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Solve the equation for all values of 
x.

|2x-10|-4=4x
Answer: 
x=

Solve the equation for all values of x x .\newline2x104=4x |2 x-10|-4=4 x \newlineAnswer: x= x=

Full solution

Q. Solve the equation for all values of x x .\newline2x104=4x |2 x-10|-4=4 x \newlineAnswer: x= x=
  1. Isolate absolute value: We have the equation 2x104=4x|2x - 10| - 4 = 4x. To solve for xx, we first need to isolate the absolute value expression on one side of the equation.\newlineAdd 44 to both sides of the equation to isolate the absolute value.\newline2x104+4=4x+4|2x - 10| - 4 + 4 = 4x + 4\newline2x10=4x+4|2x - 10| = 4x + 4
  2. Positive case equation: Now we have 2x10=4x+4|2x - 10| = 4x + 4. The absolute value equation can be split into two separate equations, one for the positive case and one for the negative case.\newlineFor the positive case:\newline2x10=4x+42x - 10 = 4x + 4
  3. Solve positive case: Solve the positive case equation from Step 22.\newlineSubtract 2x2x from both sides:\newline2x102x=4x+42x2x - 10 - 2x = 4x + 4 - 2x\newline10=2x+4-10 = 2x + 4\newlineNow, subtract 44 from both sides:\newline104=2x+44-10 - 4 = 2x + 4 - 4\newline14=2x-14 = 2x\newlineFinally, divide by 22:\newline14/2=2x/2-14 / 2 = 2x / 2\newline7=x-7 = x
  4. Negative case equation: For the negative case, we consider the expression inside the absolute value as negative: \newline(2x10)=4x+4- (2x - 10) = 4x + 4\newlineSimplify the negative sign:\newline2x+10=4x+4-2x + 10 = 4x + 4
  5. Solve negative case: Solve the negative case equation from Step 44.\newlineAdd 2x2x to both sides:\newline2x+10+2x=4x+4+2x-2x + 10 + 2x = 4x + 4 + 2x\newline10=6x+410 = 6x + 4\newlineNow, subtract 44 from both sides:\newline104=6x+4410 - 4 = 6x + 4 - 4\newline6=6x6 = 6x\newlineFinally, divide by 66:\newline66=6x6\frac{6}{6} = \frac{6x}{6}\newline1=x1 = x
  6. Check x=7x = -7: We have found two potential solutions for xx: x=7x = -7 and x=1x = 1. However, we must check these solutions in the original equation to ensure they do not result from extraneous solutions introduced by the absolute value.\newlineCheck x=7x = -7:\newline2(7)104=4(7)|2(-7) - 10| - 4 = 4(-7)\newline14104=28|-14 - 10| - 4 = -28\newline244=28| -24 | - 4 = -28\newline244=2824 - 4 = -28\newline202820 \neq -28\newlineThe solution x=7x = -7 does not satisfy the original equation, so it is an extraneous solution.
  7. Check x=1x = 1: Check x=1x = 1:
    |2(1)102(1) - 10| - 4=4(1)4 = 4(1)
    |2102 - 10| - 4=44 = 4
    |8-8| - 4=44 = 4
    84=48 - 4 = 4
    4=44 = 4
    The solution x=1x = 1 satisfies the original equation, so it is a valid solution.

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