How to solve $|4x+1|+|4-x|=x+19$?

Question

I need some help solving the equation $|4x+1|+|4-x|= x+19$.

I know I have to divide it into different cases. I thought I had two cases.

First case: $4x+1 = -2x-15$. The first case is correct, and I get the correct solution.

However, the second case I thought of was: $4x+1 = 2x+15$. However, this gives me $x=7$, which is the incorrect solution. I am not quite sure how to find the other solution.

I tried googling it, but mostly it only shows examples of equations using just one absolute value expression.

Answer 1

Hint: You can have $4$ cases for two absolute values

1. $+\mid+$
2. $+ \mid -$
3. $- \mid +$
4. $- \mid -$

• First case is when $x \ge -1/4$ and $x \le 4$.
• Second case is when $x \ge -1/4$ and $x > 4$.
• Third case is when $x < -1/4$ and $x \le 4$.
• Fourth case is when $x < -1/4$ and $x > 4$.

Consider each case, as you have tried in your attempt. After finishing each case, you should check if the root you found belongs to the range of that case.

Answer 2

An alternative purely arithmetic method, if you don't want to discuss all the inequalities, is to use variables for the signs.

$a(4x+1)+b(4-x)=x+19\iff x=\dfrac{19-a-4b}{4a-b-1}\ $ for $\ a,b\in \{-1,+1\}$

You get $4$ possible values and plugging them eliminates two of them.

Answer 3

Here's a "critical-points" method, which is more efficient than taking cases (in general, there are $2^m$ separate cases, where $m$ is the number of absolute-value terms). $$|4x+1|+|4-x|= x+19$$

Let $f(x)=|4x+1|+|4-x|- x-19.$

	$x<-\frac14$	$-\frac14<x<4$	$x>4$
$f(x)$	$-6x-16$	$2x-14$	$4x-22$
$f(x)=0$	$x=-\frac83$	$$x=7\\\text{(contradiction; rejected)}$$	$x=\frac{11}2$

Answer 4

$$ 1) \begin {cases} x< -\frac {1} {4} \\-4x-1+4-x=x+19 \end {cases}$$

$$ 2) \begin {cases} -\frac {1} {4} \le x<4 \\4x+1+4-x=x+19 \end {cases}$$

$$ 3) \begin {cases} x \ge 4 \\ 4x+1+x-4=x+19 \end {cases}$$