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3. Explain why the equation 6lx] + 25 = 15 has no solution. O When one solves, they arrive at a step where x is equal to a negative number. Since x can never be negative inside of the absolute value bars, there is no solution. O The statement is false. There is a solution. O When one solves, they arrive at a step where Ixl is equal to a negative number. Since I xl can never be negative, there is no solution. O When one solves, they arrive at a step where [x] is equal to a fraction that may not be represented as an integer. Since | xl must be an integer, there is no solution.

3. Explain why the equation 6lx] + 25 = 15 has no solution. O When one solves, they arrive at a step where x is equal to a negative number. Since x can never be class=

Answer :

To get the solution to the question, we will attempt to solve the absolute value equation:

[tex]6|x|+25=15[/tex]

Step 1: Subtract 25 from both sides of the equation

[tex]\begin{gathered} 6\lvert x\rvert+25-25=15-25 \\ 6|x|=-10 \end{gathered}[/tex]

Step 2: Divide both sides of the equation by 6

[tex]\begin{gathered} \frac{6|x|}{6}=-\frac{10}{6} \\ |x|=-\frac{5}{3} \end{gathered}[/tex]

Step 3: Recall that an absolute value is nonnegative, meaning it is either zero or positive. The output of the absolute value operator is never negative. Therefore, there is no solution

[tex]\mathrm{No\:Solution\:for}\:x\in \mathbb{R}[/tex]

ANSWER: The THIRD OPTION is correct.

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