Problem of the Day

Solution

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For what values of $x$ is $|x-4|-|x-3|\geq7$?

Algebraic expressions with absolute value signs are difficult to work with, so one approach is to rewrite the expression in an equivalent way that does not require absolute value signs. To do this, we’ll need to break the problem into three cases: $x\leq3$, $3<x<4$, and $x\geq4$.

Case 1: $x\leq3$

If we are only considering values of $x$ that are less than or equal to 3, then we can rewrite our expression as $(4-x)-(3-x)\geq7$. This simplifies to $1\geq7$, which is definitely not true. This tells us that there is no solution to this inequality where $x\leq3$.

Case 2: $3<x<4$

In this case we can rewrite the expression as $(4-x)-(x-3)\geq7$. This simplifies to $-2x+7\geq7$ or $-2x\geq0$ or $x\leq0$. This may appear to provide a solution set, but keep in mind that the way we rewrote the expression in this case is ONLY valid if $x$ is between 3 and 4. None of the values in $x\leq0$ are between 3 and 4, so we cannot include any of them in our solution set based on the work from this case. So, no $x$ values between 3 and 4 solve this inequality.

Case 3: $x\geq4$

In this case we can rewrite the expression as $(x-4)-(x-3)\geq7$ which simplifies to $-7\geq7$. This is not true, which means there is no solution to this inequality where $x\geq4$.

All three of our cases failed to produce a solution, and all real numbers were accounted for between all three cases. Therefore, no solution to this inequality exists.