Linear equations with absolute value

An absolute value is a function that does not change a non-negative number and makes a negative number positive. Absolute value can occur when solving linear equations.

Repetition

Recall how absolute value is calculated and written. An expression in absolute value is written using the vertical bars as follows: |a|. We say "the absolute value of the number a". It is then true that the absolute value of a non-negative number is the same number. Examples: |3| = 3, |64| = 64, |π| = π. However, the absolute value of a negative number is a positive number: $|-7|=7, |-\pi|=\pi, |-\frac12|=\frac12$. The absolute value of zero is zero.

A simple example

An example of a simple linear function with an absolute value is the equation: |x| = 3. The question now is when does the absolute value return three as the result? There will be two x - one positive and one negative; so the equation has two solutions. The first is equal to x₁ = 3, because |3| = 3. The second solution is equal to x₂ = −3, because |−3| = 3.

Let's try to modify the equation and make it harder. |2x + 1| = 5 Inside the absolute value, we already have a more complicated expression. But we can proceed in exactly the same way as before. What numbers are equal to five in absolute value? Again, 5 and −5. What does this imply? That the inside of the absolute value, the expression 2x + 1, must be equal to five or minus five. Only then does the equation have a solution. So we solve 2x + 1 = 5 and 2x + 1 = −5. We solve the equations as classical linear equations. We get:

$$\begin{eqnarray} 2x+1&=&5\\ 2x&=&4\\ x&=&2 \end{eqnarray}$$

And the second result:

$$\begin{eqnarray} 2x+1&=&-5\\ 2x&=&-6\\ x&=&-3 \end{eqnarray}$$

So we have two results: x₁ = 2 and x₂ = −3. We can try to fit them into the original equation. In the case of x = 2 we get: |2 · 2 + 1| = |4 + 1| = |5| = 5 and in the case of x = −3 we have: |2 · (−3)+1| = |−6 + 1| = |−5| = 5.

Removing the absolute value

But the previous procedure is hard to apply when you have a more complex equation. For example:

$$|4x+2|+|x-1|=6$$

Here we already have two expressions in absolute value and so we have to choose a different procedure. First a little simpler example: what will be true for the expression |x − 2|, if we take x from the interval (2, ∞)? The expression in absolute value will always be positive. For example, if we add a three, we get: 3 − 2 = 1. If we add a ten, we get: 10 − 2 = 8 etc. For any number in this interval, the expression is positive in absolute value. How does the absolute value then bend the value for us? Not at all. This is an important observation. If the expression below the absolute value is always positive, then we can remove the absolute value - it's unnecessary there. If we take x from the interval (2, ∞), then the equality of |x − 2| = x − 2 holds (i.e. we have removed the absolute value).

Now the opposite example - what if we take x from the interval (−∞, 2)? What will be the value of the expression in absolute value, i.e. the value of the expression x − 2? It will always be negative! Let's try substituting one: 1 − 2 = −1 or zero: 0 − 2 = −2. What will the absolute value do to this value? It will flip it into positive form, change the sign. We can change the sign by multiplying minus one, because 5 · (−1) = −5, but also −7 · (−1) = 7 and −15 · (−1) = 15.

So if we take x from the interval (−∞, 2), then the expression will always be negative in absolute value, and so the absolute value will work and change the sign of the expression - it will multiply it by −1. So we can write, for x of (−∞, 2), the equality is: |x − 2| = −(x − 2). Let's try putting a zero in there, for example. First in the expression with the absolute value: |0 − 2| = |−2| = 2 and now in the right-hand side: −(0 − 2) = −(−2) = 2.

As you can see, similar intervals allow us to split the expression with the absolute value into two different expressions that we can solve completely separately. This is convenient. The question now is, how to find these intervals?

The zero-point method

When did the expression x − 2 change sign? From the previous section, we know that the expression is negative throughout the interval (−∞, 2) and positive throughout the interval (2, ∞). What is it at the point x = 2? By addition, we find that 2 − 2 = 0 is zero. This is not a coincidence. The reason is best seen in the graph of the linear function:

The graph of a linear function is a line, it can only cross the x axis once, and when it crosses the x axis , it changes sign. Therefore, if a linear function crosses the x axis at a, then it will have a different sign in the (−∞, a) interval than in the (a, ∞) interval.

