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Which option of linear equation is inconsistent?

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x + y = 1

2 x + y = 0

x - y = 2

2 x - 2 y = 1

2 x - 3 y = 0

2 y + x = 1

2 x - 4 y = 19

- 2 x + y = 0

answer is B.

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Detailed Solution

Given a pair of linear equation,
The general form for a pair of linear equations in two variables

x

and

y

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

and

a_{2} x + b_{2} y + c_{2} = 0

.
Considering pair of equations

x + y = 1

2 x + y = 0

x + y - 1 = 0

\Rightarrow a_{1} = 1, b_{1} = 1, c_{1} = - 1

2 x + y = 0

\Rightarrow a_{2} = 2, b_{2} = 1, c_{2} = 0

Now,

\frac{a_{1}}{a_{2}} = \frac{1}{2}

\frac{b_{1}}{b_{2}} = \frac{1}{1}

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

System of equations

x + y = 1

2 x + y = 0

is consistent.
Considering pair of equations

x - y = 2

2 x - 2 y = 1

x - y = 2

…….(1)

2 x - 2 y = 1

…….(2)
Comparing equations (1) and (2) with the general form of equation to consider their co-efficient,

a_{1} = 1

a_{2} = 2

b_{1} = - 1

b_{2} = - 2

c_{1} = 2

c_{2} = 1

So,

\frac{1}{2} = \frac{- 1}{- 2} \neq \frac{2}{1}

We know that

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

is said to be inconsistent.
Hence, the equations are inconsistent.
Therefore, option (2) is correct.

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