Question

plz help me
simplify this

Answer 1

Answer:

Step-by-step explanation:

$\frac{1}{x+2} +\frac{1}{x+3} =\frac{7}{12} \\\\Taking LCM,\\\\\frac{x + 3 + x +2}{(x + 2)(x + 3)} =\frac{7}{12} \\\\\frac{2x+5}{(x + 2)(x + 3)} =\frac{7}{12}\\\\12(2x+5)=7(x+2)(x+3)\\\\24x+60=(7x+14)(x+3)\\\\24x+60=7x^{2} +21x+14x+52\\\\0=7x^{2} +21x+14x+52-24x-60\\\\7x^{2} +11x-8=0\\\\Quadratic formula,$

The further solution in the attachment

Answer 2

Question :-

$\frac{1}{x + 2} + \frac{1}{x + 3} = \frac{7}{12}$

To Find :-

The value of x.

Process :-

Given that,

$\frac{1}{x + 2} + \frac{1}{x + 3} = \frac{7}{12}$

By doing cross multiplication on LHS,

$\frac{x + 3 + x + 2}{(x + 2)(x + 3)} = \frac{7}{12}$

$\frac{2x + 6}{ {x}^{2} + 3x + 2x + 6 } = \frac{7}{12}$

$\frac{2x + 6}{ {x}^{2} + 5x + 6 } = \frac{7}{12}$

By doing cross multiplication on Both LHS and RHS,

$24x + 72 = 7 {x}^{2} + 35x + 42$

$7 {x}^{2} + 35x - 24x + 42 - 72 = 0$

$7 {x}^{2} - 10x + 21x - 30 = 0$

$(7x - 10)( x + 3) = 0$

Therefore,

$7x - 10 = 0 \: \: \: \:and \: \: \: x + 3 = 0$

we have two solutions for this equation.

they are,

10/7, -3.

HOPE IT HELPS