1) Find all real numbers x that satisfy the equation |x+4| + |x-7| = |2x-1|.
2) Find all x for which \(\left| x - \left| x-1 \right| \right| = \lfloor x \rfloor.\) Express your answer in interval notation.
+26367 Help with absolute value
1)
Find all real numbers x that satisfy the equation \(\large{\mathbf{|x+4| + |x-7| = |2x-1|}}\).
There are three discontinuities at \(x=-4,\ x=7,\ x=\dfrac{1}{2}\).
These digits divide the number range from \(-\infty\) to \(+\infty\) into the four ranges \(x<-4\), \(-4\leq x < \dfrac{1}{2}\), \(\dfrac{1}{2}\leq x <7\) and \(x\geq 7\).
\(\begin{array}{lrcll} x<-4 :& -(x+4)-(x-7)+(2x-1) &=& 0 \\ & -x-4-x+7+2x -1 &=& 0 \\ &6 &=& 0 \quad | \quad \text{no solution!} \\ -4\leq x < \dfrac{1}{2} : & (x+4)-(x-7)+(2x-1) &=& 0 \\ & x+4-x+7+2x-1 &=& 0 \\ & 2x+10 &=& 0 \\ & 2x &=& -10 \\ & x &=& -5 \quad | \quad \text{no solution! Because }-4\leq x < \dfrac{1}{2} \\ \\ \dfrac{1}{2}\leq x <7 : & (x+4)-(x-7)-(2x-1) &=& 0 \\ & x+4-x+7-2x+1 &=& 0 \\ &-2x+12 &=& 0 \\ & 2x &=& 12 \\ & \mathbf{x} &=& \mathbf{6} \\ \\ x\geq 7 : & (x+4)+(x-7)-(2x-1) &=& 0 \\ & x+4+x-7-2x+1 &=& 0 \\ & -2 &=& 0 \quad | \quad \text{no solution!} \\ \end{array}\)
\(\mathbf{x = 6}\) satisfy the equation \(\large{\mathbf{|x+4| + |x-7| = |2x-1|}}\).
