How do you solve $7-3abs(4d-7)=4$ ?

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Answer 1 · 2015-08-02T18:17:05

$d=2$ or $d=3/2$

$7 - 3|4d-7| = 4$

Start by isolating the modulus on one side of the equation

$color(red)(cancel(color(black)(7))) - color(red)(cancel(color(black)(7))) - 3|4d-7| = 4 - 7$

$(color(red)(cancel(color(black)(-3))) * |4d-7|)/color(red)(cancel(color(black)(-3))) = (-3)/(-3)$

$|4d-7| = 1$

Now, this equation will produce two values for $d$ , depending on which condition is true

$|4d-7| = 4d-7$

and the equation becomes

$4d-7 = 1 => d = 8/4 = color(green)(2)$

$|4d-7| = -(4d-7) = -4d + 7$

This will get you

$-4d+7 = 1 => d = 6/4 = color(green)(3/2)$

How do you solve 7-3abs(4d-7)=47-3abs(4d-7)=4?