Question

x^2-(x+2)(x+3)/7x+1=2/3
give x value plz but do the verification . value i have got plz verify clearly

Answer 1

Answer:-

$\frac{ {x }^{2} - (x + 2)(x + 3) }{7x + 1} = \frac{2}{3} \\ \\ = > \frac{ {x}^{2} - ( {x}^{2} + 5x + 6) }{7x + 1} = \frac{2}{3} \\ \\ = > \frac{ {x }^{2} - {x}^{2} - 5x - 6 }{7x + 1} = \frac{2}{3} \\ \\ = > \frac{ - 5x - 6}{7x + 1} = \frac{2}{3} \\ \\ = > 3 \times (- 5x - 6) = 2 \times (7x + 1) \\ \\ = > - 15x - 18 = 14x + 2 \\ \\ = > 29x = - 20 \\ \\ = > x = \frac{ - 20}{29}$

Now verification

Putting the value of X

$= > \frac{ { \frac{ - 20}{29} }^{2} - { \frac{ - 20}{29} }^{2} - 5 \times \frac{ - 20}{29} - 6 }{7 \times \frac{ - 20}{29} + 1} = \frac{2}{3} \\ \\ = > \frac{ \frac{ 100}{29} - 6 }{ \frac{ - 140}{29} + 1} = \frac{2}{3} \\ \\ = > \frac{ \frac{100 - 174}{29} }{ \frac{ - 140 + 29}{29} } = \frac{2}{3} \\ \\ = > \frac{ \frac{ - 74}{29} }{ \frac{ - 111}{29} } = \frac{2}{3} \\ \\ = > \frac{ - 74}{ - 111} = \frac{2}{3} \\ \\ = > \frac{74}{111} = \frac{2}{3} \\ \\ = > \frac{37 \times 2}{37 \times 3} = \frac{2}{3} \\ \\ = > \frac{2}{3} = \frac{2}{3}$

Hence LHS = RHS verified

Answer 2

Given that,

The equation is

$\dfrac{x^2-(x+2)(x+3)}{7x+1}=\dfrac{2}{3}$

We need to calculate the value of x

Using given equation

$\dfrac{x^2-(x+2)(x+3)}{7x+1}=\dfrac{2}{3}$

$\dfrac{x^2-(x^2+3x+2x+6}{7x+1}=\dfrac{2}{3}$

$\dfrac{x^2-x^2-5x-6}{7x+1}=\dfrac{2}{3}$

$-15x-18=14x+2$

$-14x-15x=2+18$

$29x=-20$

$x=\dfrac{-20}{29}$

We need to verification of the value of x

Using given equation

$\dfrac{x^2-(x+2)(x+3)}{7x+1}=\dfrac{2}{3}$

Using left hand side

$=\dfrac{x^2-(x+2)(x+3)}{7x+1}$

Put the value of x

$=\dfrac{(\dfrac{-20}{29})^2-(\dfrac{-20}{29}+2)(\dfrac{-20}{29}+3)}{7\times\dfrac{-20}{29}+1}$

$=\dfrac{2}{3}$

= right hand side

The proof is that left hand side is equal to the right hand side.

Hence, The value of x is $\dfrac{-20}{29}$

That is proved.