Mr hassan opened an account with Rs 45,687. He deposits Rs 5800 in the first month and withdraws Rs6981 in the second month .How
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<h3>Answer: Choice C. |x+3| < 5</h3>
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Explanation:
Let's go through each answer choice and solve for x
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Choice A
|x+3| < -5
This has no solutions because |x+3| is never negative. It is either 0 or positive. Therefore, it can never be smaller than -5. So we can rule this out right away.
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Choice B
|x+8| < 2
-2 < x+8 < 2 .... see note below
-2-8 < x+8-8 < 2-8 ... subtract 8 from all sides
-10 < x < -6
We will have a graph where the open circles are at -10 and -6, with shading in between. This does not fit the original description. So we can rule this out too.
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Choice C
|x+3| < 5
-5 < x+3 < 5 .... see note below
-5-3 < x+3-3 < 5-3 .... subtracting 3 from all sides to isolate x
-8 < x < 2
We found our match. This graph has open circles at -8 and 2, with shading in between. The open circles indicate to the reader "do not include this value as a solution".
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note: For choices B and C I used the rule that turns into
where k is some positive number. For choice A, we have k = -5 which is negative so this formula would not apply.
Looks like the system is
x + 4y - z = -14
5x + 6y + 3z = 4
-2x + 7y + 2z = -17
or in matrix form,
Cramer's rule says that
where is the solution for i-th variable, and
is a modified version of
with its i-th column replaced by
.
We have 4 determinants to compute. I'll show the work for det(A) using a cofactor expansion along the first row.
The modified matrices and their determinants are
Then by Cramer's rule, the solution to the system is