Bytelearn - cat image with glassesAI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Solve the equation for all values of 
x.

|2x-1|+5=4x
Answer: 
x=

Solve the equation for all values of x x .\newline2x1+5=4x |2 x-1|+5=4 x \newlineAnswer: x= x=

Full solution

Q. Solve the equation for all values of x x .\newline2x1+5=4x |2 x-1|+5=4 x \newlineAnswer: x= x=
  1. Consider Non-Negative Case: We have the equation 2x1+5=4x|2x-1| + 5 = 4x. To solve for xx, we need to consider two cases because of the absolute value: one where 2x12x - 1 is non-negative, and one where 2x12x - 1 is negative.
  2. Simplify Equation: First, let's consider the case where 2x12x - 1 is non-negative, which means 2x102x - 1 \geq 0. In this case, the absolute value expression 2x1|2x - 1| is just 2x12x - 1. So the equation becomes:\newline2x1+5=4x2x - 1 + 5 = 4x
  3. Isolate xx: Simplify the equation by combining like terms: 2x+4=4x2x + 4 = 4x
  4. Check Validity for x=2x=2: To isolate xx, we subtract 2x2x from both sides of the equation:\newline4=2x4 = 2x
  5. Consider Negative Case: Divide both sides by 22 to solve for xx: x=2x = 2
  6. Distribute Negative Sign: Now, let's consider the second case where 2x12x - 1 is negative, which means 2x - 1 < 0. In this case, the absolute value expression 2x1|2x - 1| is (2x1)-(2x - 1). So the equation becomes:\newline(2x1)+5=4x-(2x - 1) + 5 = 4x
  7. Combine Like Terms: Distribute the negative sign inside the parentheses: 2x+1+5=4x-2x + 1 + 5 = 4x
  8. Divide by 66: Combine like terms:\newline6=6x6 = 6x
  9. Check Validity for x=1x=1: Divide both sides by 66 to solve for xx:x=1x = 1
  10. Check Validity for x=1x=1: Divide both sides by 66 to solve for xx:x=1x = 1We need to check if our solutions satisfy the original equation. For x=2x = 2, the original equation 2x1+5=4x|2x-1|+5=4x becomes 41+5=8|4-1|+5=8, which simplifies to 3+5=83+5=8. This is true, so x=2x = 2 is a valid solution.
  11. Check Validity for x=1x=1: Divide both sides by 66 to solve for xx:x=1x = 1We need to check if our solutions satisfy the original equation. For x=2x = 2, the original equation 2x1+5=4x|2x-1|+5=4x becomes 41+5=8|4-1|+5=8, which simplifies to 3+5=83+5=8. This is true, so x=2x = 2 is a valid solution.For x=1x = 1, the original equation 2x1+5=4x|2x-1|+5=4x becomes 6611, which simplifies to 6622. This is not true, so x=1x = 1 is not a valid solution.

More problems from Evaluate absolute value expressions