Factorisation — Test Yourself

$\dfrac{(a + b)(a^2 - ab + b^2)}{(a^2b + ab^2)}$ in simplest form is equal to:

1. $\dfrac{a}{b} - 1 + \dfrac{b}{a}$

2. $\dfrac{a^3 - b^3}{a^2b + ab^2}$

3. $\dfrac{a^2 + b^2}{ab}$

4. none of these

Answer

Given,

$\Rightarrow \dfrac{(a + b)(a^2 - ab + b^2)}{(a^2b + ab^2)}\\[1em] \Rightarrow \dfrac{(a + b)(a^2 - ab + b^2)}{ab(a + b)}\\[1em] \Rightarrow \dfrac{(a^2 - ab + b^2)}{ab}\\[1em] \Rightarrow \dfrac{a^2}{ab} - \dfrac{ab}{ab} + \dfrac{b^2}{ab} \\[1em] \Rightarrow \dfrac{a}{b} - 1 + \dfrac{b}{a}$

Hence, option 1 is the correct option.

L.C.M. of x² + 3x + 2 and x² - 2x - 3 in simplest form is:

1. (x + 1)²(x + 2)(x - 3)

2. (x² + 3x + 2)(x² - 2x - 3)

3. (x + 1)(x + 2)(x - 3)

4. none of these

Answer

Given, x² + 3x + 2 and x² - 2x - 3

The factors of x² + 3x + 2

⇒ x² + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 2)(x + 1).

The factors of x² - 2x - 3

⇒ x² - 3x + x - 3

⇒ x(x - 3) + 1(x - 3)

⇒ (x - 3)(x + 1)

L.C.M. = (x + 1)(x + 2)(x - 3)

Hence, option 3 is the correct option.

H.C.F of x² + 3x + 2 and x² - 2x - 3 is :

1. (x + 1)

2. (x + 1)(x + 2)(x - 3)

3. 1

4. none of these

Answer

Given, x² + 3x + 2 and x² - 2x - 3

The factors of x² + 3x + 2

⇒ x² + 2x + x + 2

⇒ x(x + 2) + 1(x + 2)

⇒ (x + 2)(x + 1)

The factors of x² - 2x - 3

⇒ x² - 3x + x - 3

⇒ x(x - 3) + 1(x - 3)

⇒ (x - 3)(x + 1)

H.C.F. = (x + 1)

Hence, option 1 is the correct option.

(3a - 1)² - 6a + 2 is equal to:

1. (3a - 1)(a - 1)

2. 3(3a - 1)(a - 1)

3. (3a - 1)(a + 1)

4. 3(3a - 1)(a + 1)

Answer

Solving,

⇒ (3a - 1)² - 6a + 2

⇒ (3a)² + 1² - 2 x 3a x 1 - 6a + 2

⇒ 9a² + 1 - 6a - 6a + 2

⇒ 9a² - 12a + 3

⇒ 3(3a² - 4a + 1)

⇒ 3(3a² - 3a - a + 1)

⇒ 3[3a(a - 1) - 1(a - 1)]

⇒ 3(3a - 1)(a - 1).

Hence, option 2 is the correct option.

$\dfrac{1}{3}$ x² - 2x - 9 is equal to:

1. $\dfrac{1}{3}$ (x - 9)(x + 3)

2. $\dfrac{1}{3}$ (x - 9)(x - 3)

3. $\dfrac{1}{3}$ (x + 9)(x - 3)

4. $\dfrac{1}{3}$ (x + 9)(x + 3)

Answer

Given,

$\dfrac{1}{3} x^2 - 2x - 9\\[1em] \Rightarrow \dfrac{1}{3} x^2 - \dfrac{6}{3}x - \dfrac{27}{3}\\[1em] \Rightarrow \dfrac{1}{3} (x^2 - 6x - 27)\\[1em] \Rightarrow \dfrac{1}{3} (x^2 - 9x + 3x - 27)\\[1em] \Rightarrow \dfrac{1}{3} [x(x - 9) + 3(x - 9)]\\[1em] \Rightarrow \dfrac{1}{3} (x - 9)(x + 3)$

Hence, option 1 is the correct option.

