How do you solve $\frac{1}{2 x + 3} + \frac{1}{4 x - 3} = 5$ $1/ (2x+ 3) + 1/ (4x-3) = 5$ ?

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How do you solve $\frac{1}{2 x + 3} + \frac{1}{4 x - 3} = 5$ $1/ (2x+ 3) + 1/ (4x-3) = 5$ ?

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Answer 1 · 2017-09-28T16:20:01

$x = \frac{- 3 \pm 3 \sqrt{6}}{10}$ $x=(-3+-3sqrt(6))/10$

Explanation:

$\frac{1}{2 x + 3} + \frac{1}{4 x - 3} = 5$ $1/(2x+3)+1/(4x-3)=5$
Take LCM of denominators $(2 x + 3) (4 x - 3)$ $(2x+3)(4x-3)$
Multiply both sides by $(2 x + 3) (4 x - 3)$ $(2x+3)(4x-3)$
$(2 x + 3) (4 x - 3) (\frac{1}{2 x + 3} + \frac{1}{4 x - 3}) = 5 (2 x + 3) (4 x - 3)$ $(2x+3)(4x-3)(1/(2x+3)+1/(4x-3))=5(2x+3)(4x-3)$
$\frac{(2 x + 3) (4 x - 3)}{2 x + 3} + \frac{(2 x + 3) (4 x - 3)}{4 x - 3} = 5 (2 x + 3) (4 x - 3)$ $((2x+3)(4x-3))/(2x+3)+((2x+3)(4x-3))/(4x-3)=5(2x+3)(4x-3)$

Cancel Equal like terms
$(4 x - 3) + (2 x + 3) = 5 (2 x + 3) (4 x - 3)$ $(4x-3)+(2x+3)=5(2x+3)(4x-3)$
$6 x = 5 (8 x^{2} - 6 x + 12 x - 9)$ $6x=5(8x^2-6x+12x-9)$
$6 x = 5 (8 x^{2} + 6 x - 9)$ $6x=5(8x^2+6x-9)$
$6 x = 40 x^{2} + 30 x - 45$ $6x=40x^2+30x-45$
Subtract $6 x$ $6x$ from both side
$6 x - 6 x = 40 x^{2} + 30 x - 45 - 6 x$ $6x-6x=40x^2+30x-45-6x$
$0 = 40 x^{2} + 24 x - 45$ $0=40x^2+24x-45$

$40 x^{2} + 24 x - 45 = 0$ $40x^2+24x-45=0$
comparing with $a x^{2} + b x + c = 0$ $ax^2+bx+c=0$
$a = 40, b = 24, c = - 45$ $a=40, b=24, c=-45$
Solution $x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $x=(-b+-sqrt(b^2-4ac))/(2a)$
$x = \frac{- 24 \pm \sqrt{{(24)}^{2} - 4 (40) (- 45)}}{2 (40)}$ $x=(-24+-sqrt((24)^2-4(40)(-45)))/(2(40))$
$x = \frac{- 24 \pm \sqrt{576 + 7200}}{80}$ $x=(-24+-sqrt(576+7200))/80$
$x = \frac{- 24 \pm \sqrt{7776}}{80}$ $x=(-24+-sqrt(7776))/80$
$x = \frac{- 24 \pm \sqrt{1296 \times 6}}{80}$ $x=(-24+-sqrt(1296xx6))/80$
$x = \frac{- 24 \pm 36 \sqrt{6}}{80}$ $x=(-24+-36sqrt(6))/80$
$x = \frac{- 3 \pm 3 \sqrt{6}}{10}$ $x=(-3+-3sqrt(6))/10$

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How do you solve $\frac{1}{2 x + 3} + \frac{1}{4 x - 3} = 5$ $1/ (2x+ 3) + 1/ (4x-3) = 5$ ?

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How do you solve $\frac{1}{2 x + 3} + \frac{1}{4 x - 3} = 5$ $1/ (2x+ 3) + 1/ (4x-3) = 5$ ?