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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
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Solution
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.
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