Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


The answer is $0$ or $-12$, but how would I solve it by algebraically solving it as opposed to sketching a graph?

$|x-5|=|2x+6|-1\\ (|x-5|)^2=(|2x+6|-1)^2\\ ...\\ 9x^4+204x^3+1188x^2+720x=0?$

share|cite|improve this question
One hard way is to determine for what values of $x$ the absolute value is positive or negative. For example, if $x\geq 5$ then $|x-5|=x-5$. Do the same to the other and take the intersections of intervals. – Sigur Jan 11 '13 at 17:54
up vote 9 down vote accepted

Consider different cases:

Case 1: $x>5$ In this case, both $x-5$ and $2x+6$ are positive, and you can resolve the absolute values positively. hence $$ x-5=2x+6-1 \Rightarrow x = -10, $$ which is not compatible with the assumption that $x>5$, hence no solution so far.

Case 2: $-3<x\leq5$ In this case, $x-5$ is negative, while $2x+6$ is still positive, so you get $$ -(x-5)=2x+6-1\Rightarrow x=0; $$ Since $0\in[-3,5]$, this is our first solution.

Case 3: $x\leq-3$ In this final case, the arguments of both absolute values are negative and the equation simplifies to $$ -(x-5) = -(2x+6)-1 \Rightarrow x = -12, $$ in agreement with your solution by inspection of the graph.

share|cite|improve this answer
I was really preparing an answer like it. Nice and understandable one. +1 – Babak S. Jan 11 '13 at 18:00

That's not the way, because you're creating extra solutions (new roots) when you put square on both sides. Just separate the options:

$$x-5=2x+6-1\\ 5-x=2x+6-1\\ x-5=-2x-6-1\\ 5-x=-2x-6-1$$

Solve all of them, and you have the solutions

Take into acount that once you get your solutions, you have to check if they're possible, for example, in the first one, you have supposed that bot things inside || are positive, so if you get something for which x-5 or 2x+6 is negative, then you have to throw it away

share|cite|improve this answer

Look for where the expressions inside the absolute values change sign: $x-5$ changes sign at $x=5$, and $2x+6$ changes sign at $x=-3$. Thus, when $x<-3$, $x-5$ and $2x+6$ are both negative, and the equation is $$-(x-5)=-(2x+6)-1\;.$$

When $-3\le x<5$, $x-5$ is negative and $2x+3$ is non-negative, so the equation is


And when $x\ge 5$, both expressions are non-negative, and the equation is


Thus, you need to solve

$$\left\{\begin{align*} &-x+5=-2x-7&&\text{when }x<-3\\ &-x+5=2x+5&&\text{when }-3\le x<5\\ &x-5=2x+5&&\text{when }x\ge 5\;. \end{align*}\right.\tag{1}$$

$(1)$ reduces to

$$\left\{\begin{align*} &x=-12&&\text{and }x<-3\\ &x=0&&\text{and }-3\le x<5\\ &x=-10&&\text{and }x\ge 5\;. \end{align*}\right.\tag{2}$$

The first two solutions in $(2)$ fall within the intervals on which they are valid; the third does not and therefore is not a solution.

share|cite|improve this answer

I think the sketch approach suggests a solution, which is to partition the domain.

enter image description here

Consider the function $f(x) = |x-5|-|2x+6|+1$. Look at the absolute value part and see that you can split the domain into $I_1= (-\infty, -3]$, $I_2 = (-3,5]$ and $I_3 = (5,\infty)$.

On $I_1$, $f(x) = x+12$, on $I_2$, $f(x) = -3x$, and on $I_3$, $f(x) = -(x+10)$.

Now look for solutions to $f(x) = 0$ on each of these intervals.

For example, on $I_3$, solving $-(x+10) = 0$ yields $x=-10$, but $-10 \notin I_3$, hence $f(x) \neq 0$ for $x \in I_3$.

share|cite|improve this answer
Nice illustration. – Babak S. Jan 11 '13 at 18:43

You have $|x-5|=|2x+6|-1$.

1)If $x\geq5$, $|x-5|=x-5$ and, $|2x+6|=2x+6$, so you have



$x=-10$ (but is not valid)

2)If $-3/2\leq x<5$, $|x-5|=-(x-5)$ and $|2x+6|=2x+6$, then





3)If $x<-3/2$, $|x-5|=-(x-5)$ and $|2x+6|=-(2x+6)$, then






So, the answer for $x$ is $x=0$ or $x=-12$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.