Question

1. Solve the following system of equations by substitution method.

Problem in attachment ​

Answer 1

Answer:

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Answer 2

Given :

• Equation 1 : $\sf\:\frac{x}{6}+y=1$

• Equation 2 : $\sf\:3x-8y=5$

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To find :

• The value of x and y.

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Concept of Substitution method :

$\bf\multimap$ Solving the first equation we will get the value of any variable.

$\bf\multimap$ Then after substituting this value in equation 2 and solving it.

$\bf\multimap$ Finally re - substituting the value in original equation to calculate the value of other variable.

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Solution :

Let -

⠀⠀⠀⌬ $\sf\:\frac{x}{6}+y=1$ -- $\small{\mathfrak\red{1}}$

⠀⠀⠀⌬ $\sf\:3x-8y=5$ -- $\small{\mathfrak\red{2}}$

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From equation 1 calculating the value of x -

$\sf\:\frac{x}{6}+y=1$

$\sf\implies\:\frac{x+6y}{6}=1$

$\sf\implies\:x+6y=6$

$\sf\color{gray}{\implies\:x=6-6y}$

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Substituting the value of x in equation 2 -

$\sf\:3x-8y=5$

$\sf\implies\:3(6-6y)-8y=5$

$\sf\implies\:18-18y-8y=5$

$\sf\implies\:18-26y=5$

$\sf\implies\:-26y=5-18$

$\sf\implies\:-26y=-13$

$\sf\implies\:\cancel{-}26y=\cancel{-}13$

$\sf\implies\:y=\cancel{\frac{13}{26}}$

$\small{\underline{\boxed{\mathrm{\implies\:y=\frac{1}{2}}}}}$ $\color{orange}{⋆}$

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Now again substituting the value of y in 1 -

$\sf\:\frac{x}{6}+y=1$

$\sf\implies\:\frac{x}{6}+\frac{1}{2}=1$

$\sf\implies\:\frac{x+3}{6}=1$

$\sf\implies\:x+3=6$

$\sf\implies\:x=6-3$

$\small{\underline{\boxed{\mathrm{\implies\:x=3}}}}$ $\color{orange}{⋆}$

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$\sf\therefore\:Value~of~x~is~3~and~that~of~y~is~\frac{1}{2}$.

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Verification :

Plugging the value of x and y in equation 1 -

$\sf\:\frac{x}{6}+y=1$

$\sf\leadsto\:\frac{3}{6}+\frac{1}{2}=1$

$\sf\leadsto\:\frac{3+3}{6}=1$

$\sf\leadsto\:\frac{6}{6}=1$

$\sf\leadsto\:\cancel{\frac{6}{6}}=1$

$\sf\leadsto\:1=1$

Hence Verified !~

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