y>12∣x∣−4y>\frac{1}{2}|x|-4y>21∣x∣−4
⟹ 12∣x∣<y+4\implies \frac{1}{2}|x|<y+4⟹21∣x∣<y+4
⟹ ∣x∣<2(y+4)\implies |x|<2(y+4)⟹∣x∣<2(y+4)
⟹ −2(y+4)<x<2(y+4)\implies-2(y+4)<x<2(y+4)⟹−2(y+4)<x<2(y+4)
The shaded portion is the required answer.
(Please note that it is an open region towards positive y-axis.)
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