Question

show that 2/(2x+11) - 1/(x-4) = 2x²+ 3x - 6 = 0​

Answer 1

$\bold{\huge{\purple{\underline{ Solution }}}}$

{Correct Question :-

Show that,

$\bold{\dfrac{ 2}{2x+11}}{\bold{-}}{\bold{\dfrac{ 1}{x - 4}}}{\bold{=}}{\bold{\dfrac{1}{2}}}$

Simplifies to,

$\bold{ 2x² + 3x - 6 = 0 }$}

Here, We have to show that,

$\bold{\blue{\dfrac{ 2}{2x+11}}}{\bold{\blue{-}}}{\bold{\blue{\dfrac{ 1}{x - 4}}}}{\bold{\blue{ = }}}{\bold{\blue{\dfrac{1}{2}}}}{\bold{\blue{= 2x² + 3x - 6 = 0}}}$

Let's Begin :-

$\bold{\dfrac{ 2}{2x+11}}{\bold{-}}{\bold{\dfrac{ 1}{x - 4}}}{\bold{=}}{\bold{\dfrac{1}{2}}}$

• By taking LCM of the given Denominators

$\sf{\dfrac{ 2(x-4) - 1(2x + 11)}{(2x+11)(x-4)}}{\sf{=}}{\sf{\dfrac{1}{2}}}$

• By solving the like terms

$\sf{\dfrac{ 2x - 8 - 2x - 11)}{2x(x-4)+11(x - 4)}}{\sf{=}}{\sf{\dfrac{1}{2}}}$

$\sf{\dfrac{ - 19 }{2x² -8x + 11x - 44}}{\sf{=}}{\sf{\dfrac{1 }{2}}}$

$\sf{\dfrac{ - 19 }{2x² + 3x - 44}}{\sf{= }}{\sf{\dfrac{1}{2}}}$

• By cross multiplying the numerator and denominator

$\sf{ - 19 × 2 = 2x² + 3x - 44}$

$\sf{ - 38 = 2x² + 3x - 44 }$

$\sf{ 0 = 2x² + 3x - 44 + 38}$

• After, Solving the given equation we proved that 2x² + 3x - 6 = 0

$\bold{ 2x² + 3x - 6 = 0 }$

Hence, It is proved

$\bold{\dfrac{ 2}{2x+11}}{\bold{-}}{\bold{\dfrac{ 1}{x - 4}}}{\bold{ = }}{\bold{\dfrac{1}{2}}}{\bold{= 2x² + 3x - 6 = 0 }}$.

Answer 2

The required equation has been proved as shown

Expressions and equations

We are to show that 2/(2x+11) - 1/(x-4) = 1/2

Find the LCM of the expression

$\frac{2(x-4)-(2x+11)}{(2x+11)(x-4)}=\frac{1}{2}\\ \frac{2x-8-2x-11}{(2x+11)(x-4)}=\frac{1}{2}$

Cross multiply

(2x+11)(x-4) = -19(2)

2x^2 -8x + 11x - 44 = -38

2x^2 + 3x = -38 + 44

2x^2 + 3x = 6

2x^2 + 3x - 6 = 0 (prove)

The required equation has been proved

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