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Question
Solve the following pair of linear equation by the elimination method and the substitution method.
`x/2 + (2y)/3 = -1 and x - y /3 = 3`
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Solution
`x/2 + (2y)/3 = - 1` and `x - y/3 = 3`
By elimination method
`x/2 + (2y)/3 = -`1 ...(i)
`x-y/3 = 3` ...(ii)
Multiplying equation (i) by 2, we get
`x + (4y)/3 = - 2` ...(iii)
Subtracting equation (ii) from equation (iii), we get
`(5y)/3 = -5`
Dividing by 5 and multiplying by 3, we get
`y = -15/5`
y = -3
Putting this value in equation (ii), we get
`x - y/3 = 3` ...(ii)
`x-(-3)/3 = 3`
x + 1 = 3
x = 2
Hence, our answer is x = 2 and y = −3.
By substitution method
`x - y/3 = 3` ...(ii)
Add `y/3` both side, we get
`x = 3 + y/3` ...(iv)
Putting this value in equation (i), we get
`x/2 + (2y)/3 = - 1` ...(i)
`(3+ y/3)/2 + (2y)/3 = -1 `
`3/2 + y/6 + (2y)/3 = - 1`
Multiplying by 6, we get
9 + y + 4y = - 6
5y = -15
y = - 3
Hence, our answer is x = 2 and y = −3.
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