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Solve each inequality, express the solution set in interval notation, then graph the solution set.
2x/x−7 ≤ 5
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The solution by Alan is incorrect.
It is incorrect because multiplying by (x-7) both sides of the given inequality
without changing the inequality sign is valid ONLY if x-7 > 0.
So, the Alan's solution is valid only if x > 7 and it produces then the solution set x >= 35/3.
Thus, when x > 7, the solution set in this domain is { x >= }.
It is the part, that Alan found out.
But it is only one part of the work which should be done.
In addition, the case x-7 < 0 should be (and MUST be) analyzed separately.
I do it below.
So, we assume now that x-7 < 0, which means x < 7.
Multiply both sides of the given inequality by (x-7).
Since this factor (x-7) is NEGATIVE now, we should change the sign of the inequality.
Thus after multiplying we have
2x >= 5*(x-7).
Simplify and solve for x
2x >= 5x - 35
35 >= 5x - 2x
35 >= 3x
x <=
Thus in the case when x < 7, we have second inequality x <= .
Thus, the second solution subset is {x < 7}.
Finally, from two basic cases, the solution set
consists of two subsets x >= and x < 7.
In other words, the solution set is the union (-oo,7) U [,oo).
This solution set is plotted below
==============|============)--------------|==================
-oo 0 7 oo
Solved, and fully explained.
------------------
That mistake that Alan did, the beginner students make every day,
until a teacher or an expert will show them a right way solving such inequalities.
So, I am glad to have this opportunity to explain the things to you.
Solve each inequality, express the solution set in interval notation, then graph the solution set.
1.
2x/x−7 ≤ 5Solve each inequality, express the solution set in interval notation, then graph the solution set.
1.
2x/x−7 ≤ 5
It's very like that it's , with , since 7 will make the denominator UNDEFINED.
2x ≤ 5(x - 7) ------ Multiplying by LCD, (x - 7)
2x ≤ 5x - 35
2x - 5x ≤ - 35
- 3x ≤ - 35
Now, we have 2 critical points: and 3 TEST POINTS for x:
When the intervals are tested, we see that are the 2 solution-intervals.
Therefore, in INTERVAL NOTATION, we get: (
)
[
)