Question

Find all numbers x satisfying the given equation.
abs(x-3) + abs(x-4) = 1

Answer 1

if you are looking for integer solutions then, as abs is always positive, you can only make 1 in these ways:

0 + 1 = 1
1 + 0 = 1

that should help you find suitable values for x

Answer 2

how should i start this?

Answer 3

from the 1st equation I listed (i.e. 0 + 1 = 1) you get:\[|x-3|=0\]and\[|x-4|=1\]can you solve that 1st?

Answer 4

There are three intervals:- \( x > 4\), \(3 \le x \le 4\) and \(x < 3\)

Note that \(|x| = x,\;\;\; x \ge 0\)
\(= -x, \;x \le 0\)

So, in the interval \(x > 4\), \(x-4 > 0\) and \(x -3 > 0\)
i.e. \(x-3 + x-4 = 1 \implies 2x = 8\) or \(x = 4\)

In the interval \(3 \le x \le 4\), \(x-4 < 0\) and \(x - 3 > 0\)
i.e. \(x-3 + 4-x = 1 \implies 1 = 1\) which is always true.

Thus all values of \(x \in [3,4]\) satisfy the equation.

In the final interval \(x < 3\), \(x - 4 < 0\) and \(x - 3 < 0\)
i.e. \(3-x + 4-x = 1 \implies 2x = 6\) or \(x=3\)

So, the final solution is \(x \in [3,4]\)
Any value between 3 and 4 (both inclusive) will work.