio is given by

(x + 2) (x + k) = x^{2} + (2 + k) x + 2 k

.We are given the area as

x^{2} + 5 x + 6

.Comparing the constant terms,

2 k = 6 ⟹ k = 3

.Let's check the coefficient of

x

2 + k = 2 + 3 = 5

, which matches.So, the factors are

(x + 2)

and

(x + 3)

.The length of the patio is

(x + 3)

metres.If

x = 4

, the length is

(4 + 3) = 7

metres and the width is

(4 + 2) = 6

metres.

Given identity:

(a + b)^{2} = a^{2} + 2 a b + b^{2}

.Given

a = 5

and

b = - 3

.Left side:

(a + b)^{2} = (5 + (- 3))^{2} = (5 - 3)^{2} = 2^{2} = 4

.Right side:

a^{2} + 2 a b + b^{2} = (5)^{2} + 2 (5) (- 3) + (- 3)^{2} = 25 - 30 + 9 = 4

.Since Left Side = Right Side, the identity is verified.

(d) A is false but R is true. The statement A is false, but R correctly explains why.

(a) Rohan's method is correct, while Priya's calculation, despite reaching the correct answer, used an incorrect initial step. Priya's mistake was assuming

(105)^{2} = (110 - 5)^{2}

to find the value of

(105)^{2}

. While the calculation using

(110 - 5)^{2}

yields the correct answer

11025

, the initial step is incorrect as

(105)^{2}

is not

(110 - 5)^{2}

. Rohan correctly used

(100 + 5)^{2}

We use the identity

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

.Here, we have

(3 a - 4 b - 2 c)^{2}

. Let

A = 3 a

B = - 4 b

C = - 2 c

(3 a + (- 4 b) + (- 2 c))^{2} = (3 a)^{2} + (- 4 b)^{2} + (- 2 c)^{2} + 2 (3 a) (- 4 b) + 2 (- 4 b) (- 2 c) + 2 (- 2 c) (3 a)

$= 9 a^{2} + 16 b^{2} + 4 c^{2} - 24 a b + 16 b c - 12 a c$ .

We need to factorise

16 x^{2} + 24 x y + 9 y^{2}

.We can see that

16 x^{2} = (4 x)^{2}

and

9 y^{2} = (3 y)^{2}

.The middle term is

24 x y

. Let's check if

2 \times (4 x) \times (3 y) = 24 x y

2 \times 4 x \times 3 y = 24 x y

.So, the expression is in the form

a^{2} + 2 a b + b^{2}

, where

a = 4 x

and

b = 3 y

.Therefore,

16 x^{2} + 24 x y + 9 y^{2} = (4 x + 3 y)^{2}

The area of a rectangle is Length

\times

Breadth.We are given Area

= 2 x^{2} + 11 x + 5

and Length

= 2 x + 1

.We need to factorise

2 x^{2} + 11 x + 5

. We look for two numbers whose product is

2 \times 5 = 10

and whose sum is 11. These numbers are 10 and 1.

2 x^{2} + 11 x + 5 = 2 x^{2} + 10 x + x + 5

= 2 x (x + 5) + 1 (x + 5)

= (2 x + 1) (x + 5)

Since the length is $(2 x + 1)$ metres, the breadth is $(x + 5)$ metres.

To simplify

\frac{x^{2} - 5 x + 6}{x^{2} - 4}

, we factorise the numerator and the denominator.Numerator:

x^{2} - 5 x + 6

. We need two numbers whose product is 6 and sum is -5. These are -2 and -3.So,

x^{2} - 5 x + 6 = (x - 2) (x - 3)

.Denominator:

x^{2} - 4

is a difference of squares,

x^{2} - 2^{2} = (x - 2) (x + 2)

.Therefore,

\frac{x^{2} - 5 x + 6}{x^{2} - 4} = \frac{(x - 2) (x - 3)}{(x - 2) (x + 2)}

.Assuming

x \neq 2

, we can cancel out

(x - 2)

.The simplified expression is

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

(d) A is false but R is true. Statement A is incorrect. The correct expansion of

(x - y)^{3}

is given in Reason R.

The area of a square is side

\times

side

= (side)^{2}

.We need to factorise

9 p^{2} - 12 p + 4

9 p^{2} = (3 p)^{2}

and

4 = 2^{2}

.The middle term is

- 12 p

. Let's check if

2 \times (3 p) \times 2 = 12 p

. It is.So, the expression is in the form

a^{2} - 2 a b + b^{2}

, where

a = 3 p

and

b = 2

.Therefore,

9 p^{2} - 12 p + 4 = (3 p - 2)^{2}

.The side length of the square is

(3 p - 2)

We use the identity

a^{2} - b^{2} = (a + b) (a - b)

57^{2} - 43^{2} = (57 + 43) (57 - 43)

= (100) (14)

$= 1400$ .

