Question

1
If x + - = 4, find the values of : (i)
2
+
(ii)
X
r​

Answer 1

: \implies{\sf\bigg({\dfrac{x}{2} - 6}\bigg) = \bigg({8 - \dfrac{2x}{3}} \bigg)}:⟹(2x−6)=(8−32x)

\begin{gathered}\end{gathered}

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

: \implies{\sf\bigg({\dfrac{x}{2} - 6}\bigg) = \bf\bigg({8 - \dfrac{2x}{3}} \bigg)}:⟹(2x−6)=(8−32x)

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

{: \implies{\sf\bigg({\dfrac{x - 12}{2}}\bigg) = \bf\bigg({\dfrac{24 - 2x}{3}} \bigg)}}:⟹(2x−12)=(324−2x)

By cross multiplication

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

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

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

: \implies\sf{x = \bf{12}}:⟹x=12

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

Hence, The value of x is 12.

\begin{gathered}\end{gathered}

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

: \implies{\sf\bigg({\dfrac{x}{2} - 6}\bigg) = \bf\bigg({8 - \dfrac{2x}{3}} \bigg)}:⟹(2x−6)=(8−32x)

Substituting the value of x

: \implies{\sf\bigg({\dfrac{12}{2} - 6}\bigg) = \bf\bigg({8 - \dfrac{2 \times 12}{3}} \bigg)}:⟹(212−6)=(8−32×12)

: \implies{\sf\bigg({\dfrac{12}{2} - 6}\bigg) = \bf\bigg({8 - \dfrac{24}{3}} \bigg)}:⟹(212−6)=(8−324)

{: \implies{\sf\bigg({\dfrac{12 - (6 \times 2)}{2}}\bigg) = \bf\bigg({\dfrac{(8 \times 3) - 24}{3}} \bigg)}}:⟹(212−(6×2))=(3(8×3)−24)

{: \implies{\sf\bigg({\dfrac{12 -12}{2}}\bigg) = \bf\bigg({\dfrac{24 - 24}{3}} \bigg)}}:⟹(212−12)=(324−24)

{: \implies{\sf\bigg({\dfrac{0}{2}}\bigg) = \bf\bigg({\dfrac{0}{3}} \bigg)}}:⟹(20)=(30)

: \implies\sf{0} = \bf{0}:⟹0=0

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

Hence Verified!!