Question

when an absolute - value sign appears in an equation or inequality, you should not use the addition, multiplication or division principle to simplify an expression inside the absolute - value sign. instead, think about what the absolute - value sign means.
the number inside the || sign could be either b or - b.
|x + 1| = 6
x + 1 = 6 or x + 1=-6
x = 5 or x=-7
to make sure that both 5 and - 7 are solutions, we can substitute each for x.
check: |5 + 1| = |6| = 6

-7 + 1

=

-6

= 6

solve each equation, and check your solutions.
|x + 5| = 7
check:
|\frac{x}{5}| = 2
check:
|3x + 4| = 10
check:
|x - 10| = 2
check:
|-3x| = 4
check:
|12x - 6| = 6
check:

Explanation:

Step1: Recall absolute - value property

If \(|a|=b\) (\(b\geq0\)), then \(a = b\) or \(a=-b\).

Step2: Solve each resulting equation

For example, when \(|x + 5|=7\), we have \(x+5 = 7\) and solve for \(x\) by subtracting 5 from both sides (\(x=7 - 5\)), and \(x+5=-7\) and solve for \(x\) by subtracting 5 from both sides (\(x=-7 - 5\)).

Step3: Check solutions

Substitute the found values of \(x\) back into the original absolute - value equations to verify they are valid. For \(|x + 5|=7\), when \(x = 2\), \(|2 + 5|=7\) and when \(x=-12\), \(|-12 + 5|=7\). Repeat this process for all equations.

Answer:

1. For \(|x + 5|=7\):

• \(x+5 = 7\) or \(x + 5=-7\).
• Solving \(x+5 = 7\), we get \(x=7 - 5=2\).
• Solving \(x + 5=-7\), we get \(x=-7 - 5=-12\).
• Check:
• When \(x = 2\), \(|2 + 5|=|7| = 7\).
• When \(x=-12\), \(|-12 + 5|=|-7| = 7\).

1. For \(|x-10| = 2\):

• \(x-10 = 2\) or \(x-10=-2\).
• Solving \(x-10 = 2\), we get \(x=2 + 10=12\).
• Solving \(x-10=-2\), we get \(x=-2+10 = 8\).
• Check:
• When \(x = 12\), \(|12-10|=|2| = 2\).
• When \(x = 8\), \(|8 - 10|=|-2| = 2\).

1. For \(|\frac{x}{5}|=2\):

• \(\frac{x}{5}=2\) or \(\frac{x}{5}=-2\).
• Solving \(\frac{x}{5}=2\), we get \(x=2\times5 = 10\).
• Solving \(\frac{x}{5}=-2\), we get \(x=-2\times5=-10\).
• Check:
• When \(x = 10\), \(|\frac{10}{5}|=|2| = 2\).
• When \(x=-10\), \(|\frac{-10}{5}|=|-2| = 2\).

1. For \(|-3x|=4\):

• \(-3x = 4\) or \(-3x=-4\).
• Solving \(-3x = 4\), we get \(x=-\frac{4}{3}\).
• Solving \(-3x=-4\), we get \(x=\frac{4}{3}\).
• Check:
• When \(x =-\frac{4}{3}\), \(|-3\times(-\frac{4}{3})|=|4| = 4\).
• When \(x=\frac{4}{3}\), \(|-3\times\frac{4}{3}|=|-4| = 4\).

1. For \(|3x + 4|=10\):

• \(3x+4 = 10\) or \(3x + 4=-10\).
• Solving \(3x+4 = 10\), we get \(3x=10 - 4 = 6\), then \(x = 2\).
• Solving \(3x + 4=-10\), we get \(3x=-10 - 4=-14\), then \(x=-\frac{14}{3}\).
• Check:
• When \(x = 2\), \(|3\times2+4|=|6 + 4|=|10| = 10\).
• When \(x=-\frac{14}{3}\), \(|3\times(-\frac{14}{3})+4|=|-14 + 4|=|-10| = 10\).

1. For \(|12x-6|=6\):

• \(12x-6 = 6\) or \(12x-6=-6\).
• Solving \(12x-6 = 6\), we get \(12x=6 + 6 = 12\), then \(x = 1\).
• Solving \(12x-6=-6\), we get \(12x=-6 + 6=0\), then \(x = 0\).
• Check:
• When \(x = 1\), \(|12\times1-6|=|12 - 6|=|6| = 6\).
• When \(x = 0\), \(|12\times0-6|=|-6| = 6\).