How do you solve #3+5abs(8-2x)=63#?

Answer 1

See the entire solution process below:

First, subtract #color(red)(3)# from each side of the equation to isolate the absolute value term while keeping the equation balanced:
#- color(red)(3) + 3 + 5abs(8 - 2x) = -color(red)(3) + 63#
#0 + 5abs(8 - 2x) = 60#
#5abs(8 - 2x) = 60#
Next, divide each side of the equation by #color(red)(5)# to isolate the absolute value function while keeping the equation balanced:
#(5abs(8 - 2x))/color(red)(5) = 60/color(red)(5)#
#(color(red)(cancel(color(black)(5)))abs(8 - 2x))/cancel(color(red)(5)) = 12#
#abs(8 - 2x) = 12#

The absolute value function takes any negative or positive term and transforms it to its positive form. Therefore, we must solve the term within the absolute value function for both its negative and positive equivalent.

First Solution

#8 - 2x = -12#
#-color(red)(8) + 8 - 2x = -color(red)(8) - 12#
#0 - 2x = -20#
#-2x = -20#
#(-2x)/color(red)(-2) = (-20)/color(red)(-2)#
#(color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2)) = 10#
#x = 10#

Option 2)

#8 - 2x = 12#
#-color(red)(8) + 8 - 2x = -color(red)(8) + 12#
#0 - 2x = 4#
#-2x = 4#
#(-2x)/color(red)(-2) = 4/color(red)(-2)#
#(color(red)(cancel(color(black)(-2)))x)/cancel(color(red)(-2)) = -2#
#x = -2#
The solution is: #x = 10# and #x = -2#
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Answer 2

To solve the equation (3 + 5|8 - 2x| = 63), follow these steps:

  1. Subtract 3 from both sides of the equation.
  2. Simplify the equation.
  3. Divide both sides by 5.
  4. Isolate the absolute value expression.
  5. Set up two equations: one where the expression inside the absolute value bars is positive and another where it's negative.
  6. Solve each equation separately.
  7. Check for extraneous solutions.

The final solutions will depend on the specific values of (x) that satisfy the equations formed when the expression inside the absolute value bars is positive and negative.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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