help! (Solve for x)
Thanks!!!
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[tex]a|bx-c|=d\ \ \ \ |:a\neq0\\\\|bx-c|=\dfrac{d}{a}\to bx-c=\pm\dfrac{d}{a}\ \ \ \ |+c\\\\bx=c\pm\dfrac{d}{a}\ \ \ \ \ |:b\neq0\\\\x=\dfrac{c}{b}\pm\dfrac{d}{ab}\\\\x=\dfrac{ac}{ab}\pm\dfrac{d}{ab}\\\\\boxed{x=\dfrac{ac-d}{ab}\ \vee\ x=\dfrac{ac+d}{ab}}[/tex]