Question

Determine by substitution if:
3 is the root of
$\frac{x + 1}{2} + \frac{x - 1}{2} = 3$
can you give me fourmula​

Answer 1

Answer:

$\frac{x + 1}{2} + \frac{x - 1}{2} = 3$

$\frac{x + 1 + x - 1}{2} = 3$

$\frac{2x}{2} = 3$

$x = 3$

Answer 2

$\color{red}{\large\underline{\underline\mathtt{Question:}}}$

$\mathtt{\mapsto \dfrac{x + 1}{2} + \dfrac{x - 1}{2} = 3}$

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$\color{purple}{\large\underline{\underline\mathtt{To\:Find:}}}$

$\textit{To find whether 3 is the root the the equation}$

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$\color{blue}{\large\underline{\underline\mathtt{Concept:}}}$

$\textit{By solving the equation , or Putting the value}$$\textit{of 3 on the equation we can determine that}$$\textit{whether 3 is the root of the equation or not}$

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$\color{magenta}{\large\underline{\underline\mathtt{Solution:}}}$

$\mathtt{\mapsto \dfrac{x + 1}{2} + \dfrac{x - 1}{2} = 3}$

$\Rightarrow \dfrac{x + 1 + (x - 1)}{2} = 3$

$\Rightarrow \dfrac{x + 1 + x - 1}{2} = 3$

$\Rightarrow \dfrac{x + \cancel{1} + x - \cancel{ 1}}{2} = 3$

$\Rightarrow \dfrac{x + x}{2} = 3$

$\Rightarrow \dfrac{2x}{2} = 3$

$\Rightarrow 2x = 3 \times 2$

$\Rightarrow 2x = 6$

$\Rightarrow x = \dfrac{6}{3}$

$\Rightarrow x = \dfrac{\cancel{6}}{\cancel{3}}$

$\Rightarrow x = 3$

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$\mathtt{\therefore 3\:is\:the\:root\:of\:the\:equation}$