Math, asked by nikita12354, 1 year ago

\textbf{\huge{\red{Hiii..Guys!!..}}}
\textbf{\huge{\pink{Gud..Evening}}}
\mathcal{\huge{\purple{Need\: Help!!}}}

---------------------
Ques:- Solve the following equation ⬆⬆ And verify it..
---------------------

#Question is from class 8th chapter Linear Equation...

\textbf{\huge{\blue{Best\: Of\: luck}}}

Attachments:

nikita12354: Hanji
nikita12354: 8th
nikita12354: Yeah
nikita12354: okay good

Answers

Answered by Anonymous
7
\huge\mathbb\purple{HELLO\:DEAR}

⭐ THE SOLUTION TO YOUR ABOVE QUESTION IS MENTIONED IN THE ABOVE PICTURE.

⭐ I HOPE THE ABOVE SOLUTION IS CLEAR TO YOU.

⭐ FEEL FREE TO ASK ME MORE DOUBTS WHICH STILL EXISTS IN YOU.
Attachments:

nikita12354: How your studies going on??
nikita12354: Okay well your are so so intelligent so you can do 10
nikita12354: Tumne phir block kar diya
nikita12354: :-(
Answered by Mercidez
9
\large{\boxed{\boxed{\boxed{\bold{\red{Solution :\longrightarrow}}}}}}

 \bold{= > 5( 2x - \frac{3}{4} ) = 1(3x + \frac{4}{5} )} \\ \\ \bold{= > 10x - \frac{15}{4} = 3x + \frac{4}{5} }\\ \\ \bold{= > 10x - 3x = \frac{4}{5} + \frac{15}{4} }\\ \\ \bold{= > 7x = \frac{16 + 75}{20} }\\ \\ \bold{= > 7x \times 20 = 91 }\\ \\\bold{ = > 140x = 91 }\\ \\ \bold{= > x = \frac{91}{140}} \\ \\ \bold{= > x = \frac{13}{20}}
Similar questions