Question

find the solution by method of substitution ​

Answer 1

7x - 15y = 2 ...(i)

x + 2y = 3 ...(ii)

Equation (ii):

x = 3 - 2y

Substituting the value of x in equation (i) we get :

7(3 - 2y) - 15y = 2

21 - 14y - 15y = 2

- 29y = - 19

29y = 19

y = 19/29

Putting the value of y in equation (ii) we get :

$x + 2( \frac{19}{29} ) = 3 \\ \\ x + \frac{38}{29} = 3 \\ \\ x = 3 - \frac{38}{29} \\ \\ x = \frac{87 - 38}{29} \\ \\ \bf x = \frac{49}{29}$

Let us cross check if the values are correct or not:

$x + 2y = 3 \\ \\ \frac{49}{29} + 2( \frac{19}{29} ) = 3 \\ \\ \frac{49}{29} + \frac{38}{29} = 3 \\ \\ \frac{87}{29} = 3 \\ \\ 3 = 3$

Hence proved.

Answer 2

7x - 15y = 2 ...(i)

x + 2y = 3 ...(ii)

Equation (ii):

x = 3 - 2y

Substituting the value of x in equation (i) we get :

7(3 - 2y) - 15y = 2

21 - 14y - 15y = 2

- 29y = - 19

29y = 19

y = 19/29

Putting the value of y in equation (ii) we get :

$x + 2( \frac{19}{29} ) = 3 \\ \\ x + \frac{38}{29} = 3 \\ \\ x = 3 - \frac{38}{29} \\ \\ x = \frac{87 - 38}{29} \\ \\ \bf x = \frac{49}{29}$

Let us cross check if the values are correct or not:

$x + 2y = 3 \\ \\ \frac{49}{29} + 2( \frac{19}{29} ) = 3 \\ \\ \frac{49}{29} + \frac{38}{29} = 3 \\ \\ \frac{87}{29} = 3 \\ \\ 3 = 3$

Hence proved.