Need help with Absolute Value Equations for Algebra 1 Fan And Medal
posting them will help
Solve for x: −4|x + 5| = −16 x = 21/4 , x = −11/4 x = −1, x = 9 x = −1, x = −9 No solution
\(\bf -4|x + 5| = -16\implies |x + 5| =\cfrac{-16}{-4}\implies |x + 5| =4\\ \quad \\ |x + 5| =4\implies \begin{cases} +(x + 5) =4\\ \quad \\ \bf -(x + 5) =4 \end{cases}\) so you'd get 2 values for "x", just solve each scenario
I keep on getting -5 for both of them
\(\bf |x + 5| =4\implies \begin{cases} +(x + 5) =4\implies x+5=4\\ \quad \\ \bf -(x + 5) =4\implies -x-5=4 \end{cases} \)
neither one is -5 btw
I got -1 and 9 , Is that right ?
yeap
well.. -1 and ... -9
Can you help me with this one too please ? I just want to make sure that I understand it . Solve for x: |x + 2| + 16 = 14 x = −32 and x = −4 x = −4 and x = 0 x = 0 and x = 28 No solution
well. you do the same thing, you isolate the absolute value expression, and then solve the 2 scenarios \(\bf |x + 2| + 16 = 14\implies|x + 2| =14-16\implies|x + 2|=-2\\ \quad \\ |x + 2|=-2\implies \begin{cases} +(x + 2)=-2\\ \quad \\ \bf -(x + 2)=-2 \end{cases}\)
-4 and 0 ?
Two equations are listed below. Solve each equation and compare the solutions. Choose the statement that is true about both solutions. Equation 1 Equation 2 |5x − 6| = −41 |7x + 13| = 27 Equation 1 has more solutions than equation 2. Equation 1 and Equation 2 have the same number of solutions. Equation 2 has more solutions than Equation 1. The number of solutions cannot be determined.
well... based on the scenarios of the previous ones, what do you think? how many solutions do each one have?
2 ?
:)
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