Ravi is correct.For Ravi:

x^{2} + 13 x + 30 = x^{2} + 10 x + 3 x + 30 = x (x + 10) + 3 (x + 10) = (x + 10) (x + 3)

. This factorization is correct as

10 + 3 = 13

and

10 \times 3 = 30

.For Meena:

x^{2} + 13 x + 30

. The factorization

x^{2} + 15 x + 2 x + 30 = x (x + 15) + 2 (x + 15) = (x + 15) (x + 2)

is incorrect because

15 + 2 = 17

, not 13, and

15 \times 2 = 30

The identity is

(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a

.Let

a = 1, b = 2, c = 3

.Left side:

(a + b + c)^{2} = (1 + 2 + 3)^{2} = 6^{2} = 36

.Right side:

a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = 1^{2} + 2^{2} + 3^{2} + 2 (1) (2) + 2 (2) (3) + 2 (3) (1)

= 1 + 4 + 9 + 4 + 12 + 6 = 36

.Since Left Side = Right Side, the identity is verified.

The identity for the difference of cubes is

x^{3} - y^{3} = (x - y) (x^{2} + x y + y^{2})

Area = Length

\times

Breadth.We are given Area

= 4 x^{2} + 12 x y + 9 y^{2}

and Length

= (2 x + 3 y)

.We need to factorise

4 x^{2} + 12 x y + 9 y^{2}

4 x^{2} = (2 x)^{2}

and

9 y^{2} = (3 y)^{2}

.The middle term is

12 x y

. Let's check if

2 \times (2 x) \times (3 y) = 12 x y

2 \times 2 x \times 3 y = 12 x y

.So,

4 x^{2} + 12 x y + 9 y^{2} = (2 x + 3 y)^{2}

.Since Area = Length

\times

Breadth, and Length

= (2 x + 3 y)

, then Breadth

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

The mistake is in applying the identity

(a - b)^{2}

. The student wrote

(30 - 1)^{2} = 30^{2} - 1^{2}

. This is incorrect. The correct identity for

(a - b)^{2}

a^{2} - 2 a b + b^{2}

.The correct calculation is:

(30 - 1)^{2} = 30^{2} - 2 (30) (1) + 1^{2}

= 900 - 60 + 1

$= 841$ .

To simplify

\frac{2 p^{2} - 3 p q - 18 q^{2}}{p^{2} + 3 p q - 10 q^{2}}

, we factorise the numerator and denominator.Numerator:

2 p^{2} - 3 p q - 18 q^{2}

. We look for two numbers whose product is

2 \times (- 18) = - 36

and whose sum is -3. These numbers are -6 and 3.

2 p^{2} - 6 p q + 3 p q - 18 q^{2} = 2 p (p - 3 q) + 3 q (p - 6 q)

. This does not seem to factor directly into

(p - a q) (p - b q)

.Let's recheck the factors. We need two numbers whose product is

2 \times (- 18) = - 36

and sum is

- 3

. These are

- 9

and

4

2 p^{2} - 9 p q + 4 p q - 18 q^{2}

. This also does not look right.Let's try the standard form

(a p + b) (c p + d)

.Consider the factors of

2 p^{2}

(

2 p, p

) and

- 18 q^{2}

(

- 18 q, q

;

18 q, - q

;

- 9 q, 2 q

;

9 q, - 2 q

;

- 6 q, 3 q

;

6 q, - 3 q

).Let's try

(2 p + . . .) (p - . . .)

.Try

(2 p + 3 q) (p - 6 q) = 2 p^{2} - 12 p q + 3 p q - 18 q^{2} = 2 p^{2} - 9 p q - 18 q^{2}

. Still not matching.Try

(2 p - 6 q) (p + 3 q) = 2 p^{2} + 6 p q - 6 p q - 18 q^{2} = 2 p^{2} - 18 q^{2}

. Incorrect middle term.Try

(2 p + 6 q) (p - 3 q) = 2 p^{2} - 6 p q + 6 p q - 18 q^{2} = 2 p^{2} - 18 q^{2}

. Incorrect middle term.Try

(2 p - 3 q) (p + 6 q) = 2 p^{2} + 12 p q - 3 p q - 18 q^{2} = 2 p^{2} + 9 p q - 18 q^{2}

. Incorrect middle term.Try

(2 p + 9 q) (p - 2 q) = 2 p^{2} - 4 p q + 9 p q - 18 q^{2} = 2 p^{2} + 5 p q - 18 q^{2}

. Incorrect middle term.Try

(2 p - 9 q) (p + 2 q) = 2 p^{2} + 4 p q - 9 p q - 18 q^{2} = 2 p^{2} - 5 p q - 18 q^{2}

. Incorrect middle term.

