Solve the following system of linear equations by applying the method of elimination by equating the coefficients

**(i)**4x – 3y = 4

2x + 4y = 3

**(ii)**5x – 6y = 8

3x + 2y = 6

#### Solution

**(i)** We have,

4x – 3y = 4 ….(1)

2x + 4y = 3 ….(2)

Let us decide to eliminate x from the given equation. Here, the co-efficients of x are 4 and 2 respectively. We find the L.C.M. of 4 and2 is 4. Then, make the co-efficients of x equal to 4 in the two equations.

Multiplying equation (1) with 1 and equation (2) with 2, we get

4x – 3y = 4 ….(3)

4x + 8y = 6 ….(4)

Subtracting equation (4) from (3), we get

`–11y = –2 ⇒ y = \frac { 2 }{ 11 }`

Substituting y = 2/11 in equation (1), we get

`⇒ 4x – 3 × \frac { 2 }{ 11 } = 4`

`⇒ 4x – \frac { 6 }{ 11 } = 4`

`⇒ 4x = 4 + \frac { 6 }{ 11 }`

`⇒ 4x = \frac { 50 }{ 11 }`

`⇒ x = \frac { 50 }{ 44 } = \frac { 25 }{ 22 }`

Hence, solution of the given system of equation is :

`x = \frac { 25 }{ 22 }, y = \frac { 2 }{ 11 }`

**(ii)** We have;

5x – 6y = 8 ….(1)

3x + 2y = 6 ….(2)

Let us eliminate y from the given system of equations. The co-efficients of y in the given equations are 6 and 2 respectively. The L.C.M. of 6 and 2 is 6. We have to make the both coefficients equal to 6.

So, multiplying both sides of equation (1) with 1 and equation (2) with 3, we get

5x – 6y = 8 ….(3)

9x + 6y = 18 ….(4)

Adding equation (3) and (4), we get

`14x = 26 ⇒ x = \frac { 13 }{ 7 }`

Putting `x = \frac { 13 }{ 7 }` in equation (1), we get

`5 × \frac { 13 }{ 7 } – 6y = 8 ⇒ \frac { 65 }{ 7 } – 6y = 8`

`⇒ 6y = \frac { 65 }{ 7 } – 8 = \frac { 65-56 }{ 7 } = \frac { 9 }{ 7 }`

`⇒ y = \frac { 9 }{ 42 } = \frac { 3 }{ 14 }`

Hence, the solution of the system of equations is `x = \frac { 13 }{ 7} , y = \frac { 3 }{ 14 }`