How to solve $|4x+1|+|4-x|=x+19$? 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.
 A: 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.
A: Hint: You can have $4$ cases for two absolute values

*

*$+\mid+$

*$+ \mid -$

*$- \mid +$

*$- \mid -$


*

*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.
A: 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 $f(x)=|4x+1|+|4-x|- x-19.$





$x<-\frac14$
$-\frac14<x<4$
$x>4$




$f(x)$
$-6x-16$
$2x-14$
$4x-22$


$f(x)=0$
$x=-\frac83$
$$x=7\\\text{(contradiction; rejected)}$$
$x=\frac{11}2$



A: $$ 1) \begin {cases} x< -\frac {1} {4} \\-4x-1+4-x=x+19 \end {cases}$$
$$ 2) \begin {cases} -\frac {1} {4} \le x<4 \\4x+1+4-x=x+19 \end {cases}$$
$$ 3) \begin {cases} x \ge 4 \\ 4x+1+x-4=x+19 \end {cases}$$
