Question

questions of the algebra
please give correct answer​

Answer 1

Given : x² - 2x/3   = 32

x² + 4x/15  = 1/5

To Find : Solve for x

Solution:

x² - 2x/3   = 32

Multiplying both sides by 3

=> 3x² - 2x  = 96

Subtracting 96 both sides

=> 3x² - 2x  - 96 = 0

Split middle term

=> 3x² - 18x + 16x - 96 = 0

=> 3x( x - 6) + 16(x - 6) = 0

=> (3x + 16)(x - 6) = 0

=> x = -16/3  or  x = 6

x² + 4x/15  = 1/5

Multiplying both sides by 15

=> 15x²+ 4x  = 3

Subtracting 3 both sides

=> 15x²+ 4x - 3 = 0

Split middle term

=>15x² -5x + 9x - 3 = 0

=> 5x( 3x - 1) + 3(3x - 1) = 0

=> (3x - 1)(5x +3) = 0

=> x =  1 /3  or  x = -3/5

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Answer 2

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$\bf \:\large \red{AηsωeR : 19.}$

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$\tt \:  \longrightarrow \: {x}^{2} - \dfrac{2}{3} x = 32$

$\tt \:  \longrightarrow \: \dfrac{3 {x}^{2} - 2x }{3} = 32$

$\tt \:  \longrightarrow \: 3 {x}^{2} - 2x = 96$

$\tt \:  \longrightarrow \: 3 {x}^{2} - 2x - 96 = 0$

$\tt \:  \longrightarrow \: 3 {x}^{2} - 18x + 16x - 96 = 0$

$\tt \:  \longrightarrow \: 3x(x - 6) + 16(x - 6) = 0$

$\tt \:  \longrightarrow \: (3x + 16)(x - 6) = 0$

$\tt \:  \longrightarrow \: \boxed{ \red{ \bf \:x \: = \: - \dfrac{16}{3} \: or \: x \: = \: 6 }}$

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$\bf \:\large \red{AηsωeR : 20}$

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$\tt \:  \longrightarrow \: {x}^{2} + \dfrac{4}{15} x = \dfrac{1}{5}$

$\tt \:  \longrightarrow \: \dfrac{15 {x}^{2} + 4x }{15} = \dfrac{1}{5}$

$\tt \:  \longrightarrow \: {15x}^{2} + 4x = 3$

$\tt \:  \longrightarrow \: 15 {x}^{2} + 4x - 3 = 0$

$\tt \:  \longrightarrow \: {15x}^{2} + 9x - 5x - 3 = 0$

$\tt \:  \longrightarrow \: 3x(5x + 3) - 1(5x + 3) = 0$

$\tt \:  \longrightarrow \: (5x + 3)(3x - 1) = 0$

$\tt\implies \: \boxed{ \red{ \bf \: x \: = \: - \dfrac{3}{5} \: or \: x \: = \: \dfrac{1}{3} }}$

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