How many solutions does the system formed by x-y=3 and ay-ax=0 have for a nonzero number a?

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How many solutions does the system formed by x-y=3 and ay-ax=0 have for a nonzero number a?

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Answer 1 · 2016-12-04T07:13:55

We have no solution.

Explanation:

For a non-zero $a$ $a$ , dividng both sides of the equation $a y - a x = 0$ $ay-ax=0$ , it becomes

$y - x = 0$ $y-x=0$ .

As $x - y = 3$ $x-y=3$ , we have the other equation as $y - x = - 3$ $y-x=-3$ ,

which means $- 3 = 0$ $-3=0$ . But this is not possible.

Hence, we have no solution.

Note : In such cases, in Cartesian plane, the equations $x - y = 3$ $x-y=3$ and $a y - a x = 0$ $ay-ax=0$ , where $a \neq 0$ $a!=0$ , represent a pair of lines. As they do not meet, we have no solution.

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How many solutions does the system formed by x-y=3 and ay-ax=0 have for a nonzero number a?

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