Math, asked by missiondjalok5, 1 month ago


solve \: the \: equation \:  \frac{x}{2}  - 6 = 8 -  \frac{2x}{3}

Answers

Answered by Anonymous
285

Answer:

x=12

Step-by-step explanation:

\\ \sf\longmapsto \dfrac{x}{2}-6=8-\dfrac{2x}{3}

\\ \sf\longmapsto \dfrac{x-12}{2}=\dfrac{24-2x}{3}

  • Using cross multiplication

\\ \sf\longmapsto 2(24-2x)=3(x-12)

\\ \sf\longmapsto 48-4x=3x-36

\\ \sf\longmapsto 48+36=3x+4x

\\ \sf\longmapsto 84=7x

\\ \sf\longmapsto x=\dfrac{84}{7}

\\ \sf\longmapsto x=12

Answered by ItzMagician
1592

Answer:

{\underline{\underline{\maltese\textbf{\textsf{\red{Question}}}}}}

: \implies{\sf\bigg({\dfrac{x}{2} - 6}\bigg) =  \bigg({8  -  \dfrac{2x}{3}} \bigg)}

\begin{gathered}\end{gathered}

{\underline{\underline{\maltese\textbf{\textsf{\red{Solution}}}}}}

: \implies{\sf\bigg({\dfrac{x}{2} - 6}\bigg) =  \bf\bigg({8  -  \dfrac{2x}{3}} \bigg)}

{: \implies{\sf\bigg({\dfrac{x  - (6 \times 2)}{2}}\bigg) =   \bf\bigg({\dfrac{(8 \times 3)  - 2x}{3}} \bigg)}}

{: \implies{\sf\bigg({\dfrac{x  - 12}{2}}\bigg) =   \bf\bigg({\dfrac{24 - 2x}{3}} \bigg)}}

  • By cross multiplication

 : \implies\sf{3(x - 12) = \bf{2(24 - 2x)}}

 : \implies\sf{3x - 36 = \bf{48 - 4x}}

: \implies\sf{4x - 3x  = \bf{48 -36}}

 : \implies\sf{x  = \bf{12}}

{\dag{\underline{\boxed{\sf{x  =12}}}}}

  • Hence, The value of x is 12.

\begin{gathered}\end{gathered}

{{\underline{\underline{\maltese\textbf{\textsf{\red{Verification}}}}}}}

: \implies{\sf\bigg({\dfrac{x}{2} - 6}\bigg) =  \bf\bigg({8  -  \dfrac{2x}{3}} \bigg)}

  • Substituting the value of x

: \implies{\sf\bigg({\dfrac{12}{2} - 6}\bigg) =  \bf\bigg({8  -  \dfrac{2 \times 12}{3}} \bigg)}

: \implies{\sf\bigg({\dfrac{12}{2} - 6}\bigg) =  \bf\bigg({8  -  \dfrac{24}{3}} \bigg)}

{: \implies{\sf\bigg({\dfrac{12 - (6 \times 2)}{2}}\bigg) =  \bf\bigg({\dfrac{(8 \times 3)   - 24}{3}} \bigg)}}

{: \implies{\sf\bigg({\dfrac{12 -12}{2}}\bigg) =  \bf\bigg({\dfrac{24   - 24}{3}} \bigg)}}

{: \implies{\sf\bigg({\dfrac{0}{2}}\bigg) =  \bf\bigg({\dfrac{0}{3}} \bigg)}}

  :  \implies\sf{0} = \bf{0}

 \dag{\underline{\boxed{\sf{LHS=RHS}}}}

Hence Verified!!

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