# |x+3|+|x-1| = 5

|x+3|+|x-1| = 5

## Homework Equations

absolute value = distance from origin
distance between 2 points on a line = |x1-x2|

so |x+3| can be viewed as the distance from x to -3
and |x-1| can be viewed as the distance from x to 1

so we are trying to find a point x which has a distance from -3 and a distance from 1 that combine to equal 5...

## The Attempt at a Solution

The options that add up to equal 5 are (0+5), (5+0), (2+3), (3+2), (1+4), (4+1)

if |x+3| = 0, then x=-3.... that gives 4 instead of 5
if |x+3| = 5, then x=2 or -8.... that gives 6 and 14 instead of 5
if |x+3| = 3, then x=0 or -6... that gives 4 and 10 instead of 5
if |x+3| = 2, then x=-1 or -5... that gives 4 and 8 instead of 5
if |x+3| = 1, then x=-2 or -4... that gives 4 and 6 instead of 5
if |x+3| = 4, then x=1 or -7... that gives 4 and 12 instead of 5

^ not only is that a really inefficient way to try to solve it. but I also couldn't get the answer! :(

can someone please tell me a way to solve these problems?

edit: I just realized I neglected all the options where x is not an integer...

Last edited:

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Mark44
Mentor

|x+3|+|x-1| = 5

## Homework Equations

absolute value = distance from origin
distance between 2 points on a line = |x1-x2|

so |x+3| can be viewed as the distance from x to -3
and |x-1| can be viewed as the distance from x to 1

so we are trying to find a point x which has a distance from -3 and a distance from 1 that combine to equal 5...

## The Attempt at a Solution

The options that add up to equal 5 are (0+5), (5+0), (2+3), (3+2), (1+4), (4+1)

if |x+3| = 0, then x=-3.... that gives 4 instead of 5
if |x+3| = 5, then x=2 or -8.... that gives 6 and 14 instead of 5
if |x+3| = 3, then x=0 or -6... that gives 4 and 10 instead of 5
if |x+3| = 2, then x=-1 or -5... that gives 4 and 8 instead of 5
if |x+3| = 1, then x=-2 or -4... that gives 4 and 6 instead of 5
if |x+3| = 4, then x=1 or -7... that gives 4 and 12 instead of 5

^ not only is that a really inefficient way to try to solve it. but I also couldn't get the answer! :(

can someone please tell me a way to solve these problems?
So the approach is not only ineffecient, but ineffective as well.

Divide the number line into three intervals: (-∞, -3), (-3, 1), and (1, ∞)

If x is in (-∞, -3), the equation can be written as -(x + 3) + -(x - 1) = 5. Can you solve this?

Now look at the resulting equation in each of the other two intervals.

The basic idea I am using is that |x - a| = -(x -a) if x < a, and |x - a| = x - a if x > a.

SammyS
Staff Emeritus
Homework Helper
Gold Member

|x+3|+|x-1| = 5

## Homework Equations

absolute value = distance from origin
distance between 2 points on a line = |x1-x2|

so |x+3| can be viewed as the distance from x to -3
and |x-1| can be viewed as the distance from x to 1

so we are trying to find a point x which has a distance from -3 and a distance from 1 that combine to equal 5...

## The Attempt at a Solution

The options that add up to equal 5 are (0+5), (5+0), (2+3), (3+2), (1+4), (4+1)

if |x+3| = 0, then x=-3.... that gives 4 instead of 5
if |x+3| = 5, then x=2 or -8.... that gives 6 and 14 instead of 5
if |x+3| = 3, then x=0 or -6... that gives 4 and 10 instead of 5
if |x+3| = 2, then x=-1 or -5... that gives 4 and 8 instead of 5
if |x+3| = 1, then x=-2 or -4... that gives 4 and 6 instead of 5
if |x+3| = 4, then x=1 or -7... that gives 4 and 12 instead of 5

^ not only is that a really inefficient way to try to solve it. but I also couldn't get the answer! :(

can someone please tell me a way to solve these problems?
"not only is that a really inefficient way to try to solve it... " True !

An instructive way to solve it (in my opinion) is to graph y = |x+3|+|x-1| .

Alternatively,
consider the piecewise definition of the absolute value.

Consider separately the graphs: y = |x+3|, and y = |x-1| .​

So the approach is not only ineffecient, but ineffective as well.

Divide the number line into three intervals: (-∞, -3), (-3, 1), and (1, ∞)

If x is in (-∞, -3), the equation can be written as -(x + 3) + -(x - 1) = 5. Can you solve this?

Now look at the resulting equation in each of the other two intervals.

The basic idea I am using is that |x - a| = -(x -a) if x < a, and |x - a| = x - a if x > a.
Ok great!

so if x <= -3 ........... -3-x+1-x = 5 ; (ans: x=-7/2)

if 1 >= x >= -3 ............ x+3+1-x = 5 (0=1 ) I realize that this indicates there is no solution for x in this interval but is there anything else to be learned from the result or should I just accept it as indicating no answer in the interval?

if x > 1 ....... x+3+x-1 = 5 ; (ans: x=3/2)

SammyS
Staff Emeritus
if 1 >= x >= -3 ............ x+3+1-x = 5 (0=1 ) I realize that this indicates there is no solution for x in this interval but is there anything else to be learned from the result or should I just accept it as indicating no answer in the interval?