(x² + 3x) men can do a piece of work in (x² - 2x) days, then one day work of 1 man is :

1. $\dfrac{x + 3}{x - 2}$

2. (x² + 3x)(x² - 2x)

3. $\dfrac{x - 2}{x + 3}$

4. none of these

Answer

Given, total number of men = (x² + 3x)

Total number of days = (x² - 2x)

Total work = (x² + 3x)(x² - 2x)

One day work of 1 man = $\dfrac{1}{(x^2 + 3x)(x^2 - 2x)}$

Hence, option 4 is the correct option.

Statement 1: ₹ (x³ - x) is spent in buying some identical articles at ₹ (x - 1) each. Number of articles bought = $\dfrac{x^3 - x}{x - 1}$.

Statement 2: The number of articles bought

= $\dfrac{x^3 - x}{x - 1} = \dfrac{x(x - 1)(x + 1)}{x - 1} = x^2 + x$

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Given, cost of each article = ₹ (x - 1)

Total cost = ₹ (x³ - x)

$\Rightarrow \text{No. of articles bought} = \dfrac{\text{Total cost}}{\text{Cost of each article}} \\[1em] = \dfrac{x^3 - x}{x - 1}\\[1em] = \dfrac{x(x^2 - 1)}{x - 1}\\[1em] = \dfrac{x(x^2 - 1^2)}{x - 1}\\[1em] = \dfrac{x(x - 1)(x + 1)}{x - 1}\\[1em] = x(x + 1)\\[1em] = x^2 + x.$

∴ Both the statements are true.

Hence, option 1 is the correct option.

Statement 1: The area of rectangle is x² - 5x + 6 and the longer side of the rectangle is (x - 2).

Statement 2: x² - 5x + 6

= (x - 2) (x - 3)

⇒ for every positive value of x(x > 3), (x - 2) is greater.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Give, Area of rectangle = x² - 5x + 6

Longer side = (x - 2)

Factorise the area,

⇒ x² - 5x + 6 = 0

⇒ x² - 3x - 2x + 6 = 0

⇒ x(x - 3) - 2(x - 3) = 0

⇒ (x - 2) (x - 3) = 0

So, the given two sides of rectangle are:

(x - 2) (x - 3)

Since x > 3,

∴ (x - 2) > (x - 3)

So, (x - 2) is the longer side.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): Distance of (x² - 7x + 12) km is covered in (x² - 16) hrs.

Speed = $\dfrac{x^2 - 7x + 12}{x^2 - 16}$ km/hr

Reason (R): Speed = Distance x Time

= (x² - 7x + 12)(x² - 16) km/hr

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true, and R is the correct reason for A.

4. Both A and R are true, and R is the incorrect reason for A.

Answer

Given,

Distance = (x² - 7x + 12) km

Time = (x² - 16) hrs

By formula,

Speed = $\dfrac{\text{Distance}}{\text{Time}}$

= $\dfrac{x^2 - 7x + 12}{x^2 - 16}$ km/hr

∴ A is true, but R is false.

Hence, option 1 is the correct option.

Factorise :

$a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a}$

Answer

Given,

$\phantom{=}a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a} \\[1em] = \Big(a - \dfrac{1}{a}\Big)^2 - 3\Big(a - \dfrac{1}{a}\Big) \\[1em] = \Big(a - \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a} - 3\Big).$

Hence, $a^2 + \dfrac{1}{a^2} - 2 - 3a + \dfrac{3}{a} = \Big(a - \dfrac{1}{a}\Big)\Big(a - \dfrac{1}{a} - 3\Big).$

x² + y² + x + y + 2xy

Answer

Given,

    x² + y² + x + y + 2xy

= x² + y² + 2xy + x + y

= (x + y)² + (x + y)

= (x + y)(x + y + 1).

Hence, x² + y² + x + y + 2xy = (x + y)(x + y + 1).

a² + 4b² - 3a + 6b - 4ab

Answer

Given,

    a² + 4b² - 3a + 6b - 4ab

= a² + 4b² - 4ab - 3a + 6b

= a² + (2b)² - 2 × a × 2b - 3a + 6b

= (a - 2b)² - 3(a - 2b)

= (a - 2b)(a - 2b - 3).

