Question

2 into bracket start 2 x minus one upon X + 3 bracket close - 3 into X + 3 upon 2 x minus 1 bracket close is equal to 5

Answer 1

2(2 ×-1/x+3) -3x+3/2×-1)=5
2(-2/x+3)-3x-3/2)=5
-4/x+6-3x-3/2=5
6-5-3/2= 3x-4/x
-1/2=(3x^2-4)/x
-x/2=3x^2-4
-x=6x^2-8
0=6x^2+x-8
0= 6x^2+

Answer 2

Answer:

$x = \frac{-1}{5}\: Or \: x = -10$

Step-by-step explanation:

$Given \\ 2\big(\frac{2x-1}{x+3}\big)-3\big(\frac{x+3}{2x-1}\big)=5$

$\implies \big(\frac{4x-2}{x+3}\big)-\big(\frac{3x+9}{2x-1}\big)=5$

$\implies \frac{(2x-1)(4x-2)-(x+3)(3x+9)}{(x+3)(2x-1)}=5$

$\implies \frac{8x^{2}-4x-4x+2-(3x^{2}+9x+9x+27)}{2x^{2}-x+6x-3}=5$

$\implies \frac{8x^{2}-8x+2-3x^{2}-18x-27}{2x^{2}+5x-3}=5$

$\implies 5x^{2}-26x-25=5(2x^{2}+5x-3)$

$\implies 5x^{2}-26x-25=10x^{2}+25x-15$

$\implies 0=10x^{2}+25x-15-5x^{2}+26x+25$

$\implies 5x^{2}+51x+10=0$

Splitting the middle term,we get

$\implies 5x^{2}+1x+50x+10=0$

$\implies x(5x+1)+10(5x+1)=0$

$\implies (5x+1)(x+10)=0$

$\implies 5x+1=0 \: Or \: x+10=0$

$\implies 5x=-1 \: Or \: x = -10$

$\implies x = \frac{-1}{5}\: Or \: x = -10$

Therefore,

$x = \frac{-1}{5}\: Or \: x = -10$

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