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If \(x\) is a non-negative integer and \(|6 - |x + 2|| = 10,\) then find the number of values of \(x\) that satisfy the

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by Gmat_mission » Tue Nov 10, 2020 7:56 am

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If \(x\) is a non-negative integer and \(|6 - |x + 2|| = 10,\) then find the number of values of \(x\) that satisfy the given absolute value inequality?

a. 0
b. 1
c. 2
d. 3
e. 4

Answer: B

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Re: If \(x\) is a non-negative integer and \(|6 - |x + 2|| = 10,\) then find the number of values of \(x\) that satisfy

by deloitte247 » Thu Nov 12, 2020 9:36 pm
$$\left|6-\right|x+2\left|\right|=10$$
$$6-\left|x+2\right|=\pm10$$
$$If\ 6-\left|x+2\right|=10,\ then\ \left|x+2\right|=-4\ or\ 16$$
Since the absolute value is not negative
$$\ \left|x+2\right|\ne-4\ $$
$$\ So\ for\ \left|x+2\right|=16$$
$$x+2=\pm16$$
$$x+2=16\ \ \ or\ x+2=-16$$
$$x=14\ \ \ or\ x=-18$$
Since x is non-negative, x = 14
$$If\ 6-\left|x+2\right|=10,\ then\ \left|x+2\right|=16$$
Following the results derived from
$$6-\left|x+2\right|=10,\ x=14$$
Only 1 value satisfies the absolute value; hence option B is the correct answer.
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