Question #90a80

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Question #90a80

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Answer 1 · 2018-01-10T00:36:29

No solutions.

Explanation:

$6 y - 9 \leq 4 y + 2 y - 16$ $6y - 9 <= 4y + 2y - 16$

$6 y - 9 \leq 6 y - 16$ $6y - 9 <= 6y - 16$

$6 y \leq 6 y - 7$ $6y <= 6y - 7$

To more easily think about this logically, let's substitute $6 y$ $6y$ for $n$ $n$ .

$n \leq n - 7$ $n <= n - 7$

Now, is it ever possible for a number to be less than or equal to 7 less than itself? This is simply impossible, so there is no solution.

Algebraically, we can see why there is no solution by continuing to solve as normal.

$6 y \leq 6 y - 7$ $6y <= 6y - 7$

$0 \leq - 7$ $0 <= -7$

As we know, $0$ $0$ is never less than or equal to $- 7$ $-7$ , so there are no values for $y$ $y$ for which $6 y - 9 \leq 4 y + 2 y - 16$ $6y - 9 <= 4y + 2y - 16$ .

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Question #90a80

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