Answer:

Solution

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2x+1

x

≥

4

1

⟹

2x+1

x

−

4

1

≥0⟹

8x+4

4x−2x−1

≥0⟹

8x+4

2x−1

≥0

Hence, either both numerator and denominator should be positive or both should be negative.

⟹{{2x−1≥0}∩{8x+4≥0}}∪{{2x−1≤0}∩{8x+4≤0}}

⟹{x≥

2

1

}∩{x<−

2

1

} (as x=−0.5 gives undefined value)

⟹x∈(−∞,−

2

1

)∪[

2

1

,∞)

Also,

4x−1

6x

<

2

1

⟹

4x−1

6x

−

2

1

<0⟹

8x−2

12x−4x+1

<0⟹

8x−2

8x+1

<0

Hence, numerator and denominator should be of opposite sign.

⟹{{8x+1<0}∩{8x−2>0}}∪{{8x+1>0}∩{8x−2<0}}

⟹{ϕ}∪{{x>−

8

1

}∩{x<

4

1

}}

⟹x∈(−

8

1

,

4

1

)

Now, the required solution is x∈{(−∞,−

2

1

)∪[

2

1

,∞)}∩(−

8

1

,

4

1

)

⟹x∈[

2

1

,

4

1

)