Given that the following equation does not have solutions in $\mathbb{R}$, find the value of $m$: $$| 2x - 5| + | 2x - 3 | = m $$
I try to resolve this equation on cases, when $| 2x - 5|$, $| 2x - 3 |$ are positive and negative. In two cases I can reduce $2x$ and then $m = \{ -8; 2 \}$, but I don't know if it is correct.
On
We give a solution that is more detailed than necessary to find the answer.
Cases sound good. There are $3$ of them: (i) $x\lt \frac{3}{2}$; $\frac{3}{2}\le x\le \frac{5}{2}$; (iii) $x\gt \frac{5}{2}$.
In Case (i), both items inside the absolute value signs are negative, so the absolute values are $3-2x$ and $5-2x$ respectively, giving sum $8-4x$.
In Case (ii), the absolute values are $2x-3$ and $5-2x$ respectively, for sum $2$.
In Case (iii), the sum of the absolute values is $4x-8$.
Draw the graph of $y=|2x-3|+|2x-5|$. For $x\lt \frac{3}{2}$ we are looking at $y=8-4x$, a line with negative slope. Then our function is steadily $2$ for a while. Finally, past $\frac{5}{2}$, we are on the line $y=4x-8$ with positive slope. So down, then steady at $2$, then up.
Thus our function does not take on any values $\lt 2$. It does take on every value $\ge 2$.
The values of $m$ for which the equation has no solution are all $m\lt 2$.
Substitute for a moment $y=2x$. Then you need to solve $|y-5|+|y-3|= m$.
Now for every $y \in [3,5]$ we have $|y-5|+|y-3|=2$, and so, if $m=2$,then every $y\in [3,5]$ is a solution.
For $y >5$ we have $m = |y-5|+|y-3| = (y-5)+ (y-3)=2y-8 \geq 2$, and so $y=m/2+4$ Similarly, for $y<3$ we have $m = |y-5|+|y-3| = (3-y)+ (5-y)=8-2y \geq 2$, and so $y=+4-m/2$
Therefore, for $m>2$ you have two solutions $x = y/2 \in \{m/4+2, 2-m/4\}$. For $m=2$ the solutions are $x \in [1.5,2.]$. For $m < 2$ there is no solution