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

(iii) $3x - 5y -4 = 0\ \textup{and} \ 9x = 2y + 7$

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Elimination Method:

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, multiplying (1) by 3 we, get

$\\9x -15 y -12=0............(3)$

Now, Subtracting (3) from (2), we get

$9x-2y-7-9x+15y+12=0$

$\Rightarrow 13y+5=0$

$\Rightarrow y=\frac{-5}{13}$

Putting this value in (1) we, get

$3x-5(\frac{-5}{13})-4=0$

$\Rightarrow 3x=4-\frac{25}{13}$

$\Rightarrow 3x=\frac{27}{13}$

$\Rightarrow x=\frac{9}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Substitution method :

Given, equations

$\\3x - 5y -4 = 0..........(1)\ \textup{and}\ \\9x = 2y + 7$

$\\\Rightarrow 9x - 2y -7=0........(2)$

Now, from (2) we have,

$y=\frac{9x-7}{2}.......(3)$

substituting this value in (1)

$3x-5\left(\frac{9x-7}{2} \right )-4=0$

$\Rightarrow 6x-45x+35-8=0$

$\Rightarrow -39x+27=0$

$\Rightarrow x=\frac{27}{39}=\frac{9}{13}$

Substituting this value of x in (3)

$\Rightarrow y=\frac{9(9/13)-7}{2}=\frac{81/13-7}{2}=\frac{-5}{13}$

Hence,

$x=\frac{9}{13}\:and\:y=-\frac{5}{13}$

Posted by

Pankaj Sanodiya

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