Question

svlc algebra 1a - standard (15260)
solving absolute value equations
which equation has no solution?
|5 - 3x| = -8
|2x - 1| = 0
|-x - 3| = 5
|-x + 9| = 0

Explanation:

Step1: Recall absolute value property

The absolute value of a number, \(|a|\), is always non - negative, i.e., \(|a|\geq0\) for all real numbers \(a\).

Step2: Analyze each equation

• For the equation \(|5 - 3x|=-8\):

Since the absolute value of any expression \(|5 - 3x|\) must be greater than or equal to \(0\) (by the definition of absolute value), and \(- 8<0\), there is no real number \(x\) that can satisfy this equation.

• For the equation \(|2x - 1| = 0\):

We know that \(|a|=0\) when \(a = 0\). So, \(2x-1 = 0\), which gives \(2x=1\) and \(x=\frac{1}{2}\). So this equation has a solution.

• For the equation \(|-x - 3| = 5\):

By the definition of absolute value, \(|-x - 3|=5\) is equivalent to \(-x - 3=5\) or \(-x - 3=-5\).

• If \(-x - 3 = 5\), then \(-x=5 + 3=8\), so \(x=-8\).
• If \(-x - 3=-5\), then \(-x=-5 + 3=-2\), so \(x = 2\). So this equation has solutions.
• For the equation \(|-x + 9| = 0\):

Since \(|a| = 0\) when \(a=0\), we have \(-x + 9=0\), which gives \(x = 9\). So this equation has a solution.

Answer:

\(|5 - 3x|=-8\)