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Q1.    Solve the following pair of linear equations by the elimination method and the substitution method :

                (iv)    \frac{x}{2} + \frac{2y}{3} = -1\ \textup{and} \ x - \frac{y}{3} = 3

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Elimination Method:

Given, equations

\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)

Now, multiplying (2) by 2 we, get

\\2x - \frac{2y}{3} =6............(3)

Now, Adding (1) and (3), we get

\\\frac{x}{2}+\frac{2y}{3}+2x-\frac{2y}{3}=-1+6

\Rightarrow \frac{5x}{2}=5

\Rightarrow x=2

Putting this value in (2) we, get

2-\frac{y}{3}=3

\Rightarrow \frac{y}{3}=-1

\Rightarrow y=-3

Hence,

 x=2\:and\:y=-3

Substitution method :

Given, equations

\\\frac{x}{2} + \frac{2y}{3} = -1........(1)\ \textup{and} \ \\ \\x - \frac{y}{3} = 3............(2)

Now, from (2) we have,

y=3(x-3)......(3)

substituting this value in (1)

\frac{x}{2}+\frac{2(3(x-3))}{3}=-1

\Rightarrow \frac{x}{2}+2x-6=-1

\Rightarrow \frac{5x}{2}=5

\Rightarrow x=2

Substituting this value of x in (3)

\Rightarrow y=3(x-3)=3(2-1)=-3

Hence,

x=2\:and\:y=-3

Posted by

Pankaj Sanodiya

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