Bunuel wrote:
For how many integer values of x, is \(|x – 6| > |3x + 6|\)?
(A) 1
(B) 3
(C) 5
(D) 7
(E) Infinite
To solve an inequality where the absolute value of an expression is greater than (or less than) the absolute value of another expression, we need to find the values that make the absolute values of the two expressions equal to each other. That is, to solve |x – 6| > |3x + 6|, we first need to solve |x – 6| = |3x + 6|. Recall that if |A| = |B|, then either A = B (Case 1) or A = -B (Case 2).
Case 1: When \(x - 6 = 3x + 6\):
\(x - 6 = 3x + 6\)
\(-2x = 12\)
\(x = -6\)
Case 2: When \(x - 6 = -(3x + 6)\):
\(x - 6 = -(3x + 6)\)
\(x - 6 = -3x - 6\)
\(4x = 0\)
\(x = 0\)
These two values, -6 and 0, are the critical values for determining the solution to the given inequality. Now when we place these two values on the number line, we will see that they partition the number line into 3 intervals:
\(x < -6\)
\(-6 < x < 0\)
\(x > 0\)
For each of these intervals, we will pick a value in that interval (for example, for x < -6, we can pick -7) and test whether it satisfies the given inequality. If it does, then EVERY number in that interval will satisfy the inequality. If it doesn’t, then NO numbers in that interval will satisfy the inequality. Recall that our given inequality is |x – 6| > |3x + 6|. Let’s begin testing the intervals.
1) For \(x < -6\), let’s pick x = -7:
|-7 – 6| > |3(-7) + 6| ?
\(|-13| > |-15|\)?
\(13 > 15\) →
No!Thus, no numbers in the interval \(x < -6\)will satisfy \(|x – 6| > |3x + 6|.\)
2) For \(-6 < x < 0\), let’s pick x = -1:
\(|-1 – 6| > |3(-1) + 6|\)?
\(|-7| > |3|\) ?
\(7 > 3\)→
YesThus, every number in the interval \(-6 < x < 0\) will satisfy \(|x – 6| > |3x + 6|.\)
3) For \(x > 0\), let’s pick x = 1:
\(|1 – 6| > |3(1) + 6|\) ?
\(|-5| > |9|\) ?
\(5 > 9\) →
NoThus, no numbers in the interval \(x > 0\) will satisfy \(|x – 6| > |3x + 6|.\)As we can see, the only interval that satisfies \(|x – 6| > |3x + 6|\) is \(-6 < x < 0\). In other words, the solution to \(|x – 6| > |3x + 6|\) is \(6 < x < 0\). In this interval, there are 5 integers: -5, -4, -3, -2, and -1.
Answer:
CHi, Jeff - thank you for your explanation. For Case 2, why do we not also negate A in finding out the value of x?