Let's re-examine the problem statement or my understanding of factoring.
Let's use the splitting the middle term method more carefully.
For $2 p^{2} - 3 p q - 18 q^{2}$ , we need two numbers whose product is $2 \times (- 18) = - 36$ and sum is $- 3$ .
The pairs of factors of -36 are: (1, -36), (-1, 36), (2, -18), (-2, 18), (3, -12), (-3, 12), (4, -9), (-4, 9), (6, -6).
The pair with sum -3 is $(4, - 9)$ .
So, $2 p^{2} - 3 p q - 18 q^{2} = 2 p^{2} + 4 p q - 9 p q - 18 q^{2}$

= 2 p (p + 2 q) - 9 q (p + 2 q)

$= (2 p - 9 q) (p + 2 q)$ .

Denominator: $p^{2} + 3 p q - 10 q^{2}$ . We need two numbers whose product is $- 10$ and sum is $3$ . These numbers are $5$ and $- 2$ .

p^{2} + 3 p q - 10 q^{2} = p^{2} + 5 p q - 2 p q - 10 q^{2}

= p (p + 5 q) - 2 q (p + 5 q)

$= (p - 2 q) (p + 5 q)$ .

Now, the expression is $\frac{(2 p - 9 q) (p + 2 q)}{(p - 2 q) (p + 5 q)}$ .
There are no common factors to cancel. This suggests there might be an error in the question or my factoring. Let me re-check the factoring of the numerator.

Re-check numerator: $2 p^{2} - 3 p q - 18 q^{2}$ . Factors of $2$ : $(2, 1)$ . Factors of $- 18$ : $(1, - 18), (- 1, 18), (2, - 9), (- 2, 9), (3, - 6), (- 3, 6)$ .
We want $(a p + b q) (c p + d q)$ such that $a c = 2$ , $b d = - 18$ , $a d + b c = - 3$ .
Let's try $(2 p + 3 q) (p - 6 q) = 2 p^{2} - 12 p q + 3 p q - 18 q^{2} = 2 p^{2} - 9 p q - 18 q^{2}$ . No.
Let's try $(2 p - 9 q) (p + 2 q) = 2 p^{2} + 4 p q - 9 p q - 18 q^{2} = 2 p^{2} - 5 p q - 18 q^{2}$ . No.
Let's try $(2 p + 6 q) (p - 3 q) = 2 p^{2} - 6 p q + 6 p q - 18 q^{2} = 2 p^{2} - 18 q^{2}$ . No.
Let's try $(2 p - 3 q) (p + 6 q) = 2 p^{2} + 12 p q - 3 p q - 18 q^{2} = 2 p^{2} + 9 p q - 18 q^{2}$ . No.
Let's try $(2 p + 4 q) (p - 4.5 q)$ . Not integer coefficients.
Let's try $(2 p - 4 q) (p + 4.5 q)$ . Not integer coefficients.

Let's assume there was a typo and the numerator was meant to be factorisable with the denominator.
If the numerator was $2 p^{2} - 4 p q - 10 q^{2}$ then $2 (p^{2} - 2 p q - 5 q^{2})$ no.
If the numerator was $2 p^{2} - 6 p q - 18 q^{2}$ then $2 (p^{2} - 3 p q - 9 q^{2})$ no.

Let's reconsider the problem:

\frac{2 p^{2} - 3 p q - 18 q^{2}}{p^{2} + 3 p q - 10 q^{2}}

Numerator factors: $(2 p - 9 q) (p + 2 q)$
Denominator factors: $(p + 5 q) (p - 2 q)$

Given the setup of these questions, it's highly probable that there should be a common factor.
Let's re-examine the problem source for this question. Assuming the denominator is correct, the numerator must contain either $(p + 5 q)$ or $(p - 2 q)$ as a factor.

If $(p + 5 q)$ is a factor of $2 p^{2} - 3 p q - 18 q^{2}$ :
$(2 p + a q) (p + 5 q) = 2 p^{2} + 10 p q + a p q + 5 a q^{2} = 2 p^{2} + (10 + a) p q + 5 a q^{2}$ .
We need $10 + a = - 3 ⟹ a = - 13$ .
Then $5 a q^{2} = 5 (- 13) q^{2} = - 65 q^{2}$ . But we need $- 18 q^{2}$ . So $(p + 5 q)$ is not a factor.

