How would one solve the following equation over $\mathbb R$?
$$|x+2| = 4 + |x-7|$$
How would one solve the following equation over $\mathbb R$?
$$|x+2| = 4 + |x-7|$$
We can write it as $|x+2|-|x-7| = 4$
Using $$|x+2|-|x-7| = \left\{\begin{matrix} -(x+2)+(x-7)&\;, x\leq -2 \\\\ x+2+x-7 & \;, -2<x \leq 7\\\\ x+2-(x-7)& , x>7 \end{matrix}\right.$$
$\bullet\; $ If $x\leq -2\;,$ Then equation convert into $-x-2+x-7=4\Rightarrow -9=4\;\; \text{(False)}$
$\bullet\; $ If $-2<x \leq 7\;,$ Then equation convert into $x+2+x-7=4\Rightarrow \displaystyle x=\frac{9}{2}\;\; \text{(True)}$
$\bullet\; $ If $x\geq 7\;,$ Then equation convert into $x+2-x+7=4\Rightarrow 9=4\;\; \text{(False)}$
So we get only one solution which is $\displaystyle x = \frac{9}{2}$
Break it where $x\leq -2$ and $-2\leq x\leq7$ and $x\geq 7$ forexample for first one $$|x+2| - 4 - |x-7|=-(x+2) - 4 -(- (x-7))=0\Rightarrow -2-4-7=0 $$ isnot acceptable. for second interval $$|x+2| - 4 - |x-7|=(x+2) - 4 -(- (x-7))=0\Rightarrow 2x+2-4-7=0 \Rightarrow 2x-9=0 \Rightarrow x=\frac{9}{2}$$ which is acceptable.
for third interval $$|x+2| - 4 - |x-7|=(x+2) - 4 - (x-7)=0 \Rightarrow 2-4+v=0$$ whichisnot acceptable. thus answer is only $x=\frac{9}{2}$.
$$|x+2| = 4+ |x-7|$$ Statement 1: if $|x+2| > 0$, then $|x+2| = x+2$
Statement 2: if $|x+2| < 0$, then $|x+2| = -x-2$
Statement 3: if $|x-7| > 0$, then $|x-7| = x-7$
Statement 4: if $|x-7| < 0$, then $|x-7| = -x+7$
if Statement 1 and Statement 3, we will have: $$x+2 = 4 + x - 7$$ $$2 = 4 - 7$$
thus, no answer, try Statement 1 and Statement 4
$$x+2 = 4 -x+7$$ $$2x=9, x= \frac 92$$
First answer, $x= \frac 92$ or $x=4.5$
Do the same thing for Statement 2 and Statement 3, Statement 2 and Statement 4. You get $x=\frac 12$ or $x=0.5$ Statement 2 and Statement 4, but no answer for Statement 2 and Statement 3
Now check your answer, $|4.5+2| = 4 + |4.5-7|$, $6.5 = 4 + 2.5$
$|0.5+2| = 4 + |0.5+7|$, $2.5 \ne 4+7.5$
thus, the answer is 4.5
So the only solution is $ x=\frac{9}{2}.$ This method is useful when you will solve inequalities with absolute value.