Question

twice one number minus three times a second number is equal to 2 and the sum of those number is 11 find the number

Answer 1

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Answer 2

$\large\sf\underline{Question:}$

Twice one number minus three times a second number is equal to 2 and the sum of those number is 11 . Find the number .

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$\large\sf\underline{Answer:}$

Let the two numbers be ${\sf{{\pink{x}}}}$ and ${\sf{{\pink{y}}}}$ .

According to the question :

$\small\rm\purple\star\:\underline{1^{st}\: condition-}$

$\sf\:3x-2y=2--(i)$

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$\small\rm\purple\star\:\underline{2^{nd}\: condition-}$

$\sf\:x+y=11--(ii)$

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Multiplying equation (ii) by 2 :

$\sf\:x+y=11$

$\sf➞\:2(x+y)=11×2$

$\sf➞\:2x+2y=22--(iii)$

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Now adding equation (i) and (iii) :

$\sf\:3x-2y+(2x+2y)=22+2$

$\sf➞\:3x-2y+2x+2y=22+2$

$\sf➞\:3x+2x-2y+2y=22+2$

$\sf➞\:5x-2y+2y=22+2$

$\sf➞\:5x-\cancel{2y}+\cancel{2y}=22+2$

$\sf➞\:5x=24$

$\sf➞\:x=\frac{24}{5}$

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Now substituting the value of x in (i) :

$\sf➞\:3x-2y=2$

$\sf➞\:3(\frac{24}{5})-2y=2$

$\sf➞\:3×\frac{24}{5}-2y=2$

$\sf➞\:\frac{72}{5}-2y=2$

$\sf➞\:\frac{72-10y}{5}=2$

$\sf➞\:72-10y=5×2$

$\sf➞\:72-10y=10$

$\sf➞\:-10y=10-72$

$\sf➞\:-10y=-62$

$\sf➞\:\cancel{-}10y=\cancel{-}62$

$\sf➞\:10y=62$

$\sf➞\:y=\frac{62}{10}$

$\sf➞\:y=6.2$

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$\large\sf\underline{Verification\:of\:my\:answer:}$

Let's substitute the value of x and y in equation (i)

$\sf\:3x-2y=2$

$\sf➻\:3(\frac{24}{5})-2×6.2=2$

$\sf➻\:3×\frac{24}{5}-2×\frac{62}{10}=2$

$\sf➻\:\frac{72}{5}-\frac{124}{10}=2$

$\sf➻\:\frac{144-124}{10}=2$

$\sf➻\:\frac{20}{10}=2$

$\sf➻\:\cancel{\dfrac{20}{10}}=2$

$\sf➻\:2=2$

$\small{\underline{\boxed{\mathrm\purple{[\:Hence\: Verified\:]}}}}$

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Similarly we can verify our answer by substituting the value of x and y in equation (ii) .

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⠀‎⠀⠀⠀‎$\large\sf\underline{Final\:Answers}$

⠀‎⠀⠀⠀${\sf{{\red{➞\:x=\frac{24}{5}}}}}$

⠀‎⠀⠀⠀‎${\sf{{\red{➞\:y=6.2}}}}$

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!! Hope it helps !!