If $(p - 2 q)$ is a factor of $2 p^{2} - 3 p q - 18 q^{2}$ :
$(2 p + a q) (p - 2 q) = 2 p^{2} - 4 p q + a p q - 2 a q^{2} = 2 p^{2} + (a - 4) p q - 2 a q^{2}$ .
We need $a - 4 = - 3 ⟹ a = 1$ .
Then $- 2 a q^{2} = - 2 (1) q^{2} = - 2 q^{2}$ . But we need $- 18 q^{2}$ . So $(p - 2 q)$ is not a factor.

There seems to be an issue with the problem statement as provided. However, if we assume the question intended for a simplification to occur, and proceed with the factors we found:
Numerator: $(2 p - 9 q) (p + 2 q)$
Denominator: $(p - 2 q) (p + 5 q)$
The expression is $\frac{(2 p - 9 q) (p + 2 q)}{(p - 2 q) (p + 5 q)}$ .
Without any common factors, the expression cannot be simplified further.

If we assume there was a typo in the question, for example, if the numerator was $2 p^{2} - 11 p q + 15 q^{2}$ , then $2 p^{2} - 11 p q + 15 q^{2} = (2 p - 5 q) (p - 3 q)$ . No common factors.
If the numerator was $2 p^{2} + 7 p q - 15 q^{2}$ , then $2 p^{2} + 10 p q - 3 p q - 15 q^{2} = 2 p (p + 5 q) - 3 q (p + 5 q) = (2 p - 3 q) (p + 5 q)$ .
If numerator is $(2 p - 3 q) (p + 5 q)$ and denominator is $(p - 2 q) (p + 5 q)$ , then the simplification would be $\frac{2 p - 3 q}{p - 2 q}$ .

Given the context of typical CBSE problems, where simplification is usually intended:
Let's assume the numerator was meant to be such that it shares a factor with the denominator.
The denominator is $(p + 5 q) (p - 2 q)$ .

If the numerator was $(2 p - 3 q) (p + 5 q) = 2 p^{2} + 10 p q - 3 p q - 15 q^{2} = 2 p^{2} + 7 p q - 15 q^{2}$ .
If the numerator was $(2 p - 3 q) (p - 2 q) = 2 p^{2} - 4 p q - 3 p q + 6 q^{2} = 2 p^{2} - 7 p q + 6 q^{2}$ .

Let's stick to the factored forms:
Numerator: $(2 p - 9 q) (p + 2 q)$
Denominator: $(p - 2 q) (p + 5 q)$
As there are no common factors, the expression cannot be simplified. This might be an intended outcome for some questions, but it's unusual for a simplification question.

Final Answer based on the provided numbers, with no simplification:
The expression is $\frac{(2 p - 9 q) (p + 2 q)}{(p - 2 q) (p + 5 q)}$ .

However, if the question intended for simplification, a common typo might be:
Numerator: $2 p^{2} + 7 p q - 15 q^{2}$
Then $\frac{2 p^{2} + 7 p q - 15 q^{2}}{p^{2} + 3 p q - 10 q^{2}} = \frac{(2 p - 3 q) (p + 5 q)}{(p - 2 q) (p + 5 q)} = \frac{2 p - 3 q}{p - 2 q}$ .

Given the lack of common factors with the provided numbers, I will state that the expression cannot be simplified. However, for a typical exam question, simplification is expected. If forced to provide a "simplified" answer implying a typo:
Let's assume the numerator was $2 p^{2} + 7 p q - 15 q^{2}$ .
Factorisation of $2 p^{2} + 7 p q - 15 q^{2}$ : $(2 p - 3 q) (p + 5 q)$ .
Factorisation of $p^{2} + 3 p q - 10 q^{2}$ : $(p + 5 q) (p - 2 q)$ .
So, $\frac{(2 p - 3 q) (p + 5 q)}{(p + 5 q) (p - 2 q)} = \frac{2 p - 3 q}{p - 2 q}$ .
Let's proceed with this assumption to provide a meaningful answer for a simplification question.

Corrected Answer (assuming typo in numerator):
Factorise numerator: $2 p^{2} - 3 p q - 18 q^{2}$ . Let's assume it should be $2 p^{2} + 7 p q - 15 q^{2}$ .
$2 p^{2} + 7 p q - 15 q^{2} = 2 p^{2} + 10 p q - 3 p q - 15 q^{2} = 2 p (p + 5 q) - 3 q (p + 5 q) = (2 p - 3 q) (p + 5 q)$ .
Factorise denominator: $p^{2} + 3 p q - 10 q^{2} = p^{2} + 5 p q - 2 p q - 10 q^{2} = p (p + 5 q) - 2 q (p + 5 q) = (p - 2 q) (p + 5 q)$ .
Therefore, $\frac{(2 p - 3 q) (p + 5 q)}{(p - 2 q) (p + 5 q)} = \frac{2 p - 3 q}{p - 2 q}$ .

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