Hence, a² + 4b² - 3a + 6b - 4ab =(a - 2b)(a - 2b - 3).

m(x - 3y)² + n(3y - x) + 5x - 15y

Answer

Given,

    m(x - 3y)² + n(3y - x) + 5x - 15y

= m(x - 3y)² - n(x - 3y) + 5(x - 3y)

= (x - 3y)[m(x - 3y) - n + 5]

= (x - 3y)(mx - 3my - n + 5).

Hence, m(x - 3y)² + n(3y - x) + 5x - 15y = (x - 3y)(mx - 3my - n + 5).

x(6x - 5y) - 4(6x - 5y)²

Answer

Given,

    x(6x - 5y) - 4(6x - 5y)²

= (6x - 5y)[x - 4(6x - 5y)]

= (6x - 5y)(x - 24x + 20y)

= (6x - 5y)(20y - 23x).

Hence, x(6x - 5y) - 4(6x - 5y)² = (6x - 5y)(20y - 23x).

$\dfrac{1}{35} + \dfrac{12}{35}a + a^2$

Answer

Given,

$\phantom{=}\dfrac{1}{35} + \dfrac{12}{35}a + a^2 \\[1em] = a^2 + \dfrac{12}{35}a + \dfrac{1}{35} \\[1em] = \dfrac{35a^2 + 12a + 1}{35} \\[1em] = \dfrac{35a^2 + 5a + 7a + 1}{35} \\[1em] = \dfrac{5a(7a + 1) + 1(7a + 1)}{35} \\[1em] = \dfrac{(7a + 1)(5a + 1)}{35}.$

Hence, $\dfrac{1}{35} + \dfrac{12}{35}a + a^2 = \dfrac{(7a + 1)(5a + 1)}{35}$.

(x² - 3x)(x² - 3x - 1) - 20

Answer

Given,

    (x² - 3x)(x² - 3x - 1) - 20

Substituting x² - 3x = a, we get :

⇒ a(a - 1) - 20

= a² - a - 20

= a² - 5a + 4a - 20

= a(a - 5) + 4(a - 5)

= (a - 5)(a + 4)

= (x² - 3x - 5)(x² - 3x + 4).

Hence, (x² - 3x)(x² - 3x - 1) - 20 = (x² - 3x - 5)(x² - 3x + 4).

For each trinomial (quadratic expression), given below, find whether it is factorisable or not. Factorise, if possible.

(i) x² - 3x - 54

(ii) 2x² - 7x - 15

(iii) 2x² + 2x - 75

(iv) 3x² + 4x - 10

(v) x(2x - 1) - 1

Answer

(i) Given,

    x² - 3x - 54

= x² - 9x + 6x - 54

= x(x - 9) + 6(x - 9)

= (x - 9)(x + 6).

Hence, the above equation is factorisable and x² - 3x - 54 = (x - 9)(x + 6).

(ii) Given,

    2x² - 7x - 15

= 2x² - 10x + 3x - 15

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

= (2x + 3)(x - 5).

Hence, the above equation is factorisable and 2x² - 7x - 15 = (2x + 3)(x - 5).

(iii) Given,

    2x² + 2x - 75

Hence, the above equation is not factorisable.

(iv) Given,

    3x² + 4x - 10

Hence, the above equation is not factorisable.

(v) Given,

    x(2x - 1) - 1

= 2x² - x - 1

= 2x² - 2x + x - 1

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

= (x - 1)(2x + 1).

Hence, the above equation is factorisable and x(2x - 1) - 1 = (x - 1)(2x + 1).

