Question

Solve the following pair of linear (Simultaneous ) equation using method of elimination by substitution :
2( x - 3 ) + 3( y - 5 ) = 0
5( x - 1 ) + 4( y - 4 ) = 0

Answer 1

Given equations are

2( x - 3 ) + 3( y - 5 ) = 0                      ...(1)

5( x - 1 ) + 4( y - 4 ) = 0                      ...(2)

From (1), we get

2x - 6 + 3y - 15 = 0

⇒ 2x + 3y = 21

⇒ 2x = 21 - 3y

⇒ x = `[ 21 - 3y ]/2`

From (2), we get

5( x - 1 ) + 4( y - 4 ) = 0

⇒ 5x - 5 + 4y - 16 = 0

⇒ 5x + 4y = 21                             ....(3)

Substituting x = `[ 21 - 3y ]/2` in (3), we get

`5(( 21 - 3y )/2) + 4y = 21`

⇒ `[105 - 15y]/2 + 4y = 21`

⇒ `(105 - 15y + 8y)/2 =21`

105 - 7y = 42 - 105

⇒ -7y = -63 

⇒ y = 9

Substituting y = 9 in

x = `[21 - 3y]/2`, we get

 = `[ 21 - 3(9) ]/2`

`x= [ 21 - 27 ]/2`

`x = -6/2`

`x= -3`

∴ The solution is x = -3 and y = 9.