Question

$\frac{(4x + 7)(3x + 5)}{(4x + 2)(3x + 4)} = 1$
solve this​

Answer 1

Given

$\rm \to \dfrac{(4x + 7)(3x + 5)}{(4x + 2)(3x + 4)} = 1$

To find the value of x

Now Take

$\rm \to \dfrac{(4x + 7)(3x + 5)}{(4x + 2)(3x + 4)} = 1$

Now Simplify the equation

$\rm \to \: \dfrac{12 {x}^{2} + 20x + 21x + 35}{12 {x}^{2} + 16x + 6x + 8} = 1$

$\rm \to \: \dfrac{12 {x}^{2} + 41x + 35}{12 {x}^{2} + 22x + 8} = 1$

Now Using Cross multiplication method

$\rm \to \: 12 {x}^{2} + 41x + 35 = 1(12 {x}^{2} + 22x + 8)$

$\rm \to \: 12 {x}^{2} + 41x + 35 = 12 {x}^{2} + 22x + 8$

$\rm \to \: 12 {x}^{2} - 12 {x}^{2} + 41x - 22x+ 35 - 8= 0$

$\rm \to \:0 + 19x + 27 = 0$

$\rm \to \: 19x + 27 = 0$

$\rm \to \: 19x = - 27$

$\rm \to \: x = \dfrac{ - 27}{19}$

Answer

$\rm \to \: x = \dfrac{ - 27}{19}$

Answer 2

Answer:

• x = -27 / 19

Step-by-step explanation:

Given

• (4x + 7) (3x + 5 / (4x + 2) (3x + 4) = 1

To find

• x

Solution

(4x + 7) (3x + 5) = (4x)(3x) + (4x)(5) + (7)(3x) + (7)(5)

• 12x² + 20x + 21x + 35
• 12x² + 41x + 35

(4x + 2) (3x + 4) = (4x)(3x) + (4x)(4) + (2)(3x) + (2)(4)

• 12x² + 16x + 6x + 8
• 12x² + 22x + 8

Now, after multiplying we get:

• 12x² + 41x + 35 / 12x² + 22x + 8 = 1

Cross multiplication:

• 12x² + 41x + 35 = 1 (12x² + 22x + 8)
• 12x² + 41x + 35 = 12x² + 22x + 8
• 12x² - 12x² + 41x - 22x + 35 - 8 = 0
• 19x + 27 = 0
• 19x = -27
• x = -27 / 19

Hence, the value of x is -27 / 19