Question

It is sum of Quadratic Equations.
It is of class 10th?

Answer 1

Answer:

2(2x-1/x+3)-3(x+3/2x-1)=5

or, {2(2x-1)²-3(x+3)²}/(x+3)(2x-1)=5

or, {2(4x²-4x+1)-3(x²+6x+9)}/(2x²+6x-x-3)=5

or, 8x²-8x+2-3x²-18x-27=5(2x²+5x-3)

or, 5x²-26x-25=10x²+25x-15

or, 5x²-10x²-26x-25x-25+15=0

or, -5x²-51x-10=0

or, 5x²+50x+x+10=0

or, 5x(x+10)+1(x+10)=0

or, (x+10)(5x+1)=0

Either, x+10=0

or, x=-10

Or, 5x+1=0

or, 5x=-1

or, x=-1/5

∴x=10,-1/5 Ans.

Step-by-step explanation:

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Answer 2

Let , ( 2x-1/x+3) be y , thus the equation becomes

$\sf\implies2y - \dfrac{3}{y} = 5$

$\sf\implies \dfrac{2 {y}^{2} - 3}{y} = 5$

$\sf\implies {2 {y}^{2} - 3} = 5y$

$\implies \sf 2 {y}^{2} - 5y - 3 = 0$

Now Factorising it by middle term splitting method

$\implies \sf \: 2 {y}^{2} - 6y + y - 3 = 0$

$\implies \sf \: 2y(y - 3) + 1(y - 3) = 0$

$\implies \sf \: (2y + 1)(y - 3) = 0$

$\implies \sf \: y = \dfrac{ - 1}{2} \:and \: y = 3$

Now , we have

$\implies \sf \: \dfrac{2x - 1}{x + 3} = y$

$\implies \sf \: \dfrac{2x - 1}{x + 3} = \dfrac{ - 1}{2}$

$\implies \sf \: 2 \times ({2x - 1} )= - 1 \times (x + 3)$

$\sf \implies \: 4x - 2 = - x - 3$

$\implies \sf \: 5x = -1$

$\implies \sf \: x = \dfrac{-1}{5}$

Also ,

$\implies \sf \: \dfrac{2x - 1}{x + 3} = y$

$\implies \sf \: \dfrac{2x - 1}{x + 3} = 3$

$\sf \implies 2x - 1 = 3x + 9$

$\sf \implies \: x = - 10$

Thus . x=-10 and x=-1/5 is the required value

$\rule{300}{3}$