Trying to solve |2x-1|-|x+5|-3=0

  • Thread starter bonildo
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Note that the conditions of (2x-1) positive and (x+5) negative, and (2x-1) negative and (x+5) positive, are equivalent.In summary, the equation |2x-1|-|x+5|-3=0 is being solved for the interval (-∞,-5). The solutions for x>1/2 and -5<x<1/2 are x=9 and x=-7/3 respectively. However, for x<-5, the solution x=3 is incorrect and should not be considered. This is because it violates the assumption that x < -5, leading to the conclusion that there is no root in the interval (-∞,-5
  • #1
bonildo
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1

Homework Statement


trying to solve |2x-1|-|x+5|-3=0:

Homework Equations


3. The Attempt at a Solution [/B]
for x>1/2 I got x=9

for -5<x<1/2 I got x=-7/3

but for x<-5 I got -(2x-1)-(-(x+5))-3=0 => x=3 (which is incorrect)

why x=3 is incorrect , or why I shouldn't considerate the case x<-5 ?
 
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  • #2
bonildo said:

Homework Statement


trying to solve |2x-1|-|x+5|-3=0:

Homework Equations


3. The Attempt at a Solution [/B]
for x>1/2 I got x=9

for -5<x<1/2 I got x=-7/3

but for x<-5 I got -(2x-1)-(-(x+5))-3=0 => x=3 (which is incorrect)

why x=3 is incorrect , or why I shouldn't considerate the case x<-5 ?
Because for that interval, (-∞, -5), you are assuming that x < -5. Therefore x cannot equal +3.
 
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  • #3
bonildo said:

Homework Statement


trying to solve |2x-1|-|x+5|-3=0:

Homework Equations


3. The Attempt at a Solution [/B]
for x>1/2 I got x=9

for -5<x<1/2 I got x=-7/3

but for x<-5 I got -(2x-1)-(-(x+5))-3=0 => x=3 (which is incorrect)

why x=3 is incorrect , or why I shouldn't considerate the case x<-5 ?

If your assumption of x <= -5 leads to the conclusion that the root is at x = +3 (outside the starting interval), that means that there IS NO ROOT in the interval (-∞,-5). Plot a graph to see why this is so.
 
  • #4
You should try checking conditions this way although not as efficient of choices:

2x-1 and x+5 both positive;
2x-1 and x+5 both negative;
2x-1 positive and x+5 negative;
2x-1 negative and x+5 positive.

Solve the equation in each of the four conditions and check each result.
 

1. What is the first step in solving this equation?

The first step in solving this equation is to isolate the absolute value terms by using the distributive property. This will result in two separate equations, one with a positive absolute value and one with a negative absolute value.

2. How do I determine the values of x that satisfy the equation?

After isolating the absolute value terms, you will have two separate equations. Solve each equation separately for x. The values of x that satisfy the equation will be the solutions to both equations.

3. Can I use substitution to solve this equation?

Yes, substitution can be used to solve this equation. However, it may result in a more complicated equation to solve. It is often easier to use the distributive property to isolate the absolute value terms.

4. Are there any extraneous solutions to this equation?

Yes, there may be extraneous solutions to this equation. After solving for x, be sure to plug the solutions back into the original equation to check for extraneous solutions. These are solutions that may satisfy the equation but do not satisfy the original equation.

5. Is there a faster way to solve this equation?

There is not necessarily a faster way to solve this equation, but there are different approaches that may be more efficient for certain equations. It is important to practice different methods and determine which works best for you.

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