Question

find the value of x ​

Answer 1

Answer- The above question is from the chapter 'Linear Equations in One Variable'.

Given question: Find the value of x.

$\dfrac{3x}{4} - \dfrac{1}{4} (x \ - \ 20) = \dfrac{x}{4} + 32$

Solution: We have,

$\dfrac{3x}{4} - \dfrac{1}{4} (x \ - \ 20) = \dfrac{x}{4} + 32$

$\dfrac{3x}{4} - \dfrac{x}{4} - (\dfrac{-20}{4}) = \dfrac{x}{4} + 32$

$\dfrac{2x}{4} - (-5) = \dfrac{x}{4} + 32$

$\dfrac{x}{2} + 5 = \dfrac{x}{4} + 32$

$\dfrac{x}{2} -\dfrac{x}{4} = 32 - 5$

L.C.M. of 2 and 4 = 4.

$\dfrac{2x - x}{4} = 27$

$\dfrac{x}{4} = 27$

$x = 27 \times 4$

$\implies \boxed{x = 108}$

∴ Value of x = 108.

Answer 2

Step-by-step explanation:

Given:-

$\sf \dfrac {3x}{4}-\dfrac {1}{4}(x-20)=\dfrac {x}{4}+32$

To find:-

Value of x

Solution:-

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac{3x}{4}-\dfrac {1}{4}(x-20)=\dfrac{x}{4}+32$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {3x}{4}-\dfrac {x}{4}-(-\dfrac{-20}{4}=\dfrac {x}{4}+32$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {3x-x}{4}-(-5)=\dfrac {x}{4}+32$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {2x}{4}+5=\dfrac {x}{4}+32$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {2x}{4}-\dfrac {x}{4}=32-5$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow \dfrac {x}{4}=27$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=27\times 4$

$\\\qquad\quad\displaystyle\sf {:}\longrightarrow x=108$

$\therefore\sf x=108$