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What are the values of 𝑥𝑥 that satisfy the equation |x7x-7|3-3|x2x-2|+4+4|x+8x+8|++|xx|==441441 ?

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Q. What are the values of 𝑥𝑥 that satisfy the equation |x7x-7|3-3|x2x-2|+4+4|x+8x+8|++|xx|==441441 ?
  1. Understand Absolute Value Properties: Understand the properties of absolute value. Absolute value equations can have two possible cases for each absolute value expression: one where the inside is non-negative (and the absolute value acts like the identity function), and one where the inside is negative (and the absolute value acts like the negation function). We will need to consider these cases to solve the equation.
  2. Set Up Critical Point Cases: Set up different cases based on the critical points of the absolute value expressions.\newlineThe critical points are x=7x = 7, x=2x = 2, x=8x = -8, and x=0x = 0. These points divide the number line into intervals where the expressions inside the absolute values do not change sign. We will need to solve the equation in each interval separately.
  3. Solve Equations in Intervals: Solve the equation in each interval.\newlineWe will start with the interval x>7x > 7, where all the expressions inside the absolute values are non-negative. The equation simplifies to:\newlinex73(x2)+4(x+8)+x=441x - 7 - 3(x - 2) + 4(x + 8) + x = 441
  4. Simplify Equation for x>7x > 7: Simplify the equation for the interval x>7x > 7. Combine like terms: x73x+6+4x+32+x=441x - 7 - 3x + 6 + 4x + 32 + x = 441 (13+4+1)x+(67+32)=441(1 - 3 + 4 + 1)x + (6 - 7 + 32) = 441 3x+31=4413x + 31 = 441
  5. Solve for xx in x>7x > 7: Solve for xx in the interval x>7x > 7.
    Subtract 3131 from both sides:
    3x=441313x = 441 - 31
    3x=4103x = 410
    Divide by 33:
    x=4103x = \frac{410}{3}
    x136.67x \approx 136.67
  6. Check Solution in x>7x > 7: Check if the solution x136.67x \approx 136.67 is in the interval x>7x > 7.\newlineSince 136.67136.67 is greater than 77, it is in the interval we are considering.
  7. Repeat Steps for Other Intervals: Repeat steps 33 to 66 for the other intervals (xx between 22 and 77, xx between 8-8 and 22, xx between 00 and 8-8, and x<8x < -8).\newlineHowever, since we have already found a solution in the interval 2200, we need to check if there are any other solutions in the other intervals. If there are, we will include them in our final answer.
  8. Verify Solution in Original Equation: Verify if the solution x136.67x \approx 136.67 satisfies the original equation.\newlinePlug x136.67x \approx 136.67 back into the original equation:\newline136.6773136.672+4136.67+8+136.67=441|136.67 - 7| - 3|136.67 - 2| + 4|136.67 + 8| + |136.67| = 441\newline129.673(134.67)+4(144.67)+136.67=441129.67 - 3(134.67) + 4(144.67) + 136.67 = 441\newline129.67404.01+578.68+136.67441129.67 - 404.01 + 578.68 + 136.67 \approx 441\newlineThe left side does not equal 441441, which means there is a math error in our previous steps.

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