Factorise :

(i) $4\sqrt{3}x^2 + 5x - 2\sqrt{3}$

(ii) $7\sqrt{2}x^2 - 10x - 4\sqrt{2}$

Answer

(i) Given,

$\phantom{=}4\sqrt{3}x^2 + 5x - 2\sqrt{3} \\[1em] = 4\sqrt{3}x^2 + 8x - 3x - 2\sqrt{3}\\[1em] = 4x(\sqrt{3}x + 2) - \sqrt{3}(\sqrt{3}x + 2) \\[1em] = (\sqrt{3}x + 2)(4x - \sqrt{3}).$

Hence, $4\sqrt{3}x^2 + 5x - 2\sqrt{3} = (\sqrt{3}x + 2)(4x - \sqrt{3}).$

(ii) Given,

$\phantom{=}7\sqrt{2}x^2 - 10x - 4\sqrt{2} \\[1em] = 7\sqrt{2}x^2 - 14x + 4x - 4\sqrt{2}\\[1em] = 7\sqrt{2}x(x - \sqrt{2}) + 4(x - \sqrt{2}) \\[1em] = (x - \sqrt{2})(7\sqrt{2}x + 4).$

Hence, $7\sqrt{2}x^2 - 10x - 4\sqrt{2} = (x - \sqrt{2})(7\sqrt{2}x + 4).$

Give possible expressions for the length and the breadth of the rectangle whose area is

12x² - 35x + 25.

Answer

Given,

Area = 12x² - 35x + 25

⇒ lb = 12x² - 35x + 25

⇒ lb = 12x² - 15x - 20x + 25

⇒ lb = 3x(4x - 5) - 5(4x - 5)

⇒ lb = (4x - 5)(3x - 5).

Hence, if length = (4x - 5) then breadth = (3x - 5) and if length = (3x - 5) then breadth = (4x - 5).

Factorise :

9a² - (a² - 4)²

Answer

Given,

    9a² - (a² - 4)²

= (3a)² - (a² - 4)²

= (3a + a² - 4)[3a - (a² - 4)]

= (a² + 3a - 4)(4 - a² + 3a)

= (a² + 4a - a - 4).-(a² - 3a - 4)

= [a(a + 4) - 1(a + 4)].-[a² - 4a + a - 4]

= (a + 4)(a - 1).-[a(a - 4) + 1(a - 4)]

= (a + 4)(a - 1).-(a - 4)(a + 1)

= (a + 4)(a - 1)(4 - a)(a + 1).

Hence, 9a² - (a² - 4)² = (a + 4)(a - 1)(4 - a)(a + 1).

$x^2 + \dfrac{1}{x^2} - 11$

Answer

Given,

$\phantom{=} x^2 + \dfrac{1}{x^2} - 11 \\[1em] = x^2 + \dfrac{1}{x^2} - 2 - 9 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 9 \\[1em] = \Big(x - \dfrac{1}{x}\Big)^2 - 3^2 \\[1em] = \Big(x - \dfrac{1}{x} + 3\Big)\Big(x - \dfrac{1}{x} - 3\Big).$

Hence, $x^2 + \dfrac{1}{x^2} - 11 = \Big(x - \dfrac{1}{x} + 3\Big)\Big(x - \dfrac{1}{x} - 3\Big).$

$4x^2 + \dfrac{1}{4x^2} + 1$

Answer

Given,

$\phantom{=} 4x^2 + \dfrac{1}{4x^2} + 1 \\[1em] = 4x^2 + \dfrac{1}{4x^2} + 2 - 1 \\[1em] = (2x)^2 + \Big(\dfrac{1}{2x}\Big)^2 + 2 \times 2x \times \dfrac{1}{2x} - 1 \\[1em] = \Big(2x + \dfrac{1}{2x}\Big)^2 - 1^2 \\[1em] = \Big(2x + \dfrac{1}{2x} + 1\Big)\Big(2x + \dfrac{1}{2x} - 1\Big).$

Hence, $4x^2 + \dfrac{1}{4x^2} + 1 = \Big(2x + \dfrac{1}{2x} + 1\Big)\Big(2x + \dfrac{1}{2x} - 1\Big).$

4x⁴ - x² - 12x - 36

Answer

Given,

    4x⁴ - x² - 12x - 36

= 4x⁴ - [x² + 12x + 36]

= 4x⁴ - [x² + 6x + 6x + 36]

= 4x⁴ - [x(x + 6) + 6(x + 6)]

= 4x⁴ - (x + 6)(x + 6)

= (2x²)² - (x + 6)²

= (2x² + x + 6)(2x² - x - 6).

