What are the possible solutions for the inequality |2x-1|+|x+2|\geq 4x?

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The discussion revolves around solving the inequality |2x-1| + |x+2| ≥ 4x. Participants analyze the inequality by breaking it into three cases based on the values of x: x < -2, -2 ≤ x < 1/2, and x ≥ 1/2. Each case yields different conditions for x, leading to solutions of x ≤ -1/7, x ≤ 1, and x ≤ 3/5, respectively. The confusion arises in combining these solutions correctly, particularly in understanding how to interpret the intersections of the intervals. Ultimately, the correct solution is x ≤ 1, as it encompasses the valid ranges derived from the cases.
thereddevils
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Homework Statement


|2x-1|+|x+2|\geq 4x


Homework Equations





The Attempt at a Solution



For x<-2 , -(2x-1)-(x+2)\geq 4x

x\leq -\frac{1}{7}

For x\geq \frac{1}{2}

2x-1+x+2\geq 4x

x\leq 1

For -2 \leq x &lt; \frac{1}{2} ,

-(2x-1)+x+2\geq 4x

x\leq \frac{3}{5}

after combining , the solution would be x\leq -\frac{1}{7}

AM i correct ? but the answer given is x\leq 1
 
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Hi thereddevils! :wink:

Your three answers are correct, but you're not putting them together correctly.

For example, you have:

if x ≤ -2, it's true for x ≤ -1/7.

so it's true for all x ≤ -2.

ok, now try the others. :smile:
 


tiny-tim said:
Hi thereddevils! :wink:

Your three answers are correct, but you're not putting them together correctly.

For example, you have:

if x ≤ -2, it's true for x ≤ -1/7.

so it's true for all x ≤ -2.

ok, now try the others. :smile:

thanks tiny , yeah , i am confused with the last part , what i did is to put x<= -1/7 , x<=1 , x<= 3/5 on the number line and take the intersection which is what i got .

so say for x>= 1/2 , x can be 4 , x<=1 , so 4<=1 ?? this is not true

i am sorry , i still do not understand the last part .
 
There are three possibilities:

x ≤ -2, -2 ≤ x ≤ 1/2, 1/2 ≤ x.

call them A B and C.

So A or B or C.

If A, the equation is true if (say) a.

If B, the equation is true if (say) b.

If C, the equation is true if (say) c.

So the equation is true if (A and a) or (B and b) or (C and c). :wink:
 

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