Question

Answer this question ​

Answer 1

$\Large{\underbrace{\sf{\orange{Required\:Answer:}}}}$

$\displaystyle{\bf{ \dfrac{(5x-7)-(2x+3)}{6x+11} = \dfrac{8}{3}}}$

Solution - $\displaystyle{\sf{\dfrac{(5x-7)-(2x+3)}{6x+11} = \dfrac{8}{3}}}$

• Open the brαckets αnd multiply the (-) sign with +2x αnd +3.

$\longrightarrow{\sf{ \dfrac{5x-7-2x-3}{6x+11} = \dfrac{8}{3}}}$

$\longrightarrow{\sf{ \dfrac{5x-2x-7-3}{6x+11} = \dfrac{8}{3}}}$

$\longrightarrow{\sf{ \dfrac{ 3x -10}{6x+11} = \dfrac{8}{3}}}$

• Do cross multiplicαtion.

$\longrightarrow{\sf{ 8(6x+11) = 3(3x-10)}}$

• Agαin, open the brαckets αnd multiply 8 with 6x αnd 11, αnd 3 with 3x αnd -10.

$\longrightarrow{\sf{ 48x + 88 = 9x - 30}}$

• Transpose 88 to R.H.S αnd 9x to L.H.S.

$\longrightarrow{\sf{ 48x - 9x = -30-88}}$

$\longrightarrow{\sf{ 39x = -118}}$

• Trαnspose 39 to R.H.S.

$\longrightarrow{\sf{ x = \dfrac{-118}{39}}}$

$\large\longrightarrow{\boxed{\bf{\red{ x = -3.025}}}}$

Hence, the vαlue of x is -3.025.

And we αre done! :D

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Answer 2

${\large{\bf{\underline{Required \; answer}}}}$

Question -

$\; \; \; \; \; \; \; \;{\tt{\dfrac{(5x \: - 7) - \: (2x \: + 3)}{6x \: + 11} = \dfrac{8}{3}}}$

Answer -

$\; \; \; \;{\rm{\leadsto \dfrac{(5x \: - 7) - \: (2x \: + 3)}{6x \: + 11} = \dfrac{8}{3}}}$

~ Let's open the brackets and (+ = +) ; (- = -)

$\; \; \; \;{\rm{\leadsto \dfrac{5x \: - 7 - \: 2x \: - 3}{6x \: + 11} = \dfrac{8}{3}}}$

~ Writing like terms together.

$\; \; \; \;{\rm{\leadsto \dfrac{5x \: - 2x \: - 3 - 7}{6x \: + 11} = \dfrac{8}{3}}}$

~ Let's add or subtract

$\; \; \; \;{\rm{\leadsto \dfrac{3x - 10}{6x \: + 11} = \dfrac{8}{3}}}$

~ Now let's cross multiply the digits.

$\; \; \; \;{\rm{\leadsto 8(6x + 11) = 3(3x-10)}}$

~ Now let's multiply

$\; \; \; \;{\rm{\leadsto 48x + 88 = 9x - 30}}$

~ Combining like terms and (+ = +) ; (- = -)

$\; \; \; \;{\rm{\leadsto 48x - 9x = -30 -88}}$

$\; \; \; \;{\rm{\leadsto 39x = -118}}$

$\; \; \; \;{\rm{\leadsto x = \dfrac{-118}{39}}}$

$\; \; \; \;{\rm{\leadsto x = -3.025}}$

${\boxed{\boxed{\bf{x \: = \: -3.025}}}}$

${\large{\bf{\underline{Additional \; information}}}}$

Fraction rules -

$\boxed{\begin{minipage}{6 cm}\bf{\dag}\:\:\underline{\textsf{Fraction Rules :}}\\\\\bigstar\:\:\sf\dfrac{A}{C} + \dfrac{B}{C} = \dfrac{A+B}{C} \\\\\bigstar\:\:\sf{\dfrac{A}{C} - \dfrac{B}{C} = \dfrac{A-B}{C}}\\\\\bigstar\:\:\sf\dfrac{A}{B} \times \dfrac{C}{D} = \dfrac{AC}{BD}\\\\\bigstar\:\:\sf\dfrac{A}{B} + \dfrac{C}{D} = \dfrac{AD}{BD} + \dfrac{BC}{BD} = \dfrac{AD+BC}{BD} \\\\\bigstar\:\:\sf\dfrac{A}{B} - \dfrac{C}{D} = \dfrac{AD}{BD} - \dfrac{BC}{BD} = \dfrac{AD-BC}{BD}\\\\\bigstar \:\:\sf \dfrac{A}{B} \div \dfrac{C}{D} = \dfrac{A}{B} \times \dfrac{D}{C} = \dfrac{AD}{BC}\end{minipage}}$