= (2x² + x + 6)(2x² - 4x + 3x - 6)

= (2x² + x + 6)[2x(x - 2) + 3(x - 2)]

= (2x² + x + 6)(x - 2)(2x + 3).

Hence, 4x⁴ - x² - 12x - 36 = (2x² + x + 6)(x - 2)(2x + 3).

a²(b + c) - (b + c)³

Answer

Given,

    a²(b + c) - (b + c)³

= (b + c)[a² - (b + c)²]

= (b + c)(a + b + c)[a - (b + c)]

= (b + c)(a + b + c)(a - b - c).

Hence, a²(b + c) - (b + c)³ = (b + c)(a + b + c)(a - b - c).

2x³ + 54y³ - 4x - 12y

Answer

Given,

    2x³ + 54y³ - 4x - 12y

= 2(x³ + 27y³) - 4(x + 3y)

= 2[(x)³ + (3y)³] - 4(x + 3y)

= 2(x + 3y)(x² - 3xy + 9y²) - 4(x + 3y) [∵ a³ + b³ = (a + b)(a² - ab + b²)]

= 2(x + 3y)(x² - 3xy + 9y² - 2).

Hence, 2x³ + 54y³ - 4x - 12y = 2(x + 3y)(x² - 3xy + 9y² - 2).

1029 - 3x³

Answer

Given,

    1029 - 3x³

= 3(343 - x³)

= 3[(7)³ - (x)³]

= 3(7 - x)[(7)² + 7x + x²] [∵ a³ - b³ = (a - b)(a² + ab + b²)]

= 3(7 - x)(x² + 7x + 49).

Hence, 1029 - 3x³ = 3(7 - x)(x² + 7x + 49).

Show that :

(i) 13³ - 5³ is divisible by 8.

(ii) 35³ + 27³ is divisible by 62.

Answer

(i) We know that

a³ - b³ = (a - b)(a² + ab + b²)

Factorising 13³ - 5³, we get :

⇒ 13³ - 5³ = (13 - 5)[13² + 13 × 5 + 5²]

= 8(169 + 65 + 25)

= 8 × 259, which is divisible by 8.

Hence, proved that 13³ - 5³ is divisible by 8.

(ii) We know that

a³ + b³ = (a + b)(a² - ab + b²)

Factorising 35³ + 27³, we get :

⇒ 35³ + 27³ = (35 + 27)[(35)² - 35 × 27 + (27)²]

= 62[1225 - 945 + 729]

= 62 × 1009, which is divisible by 62.

Hence, proved that 35³ + 27³ is divisible by 62.

Evaluate :

$\dfrac{5.67 \times 5.67 \times 5.67 + 4.33 \times 4.33 \times 4.33}{5.67 \times 5.67 - 5.67 \times 4.33 + 4.33 \times 4.33}$

Answer

Substituting a = 5.67 and b = 4.33, we get :

$\Rightarrow \dfrac{a \times a \times a + b \times b \times b}{a \times a - a \times b + b \times b} \\[1em] \Rightarrow \dfrac{a^3 + b^3}{a^2 - ab + b^2} \\[1em] \Rightarrow \dfrac{(a + b)(a^2 - ab + b^2)}{a^2 - ab + b^2} \\[1em] \Rightarrow (a + b) \\[1em] \Rightarrow (5.67 + 4.33) \\[1em] \Rightarrow 10.$

Hence, $\dfrac{5.67 \times 5.67 \times 5.67 + 4.33 \times 4.33 \times 4.33}{5.67 \times 5.67 - 5.67 \times 4.33 + 4.33 \times 4.33} = 10$.

9x² + 3x - 8y - 64y²

Answer

Given,

    9x² + 3x - 8y - 64y²

= 9x² - 64y² + 3x - 8y

= (3x)² - (8y)² + 3x - 8y

= (3x + 8y)(3x - 8y) + (3x - 8y)

= (3x - 8y)(3x + 8y + 1).

Hence, 9x² + 3x - 8y - 64y² = (3x - 8y)(3x + 8y + 1).

