The set A satisfying the given inequality is A = (-[tex]\infty[/tex], -10].
What are some properties of an inequality relation?
Following are some facts which are true for an inequality relation:
- Equal numbers can be added or subtracted from both sides of an inequality without affecting the inequality sign.
- The Inequality sign is unchanged if both sides are multiplied or divided by a positive number, but when multiplied or divided by a negative number the inequality sign is reversed.
[tex]\frac{5x-2}{8} - \frac{3x-5}{10} &\ge& x+y\\\\\Rightarrow\;\; \frac{13}{40}x + \frac{1}{4}&\ge& x+y\\\\\Rightarrow\;\;\;\;\;\; -\frac{27}{40}x &\ge & y - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\;\; -x &\ge & \frac{40}{27}\left( y-\frac{1}{4} \right).\hspace{1cm}(1)[/tex]
Since y ∈ B, -2 ≤ y ≤ 7. So,
[tex]\;\;\;\;\;\;\;\,-2 - \frac{1}{4}\; \le\; y - \frac{1}{4} \;\le\; 7 - \frac{1}{4}\\\\\Rightarrow\;\;\;\;\;\;\;\;\; -\frac{9}{4}\; \le\; y - \frac{1}{4} \;\le\; \frac{27}{4}\\\\\Rightarrow\;\;\; -\frac{9}{4}\cdot \frac{40}{27} \;\le\; \frac{40}{27} \left( y-\frac{1}{4} \right) \;\le \;\frac{27}{4}\cdot \frac{40}{27}\\\\\Rightarrow\;\;\;\;\;\;\;\; -\frac{10}{3} \;\le\; \frac{40}{27}\left( y - \frac{1}{4} \right)\; \le\; 10.[/tex]
The set {-x | inequality (1) holds ∀ y ∈ B} is [10, [tex]\infty[/tex]) i.e.
10 ≤ -x ≤ [tex]\infty[/tex].
Multiplying -1 throughout gives
-10 ≥ x ≥ -[tex]\infty[/tex].
x, thus, lies in the range A = (-[tex]\mathbf{\infty}[/tex], -10}.
Learn more about the inequality here.
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Question
Find the set A such that for x ∈ A
[tex]\frac{5x - 2}{8} - \frac{3x - 5}{10} \ge x + y[/tex]
∀y ∈ B = {y ∈ R | -2 ≤ y ≤ 7}.
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