Question

Name Date 34bral Kuta Software - Infinite Algebra 2 Solving Absolute Value Equations Solve each equation. 1) 13x1=9 2) 1-3r1=9 R=3 and 2-3 3x = $ 37 = 24 25 x=3 and x-3 Lom =30

Answer 1

$-5|3+4k|=-115$

start by dividing both sided by -5

$\frac{-5}{-5}|3+4k|=-\frac{115}{-5}$$|3+4k|=23$

when we have an absolute we obtain two equations, one for the positive value and one for the negative value

$\begin{gathered} 3+4k=23 \\ -(3+4k)=23 \end{gathered}$

solve each of them

$\begin{gathered} 3+4k=23 \\ 4k=23-3 \\ 4k=20 \\ k=\frac{20}{4}=5 \end{gathered}$

for the second equation

$\begin{gathered} -(3+4k)=23 \\ -3-4k=23 \\ -4k=23+3 \\ -4k=26 \\ k=-\frac{26}{4}=-6.5 \end{gathered}$

the solutions for k are -6.5 and 5