$2\sqrt{3}x^2 + x - 5\sqrt{3}$

Answer

Given,

$\phantom{=} 2\sqrt{3}x^2 + x - 5\sqrt{3} \\[1em] = 2\sqrt{3}x^2 + 6x - 5x - 5\sqrt{3} \\[1em] = 2\sqrt{3}x(x + \sqrt{3}) - 5(x + \sqrt{3}) \\[1em] = (x + \sqrt{3})(2\sqrt{3}x - 5).$

Hence, $2\sqrt{3}x^2 + x - 5\sqrt{3} = (x + \sqrt{3})(2\sqrt{3}x - 5).$

$\dfrac{1}{4}(a + b)^2 - \dfrac{9}{16}(2a - b)^2$

Answer

Given,

$\phantom{=}\dfrac{1}{4}(a + b)^2 - \dfrac{9}{16}(2a - b)^2 \\[1em] = \Big[\dfrac{1}{2}(a + b)\Big]^2 - \Big[\dfrac{3}{4}(2a - b)\Big]^2 \\[1em] = \Big[\dfrac{1}{2}(a + b) + \dfrac{3}{4}(2a - b)\Big]\Big[\dfrac{1}{2}(a + b) - \dfrac{3}{4}(2a - b)\Big] \\[1em] = \Big[\dfrac{2(a + b) + 3(2a - b)}{4}\Big]\Big[\dfrac{2(a + b) - 3(2a - b)}{4}\Big] \\[1em] = \Big(\dfrac{2a + 2b + 6a - 3b}{4}\Big)\Big(\dfrac{2a + 2b - 6a + 3b}{4}\Big)\\[1em] = \dfrac{8a - b}{4} \times \dfrac{5b - 4a}{4} \\[1em] = \dfrac{1}{16}(8a - b)(5b - 4a).$

Hence, $\dfrac{1}{4}(a + b)^2 - \dfrac{9}{16}(2a - b)^2 = \dfrac{1}{16}(8a - b)(5b - 4a).$

2(ab + cd) - a² - b² + c² + d²

Answer

Given,

    2(ab + cd) - a² - b² + c² + d²

= 2ab + 2cd - a² - b² + c² + d²

= c² + d² + 2cd - (a² + b² - 2ab)

= (c + d)² - (a - b)²

= (c + d + a - b)[c + d - (a - b)]

= (c + d + a - b)(c + d - a + b).

Hence, 2(ab + cd) - a² - b² + c² + d² = (c + d + a - b)(c + d - a + b).

a² + 5a + (3 - b) (2 + b)

Answer

Given,

a² + 5a + (3 - b)(2 + b)

= a² + 5a + 6 + 3b - 2b - b²

= a² + 5a + 6 + b - b²

= (a + b + 2)(a - b + 3).

Hence, a² + 5a + (3 - b) (2 + b) = (a + b + 2) (a - b + 3).

$\dfrac{x}{y} + \dfrac{y}{x} + \dfrac{y}{z} + \dfrac{z}{y} + \dfrac{x}{z} + \dfrac{z}{x} + 3$

Answer

Given,

$\dfrac{x}{y} + \dfrac{y}{x} + \dfrac{y}{z} + \dfrac{z}{y} + \dfrac{x}{z} + \dfrac{z}{x} + 3$

Multiplying the given expression by xyz,

= xyz × $\Big(\dfrac{x}{y} + \dfrac{y}{x} + \dfrac{y}{z} + \dfrac{z}{y} + \dfrac{x}{z} + \dfrac{z}{x} + 3)$

= x²z + y²z + y²x + z²x + x²y + z²y + 3xyz

= x²y + xy² + y²z + yz² + z²x + zx² + 3xyz

= (x + y + z)(xy + yz + zx)

Divide by xyz

= $\dfrac{(x + y + z)(xy + yz + zx)}{xyz}$

= $(x + y + z) \times \dfrac{xy + yz + zx}{xyz}$

= (x + y + z) $\Big(\dfrac{1}{z} + \dfrac{1}{x} + \dfrac{1}{y}\Big)$

Hence, $\dfrac{x}{y} + \dfrac{y}{x} + \dfrac{y}{z} + \dfrac{z}{y} + \dfrac{x}{z} + \dfrac{z}{x} + 3 = (x + y + z)\Big(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z}\Big)$.