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w p davidson high school - credit bearing (tutor) algebra i with probab…

Question

w p davidson high school - credit bearing (tutor) algebra i with probability ic delange 1st quarter
what is the solution to ( 3|-3x + 9|=-18 )?
( x = -5 )
( x = 5 ) or ( x = 1 )
no solution

Explanation:

Step1: Isolate the absolute value expression

Divide both sides of the equation \(3|-3x + 9|=-18\) by \(3\).
We get \(|-3x + 9|=\frac{-18}{3}=-6\).

Step2: Analyze the property of absolute value

The absolute value of any real - number \(a\), denoted as \(|a|\), is defined as \(|a|=

$$\begin{cases}a, & a\geq0\\-a, & a < 0\end{cases}$$

\). By definition, \(|a|\geq0\) for all \(a\in R\).

Since the left - hand side \(|-3x + 9|\) represents the absolute value of an expression (which is always non - negative: \(|-3x + 9|\geq0\)) and the right - hand side is \(-6<0\).

Answer:

no solution