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Which of the following pairs of equations represent an inconsistent system?

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3 x - 2 y = 8

2 x + 3 y = 1

3 x - y = - 8

3 x - y = 24

x - y = m

x + my = 1

5 x - y = 10

10 x - 6 y = 20

answer is B.

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Detailed Solution

Given that four systems of linear equations are,
1)

3 x - 2 y = 8

2 x + 3 y = 1

3 x - y = - 8

3 x - y = 24

x - y = m

x + my = 1

5 x - y = 10

10 x - 6 y = 20

.
We want to find the pairs of equations that represent an inconsistent system.
Now let consider the system of equations,

a_{1} x + b_{1} y + c_{1} = 0

…..(1)

a_{2} x {+ b}_{2} y + c_{2} = 0

……(2).
Now, equation (1) and (2) has no solutions if

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}

….(1)
and has a unique solution if

\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}

…..(2).
Now from the given options, option (2) is,

3 x - y + 8 = 0

and

3 x - y - 24 = 0

.
It satisfy the above condition (1) i.e,

\frac{3}{3} = \frac{- 1}{- 1} \neq \frac{- 8}{24}

.
So the pair of equations represents inconsistent system.
In option 1,

\frac{3}{2} \neq \frac{- 2}{3}

.
So the system of equation has a unique solution.
In option 4,

\frac{1}{1} \neq \frac{- 1}{m}

for any value of m except -1.
If

m = - 1

then we get,

\frac{1}{1} = \frac{- 1}{- 1} \neq \frac{- 1}{1}

.
Then the equation has no solution.
This special case is ignorable since m may be any integer other than -1.
For the option (4) we have,

\frac{5}{10} \neq \frac{- 1}{- 6}

.
So they have unique solution.
So, option 2 represents the inconsistent system.
So, the correct option is 2.

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