Question

Solve the following pairs of linear equation by substitution method:3x -y-3=0
x-3y-9=0
​

Answer 1

Solution:-

:- We have

=> 3x - y - 3 = 0 .......(i) eq

=> x - 3y - 9 = 0 ........(ii)eq

Now Take (ii)eq

=> x - 3y - 9 = 0

=> x = 3y + 9 .......(iii)eq

Now substitute (iii)eq in (i)eq

=> 3( 3y + 9 ) - y - 3 = 0

=> 9y + 27 - y - 3 = 0

=> 8y + 24 = 0

=> 8y = -24

=> y = - 3

Now substitute value of y (iii)eq

=> x = 3y + 9

=> x = 3 × - 3 + 9

=> x = -9 + 9

=> x = 0

Value of x = 0 and y = - 3

Answer 2

Here,

$\rm3x - y - 3 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cdots(i) \\ \rm{x - 3y - 9 = 0\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cdots(ii)}$

Or,

$\rm3x - y = 3\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cdots(iii) \\ \rm{x - 3y = 9}\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cdots(iv)$

From Equation (iii),

$\rm{y =3x - 3 }\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \cdots(A)$

Putting Equation A in Equation (iv), we get,

$\rm{}x - 3(3x - 3) = 9 \\ \rm{}x - 9x + \cancel 9 = \cancel9 \\ \rm - 8x = 0 \\ \\ \rm \frac{ - 8x}{ - 8} = \frac{0}{ - 8} \\ \\ \huge \rm x = \pink 0$

Now, Putting the value of x in Equation A

$\huge \rm{}y = 3x - 3 \\ \\ \huge \rm{}y = 3(0) - 3 \\ \huge \rm{}y = 0 - 3 \\ \huge \rm{}y = \pink{ - 3}$

Thus,

• Value of x = 0
• Value of y = -3