Number of solutions for the system of equations $x + 2 y + z = 0, 3 x + 5 y + 2 z = 2$ and $2 x + 4 y + 2 z = 1$

0

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1

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2

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infinity

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Solution

The correct option is A $0$
Consider the first equation
$x + 2 y + z = 0 \dots (1)$

Consider the last equation
$2 x + 4 y + 2 z = 1$
$\Rightarrow x + 2 y + z = \frac{1}{2} \dots (2)$

Hence both $(1)$ and $(2)$ consist of a system of parallel planes.
Therefore, they never intersect and hence they do not have any common solution.

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Similar questions

2 x + y - z = 7

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x + 2 y - z = 7, 2 x - y + z = 2, 3 x - 5 y + 2 z = - 7

Number of solutions for the system of linear equations, will be,

x + y + z =6

x + 2y + 3z = 14

x + 4y + 2z =30

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