Question

Solve the equation. Check your result in each case...
$\small\bf{plz \: ans \: do \: koi.....}$
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Answer 1

Answer:

3x-1/5-x/7=3

Or, 3x-x/7=3+1/5

Or, (21x-x)/7=(15+1)/5

Or, 20x/7=16/5

Or, x=16×7/20×5

Or, x=4×7/5×5

Or, x=28/25

Answer 2

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

• $\mathtt{\dfrac{3x - 1}{5} - \dfrac{x}{7} = 3}$
• $\mathtt{\dfrac{y - 1}{3} - \dfrac{y - 2}{4} = 1}$

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

$\textit{The value x in first case and the value of y in other case.}$

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

$\textsf{By simple calculation we can}$

$\textsf{find the required value of variables.}$

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

• $\mathtt{\dfrac{3x - 1}{5} - \dfrac{x}{7} = 3}$

$\text{LCM is 35}$

$\Rightarrow \dfrac{21x - 7 - 5x}{35} = 3$

$\Rightarrow \dfrac{16x - 7 }{35} = 3$

$\Rightarrow 16x - 7 = 3 \times 35$

$\Rightarrow 16x - 7 = 105$

$\Rightarrow 16x = 105 + 7$

$\Rightarrow 16x = 112$

$\Rightarrow x = \dfrac{112}{16}$

$\Rightarrow x = \dfrac{\cancel{112}}{\cancel{16}}$

$\Rightarrow x = 7$

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• $\mathtt{\dfrac{y - 1}{3} - \dfrac{y - 2}{4} = 1}$

$\text{LCM is 12}$

$\Rightarrow\dfrac{4y - 4 - 3y + 6}{12} = 1$

$\Rightarrow\dfrac{y + 2}{12} = 1$

$\Rightarrow y + 2 = 1 \times 12$

$\Rightarrow y + 2 = 12$

$\Rightarrow y = 12 - 2$

$\Rightarrow y = 10$ ______________________________________

$\color{lime}{\large\underline{\underline\mathtt{Check:}}}$

$\textit{By putting the value in the equation}$

• $\mathtt{\dfrac{3x - 1}{5} - \dfrac{x}{7} = 3}$

$\Rightarrow \dfrac{3 \times 7 - 1}{5} - \dfrac{7}{7} = 3$

$\Rightarrow \dfrac{21 - 1}{5} - 1 = 3$

$\Rightarrow \dfrac{20}{5} - 1 = 3$

$\Rightarrow \dfrac{20}{5} = 3 + 1$

$\Rightarrow 4 = 4$

$\boxed{Proved}$

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• $\mathtt{\dfrac{y - 1}{3} - \dfrac{y - 2}{4} = 1}$

$\Rightarrow \dfrac{10 - 1}{3} - \dfrac{10 - 2}{4} = 1$

$\Rightarrow \dfrac{9}{3} - \dfrac{8}{4} = 1$

$\Rightarrow 3 - 2 = 1$

$\Rightarrow 1 = 1$

$\boxed{Proved}$

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