Question

$3(x - 1) - 2(3x - 1) = 4$
​

Answer 1

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

3(x-1)-2(3x-1)=4

How to solve :

Step1: First step is to open the brackets of all the terms and Multiply the constant with each and every term.

Step2:Now,arrange the terms in such a manner that like terms gets together.

Step3:Shifts all the constant terms to the constant side and let the Variable alone on the other side .

➝ 3(x-1)-2(3x-1)=4

➝3x-3-6x+2=4

➝3x-6x-3+2=4

➝-3x-1=4

➝-3x=4+1

➝-3x=5

➝x= -5/3

Check:

3(x-1)-2(3x-1)=4

Taking LHS

➝3(-5/3-1)-2(3(-5/3)-1)

➝3(-5-3/3 ) -2(-15/3-1)

➝3(-8/3)-2(-5-1)

➝-24/3-2(-6)

➝-8+12

=4

Hence,LHS = RHS (Verified)

Answer 2

$\large\textsf{ }$

$\LARGE\mathcal{\underline\textcolor{aqua}{✯\; Question :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{3 ( x - 1 ) - 2 ( 3x - 1 ) = 4}$

$\large\textsf{ }$

$\LARGE\mathcal{\underline\textcolor{aqua}{✯\; Solution :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{3 ( x - 1 ) - 2 ( 3x - 1 ) = 4}$

$\qquad\tt{:}\longrightarrow\large\textsf{3x - 3 - 6x + 2 = 4}$

$\qquad\tt{:}\longrightarrow\large\textsf{3x - 6x - 3 + 2 = 4}$

$\qquad\tt{:}\longrightarrow\large\textsf{- 3x - 1 = 4 }$

$\qquad\tt{:}\longrightarrow\large\textsf{- 3x = 4 + 1}$

$\qquad\tt{:}\longrightarrow\boxed{\large\textsf\textcolor{red}{ x = \cfrac{\large\textsf{- 5 }}{\large\textsf{3}} }}$

$\large\textsf{ }$

$\LARGE\mathcal{\underline\textcolor{aqua}{✯\; Verification :-}}$

$\large\textsf{ }$

⇝ Taking L.H.S. :-

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{3 ( x - 1 ) - 2 ( 3x - 1 )}$

$\qquad\tt{:}\longrightarrow\large\textsf{ \large\textsf{3}×\left(\cfrac{\large\textsf{- 5 }}{\large\textsf{3}} \; - \; 1 \right) \; - \; 2 \; × \left(3 \; × \cfrac{\large\textsf{ - 5 }}{\large\textsf{3}} \; - \; 1\right) }$

$\qquad\tt{:}\longrightarrow\large\textsf{ \large\textsf{3} × \left(\cfrac{\large\textsf{ - 5 - 3 }}{\large\textsf{3}}\right) \; - \; 2\left(\large\textsf{ - 5 - 1 }\right) }$

$\qquad\tt{:}\longrightarrow\large\textsf{ \left(\large\textsf{3} × \cfrac{\large\textsf{- 8 }}{\large\textsf{3}}\right)\; - \; \large\textsf{ ( 2 × - 6 )} }$

$\qquad\tt{:}\longrightarrow\large\textsf{- 8 - ( - 12 )}$

$\qquad\tt{:}\longrightarrow\large\textsf{- 8 + 12}$

$\qquad\tt{:}\longrightarrow\large\textsf{4}$

$\large\textsf{ }$

$\qquad\large\textsf{\underline\textcolor{orange}{Hence , LHS = RHS}}$

$\large\textsf{ }$

$\LARGE\mathcal{\underline\textcolor{aqua}{✯\; Algebraic \; \; Formulas :-}}$

$\large\textsf{ }$

$\qquad\tt{:}\longrightarrow\large\textsf{( a + b )² = a² + b² + 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{( a - b )² = a² + b² - 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{a² - b² = ( a + b ) ( a - b )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a² + b² = ( a + b )² - 2ab}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ + b³ = ( a + b ) ( a² - 2ab + b² )}$

$\qquad\tt{:}\longrightarrow\large\textsf{a³ - b³ = ( a - b ) ( a² + ab + b² )}$

$\qquad\tt{:}\longrightarrow\large\textsf{( a + b + c )² = a² + b² + c² + 2ab + 2bc + 2ac }$

$\large\textsf{ }$

$\large\textsf\textcolor{purple}{ \; \; \; \; \; \; \; \; \; \; \; \; ◈ ━━━━━━━ ✪ ━━━━━━━ ◈}$