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# Help with absolute value

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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.

Aug 5, 2019

#1
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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 $$\large{\mathbf{|x+4| + |x-7| = |2x-1|}}$$.

Aug 6, 2019