Absolute Value Linear Inequalities

Absolute value linear inequalities are inequalities containing one or more absolute expressions of linear terms.

A simple example of absolute value linear inequalities would be $\lvert ax+b\rvert>c.$ The universal way to solve these is to divide the absolute value expression into two cases: when the term inside is positive, or negative. The case when the expression is exactly zero can be included in either one of the two cases. Let's see how it works through an example.

When the sign is $<,$ e.g. $|f(x)|< g( x ),$ the result will be $- g(x) < f(x) < +g(x).$

When the sign is $>,$ e.g. $| f(x) |> g( x ),$ the result will be $f(x)< - g(x)$ and $f(x) > +g(x).$

The red portions are our answers.

Solve for $x:$ $\lvert x-1\rvert<4.$

---

We know that $x-1\geq0$ when $x\geq1,$ and $x-1<0$ when $x<1.$ Thus we divide it into those two cases.

(i) When $x\geq1,$ the term $\lvert x-1\rvert$ is equivalent to $x-1.$ Hence we have $$\begin{aligned} \lvert x-1\rvert&<4\\ x-1&<4\\ \Rightarrow x&<5. \end{aligned}$$ However, since we have assumed that $x\geq1,$ the values of $x$ that satisfy the given inequality under this condition is $1\leq x<5.$

(ii) When $x<1,$ the term $\lvert x-1\rvert$ is equivalent to $-x+1.$ Hence we have $$\begin{aligned} \lvert x-1\rvert&<4\\ -x+1&<4\\ \Rightarrow -3&<x. \end{aligned}$$ However, since we have assumed that $x<1,$ this time we have $-3<x<1.$

After solving each case, we find the union of the solution sets of each case, which is the answer. Therefore our answer is $-3<x<5.$ $_\square$

If there are more than one absolute value expression (e.g. $2\lvert x+1\rvert-3\lvert x-5\rvert<7$), then we do the same thing for each expression. When we have $n$ absolute value expressions, we will have to divide it into $n+1$ cases (unless an expression is the multiple of another one, like $\lvert x-1\rvert$ and $\lvert 2x-2\rvert$). Then the union of the solution sets of each case is the answer.

Inequalities with a single absolute value expression can be approached in a slightly easier way. Note that $\lvert x\rvert>5$ is equivalent to $x>5$ or $x<-5,$ and $\lvert x\rvert<5$ is equivalent to $-5<x<5.$ Using this principle, the example above can be solved as thus:

Solve for $x:$ $\lvert x-1\rvert<4.$

---

Using the principle mentioned above, we have

$$\begin{aligned} -4<x-1&<4\\ \Rightarrow -3<x&<5.\ _\square \end{aligned}$$

Now let's try solving some more example problems.

Solve for $x:$ $2\lvert x-2 \rvert + x >8.$

---

Solution 1:

If $x<-2,$ then $2\times (-(x-2))+x>8,$ which implies $x<-4.$ If $x\ge -2,$ then $2\times (x-2)+x>8,$ which implies $x>4.$ Hence, the solution is $x<-4$ or $x>4.$ $_\square$

Solution 2:

Observe that the given inequality is equivalent to $\lvert x-2\rvert>\frac{8-x}{2}.$
Then, if $x\ge 2,$ we have $$\begin{aligned} x - 2 &>\dfrac{8 - x}2\\ x &>4 - \dfrac x 2 +2\\ \dfrac{ 3x} 2 &>6\\ x&>4. \qquad (\text{this satisfies the condition }x\ge 2) \end{aligned}$$ If $x< 2,$ we have $$\begin{aligned} -(x - 2) &>\dfrac{8 - x}2\\ -x&>4 - \dfrac x 2 -2\\ - \dfrac x 2&>2\\ x&<- 4. \qquad (\text{this satisfies the condition }x<-2) \end{aligned}$$ Therefore, the answer is $x>4$ or $x<-4.$ $_\square$

How many integers $x$ satisfy $\lvert 3-4x \rvert \le 5?$$$$$

$$\begin{array}{c}\\ \\ (a)\ 1 &&(b)\ 2 &&(c)\ 3 &&(d)\ 4 &&(e)\ 5\end{array}$$

---

We can rewrite the inequality as follows to obtain:

$$\begin{aligned} \lvert 3-4x \rvert &\le 5 \\ -5 &\le 3-4x \le 5 \\ -8 &\le -4x \le 2 \\ -2 &\le 4x \le 8 \\ -\frac{1}{2} &\le x \le 2. \end{aligned}$$

Thus, three integers $0, 1, 2$ satisfy the inequality, so the answer is $(c).$ $_\square$

How many integer solutions $x$ does the following inequality have:$$$$

$$2\lvert x-1 \rvert +3\lvert x+1 \rvert <14?$$ $$$$$$$$ $$\begin{array}{c}&(a)\ 1 &&&(b)\ 2 &&&(c)\ 3 &&&(d)\ 4 &&&(e)\ 5\end{array}$$

---

If $x<-1,$ we have $$\begin{aligned} 2\times (-(x-1))+3\times (-(x+1))&<14\\ -5x-1&<14\\ x&>-3\\ -3&<x<-1. \qquad (\text{since } x<-1) \end{aligned}$$

If $-1\le x< 1,$ we have $$\begin{aligned} 2\times (-(x-1))+3\times (x+1)&<14\\ x+5&<14\\ x&<9\\ -1&\le x< 1. \qquad (\text{since } -1\le x< 1) \end{aligned}$$

If $x\ge 1,$ we have $$\begin{aligned} 2\times (x-1)+3\times (x+1)&<14\\ 5x+1&<14\\ x&<\frac{13}{5}\\ 1&\le x< \frac{13}{5}. \qquad (\text{since } x\ge 1) \end{aligned}$$

Hence, from the above three cases we have $-3<x<\frac{13}{5},$ which implies that five integers $-2, -1, 0, 1, 2$ satisfy the given inequality. So our answer is $(e).$ $_\square$

Let the solutions to the inequality $\lvert 2x+p \rvert \ge 5$ be $x\ge q$ or $x\le -1,$ where $p$ and $q$ are constants. What is $p+q?$

---

Observe that $\lvert 2x+p \rvert \ge 5$ implies $2x+p\ge 5$ or $2x+p\le -5,$ which is equivalent to

$$\begin{array}{c}&x\ge \frac{5-p}{2} &\text{or} &x\le \frac{-5-p}{2}.\end{array}$$

Then since we are given $x\ge q$ or $x\le -1,$ we have the following two equations:

$$\begin{array}{c}&\frac{5-p}{2}=q &\text{and} &\frac{-5-p}{2}=-1,\end{array}$$

which implies $p=-3$ and $q=4.$ Therefore, $p+q=-3+4=1.$ $_\square$