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Solve for x. one fifth times the absolute value of the quantity x minus 4 end quantity minus 3 equals 6 x = −49 and x = 41 x = −41 and x = 49 x = −19 and x = 11 x = −11 and x = 19
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Well first step would be to write out the equation: \[\large \frac{1}{5}\left| x-4 \right| -3=6\] \[\large \frac{1}{5}\left| x-4 \right| = 9 \] multiply by 5 \[\large 5*\frac{1}{5}\left| x-4 \right| = 9*5 \] you get: \[\large \left| x-4 \right| = 45 \] now just solve these two cases: \[\large x-4 = 45 ~and ~ x-4=-45 \]
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