1. **Problem statement:** We have two functions \(y = x - |x - 1| + 2\) and \(y = |x|\) to plot on the same graph. Also, solve the inequality \(|x| + |x - 2| \leq x + 4\) and find the range of \(x\) values that satisfy it. 2. **Understanding the functions:** - The function \(y = x - |x - 1| + 2\) involves an absolute value expression \(|x - 1|\). - The function \(y = |x|\) is the absolute value of \(x\). 3. **Breaking down \(y = x - |x - 1| + 2\):** - For \(x \geq 1\), \(|x - 1| = x - 1\), so \[y = x - (x - 1) + 2 = x - x + 1 + 2 = 3\] - For \(x < 1\), \(|x - 1| = 1 - x\), so \[y = x - (1 - x) + 2 = x - 1 + x + 2 = 2x + 1\] 4. **Summary of \(y = x - |x - 1| + 2\):** \[ y = \begin{cases} 2x + 1 & x < 1 \\ 3 & x \geq 1 \end{cases} \] 5. **Function \(y = |x|\):** \[ y = \begin{cases} x & x \geq 0 \\ -x & x < 0 \end{cases} \] 6. **Inequality to solve:** \[|x| + |x - 2| \leq x + 4\] 7. **Consider intervals based on absolute values:** - For \(x < 0\): \(|x| = -x\), \(|x - 2| = 2 - x\) \[-x + (2 - x) \leq x + 4 \implies -2x + 2 \leq x + 4\] Move terms: \[-2x + 2 \leq x + 4 \implies -2x - x \leq 4 - 2 \implies -3x \leq 2\] Divide by -3 (flip inequality): \[x \geq -\frac{2}{3}\] But this contradicts \(x < 0\), so solution in this interval is \(-\frac{2}{3} \leq x < 0\). - For \(0 \leq x < 2\): \(|x| = x\), \(|x - 2| = 2 - x\) \[x + (2 - x) \leq x + 4 \implies 2 \leq x + 4 \implies x \geq -2\] This is always true for \(0 \leq x < 2\), so entire interval satisfies. - For \(x \geq 2\): \(|x| = x\), \(|x - 2| = x - 2\) \[x + (x - 2) \leq x + 4 \implies 2x - 2 \leq x + 4 \implies 2x - x \leq 4 + 2 \implies x \leq 6\] So solution is \(2 \leq x \leq 6\). 8. **Combine all intervals:** \[ \left[-\frac{2}{3}, 0\right) \cup [0, 2) \cup [2, 6] = \left[-\frac{2}{3}, 6\right] \] 9. **Final answer:** The solution set for \(x\) is \(\boxed{\left[-\frac{2}{3}, 6\right]}\).