Therefore, if we are looking for intervals when the linear function is positive and when it is negative, we just need to find the so-called zero point, i.e. the point x₀, for which the linear function f has a functional value of zero: f(x₀) = 0. So if we are looking for intervals when the expression x − 2 is positive, the first equation we solve is x − 2 = 0. The solution is x = 2. We can then create the intervals: (−∞, 2) and (2, ∞). In these intervals we then continue to solve the equation, but with the absolute value removed.

Finally, back to the example...

Many lines ago we defined an example:

$$|4x+2|+|x-1|=6$$

How do we solve it? First, we find the zero points of each expression below the absolute value, i.e. for the expressions 4x + 2 and x − 1. The expression 4x + 2 has the zero point $x=-\frac12$ and the expression x − 1 has the zero point x = 1. Since we have two zero points, we will have three intervals in total. Each zero point will somehow split the interval (−∞, ∞). So in total we get the intervals $(-\infty, -\frac12)$, $\left<-\frac12, 1\right>$, and (1, ∞). We always choose the intervals to cover the entire interval (−∞, ∞).

In the next step, we need to determine whether the terms in the intervals are positive or negative. The table will be used for this purpose:

$$ \Large \begin{matrix} x&(-\infty, -\frac12)&\left<-\frac12, 1\right>&(1, \infty)\\\hline 4x+2&-&+&+\\ x-1&-&-&+ \end{matrix} $$

The minus sign − indicates the expression in the given interval is negative, the plus sign + indicates it is positive. How did we figure this out? We plugged in an arbitrary number from the interval. For example, in the middle interval, we could have substituted zero, which is computationally the easiest. Then for the expression 4x + 2 we get 4 · 0 + 2 = 2 and for x + 1 we get 0 − 1 = −1.

Hint: if the expression is negative in the first interval and positive in the next interval, the expression will be positive again in every possible subsequent interval. A linear function cannot change sign twice, see the picture above with the graph of the linear function.

At this point, we know in which intervals what expressions are positive or negative. Our next calculation has to be split into three steps, we have to solve separately in each interval.

The first interval, i.e. x we take from $(-\infty, -\frac12)$. In this interval both expressions are negative, so if we want to remove the absolute value, both expressions must change sign. So in this interval we solve the equation:

$$-(4x+2)-(x-1)=6$$

We just solve this equation simply by using equivalence adjustments:

$$\begin{eqnarray} -(4x+2)-(x-1)&=&6\\ -4x-2-x+1&=&6\\ -5x-1&=&6\\ -5x&=&7\\ x&=&-\frac{7}{5} \end{eqnarray}$$

Note that we have to check that the solution is from the interval we are in. We are now assuming that x is taken from the interval $(-\infty, -\frac12)$, so if we get a solution that is not from that interval, we cannot use it. Minus seven-fifths is in the interval $(-\infty, -\frac12)$, so this is a valid solution to the equation.

The second interval, x, is taken from $\left<-\frac12, 1\right>$. In this interval, the expression 4x + 2 is positive and the expression x − 1 is still negative. Thus we solve the equation

$$4x+2-(x-1)=6$$

Again, we simply modify:

$$\begin{eqnarray} 4x+2-(x-1)&=&6\\ 4x+2-x+1&=&6\\ 3x+3&=&6\\ 3x&=&3\\ x&=&1 \end{eqnarray}$$

Is the solution from the interval we are working in? Yes, it is. We have found another solution to the linear equation.

The third interval, the interval (1, ∞). In this interval, both terms are positive, so we can remove the absolute value without further change. So we solve the equation:

$$4x+2+x-1=6$$

We can easily solve this equation:

$$\begin{eqnarray} 4x+2+x-1=6\\ 5x+1=6\\ 5x=5\\ x=1 \end{eqnarray}$$

We got another result, the same as in the previous interval. However, is this a valid solution for the interval we are in? It is not! We are moving in the open interval (1, ∞), one is not part of this interval, so we cannot count this solution found in this third step as a valid solution.

A linear equation with an absolute value has two solutions, $K=\left\{-\frac75, 1\right\}$.

Graphical solution

Like the classical linear equation, this one can be solved graphically. On the left we have the function f(x) = |4x + 2|+|x − 1| and on the right g(x) = 6. Let's plot these two graphs - they intersect at two points. x-the coordinates of these points are the solutions to the equation.

Thus the solution set K contains the points $K=\left\{-\frac75, 1\right\}$.