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Solve each of the following equations \( \begin{array}{ll}\text { (a) }\left|x-\frac{1}{2}\right|=\frac{5}{2} & \text { (b) }|4 x-3|=|1+2 x| \\ \text { (c) }|x+1|-|x+2|=0 & \text { (d) }|x+1|=-3 \\ \text { (e) } \sqrt{2 x-1}=3 & \text { (f) } \sqrt{3 x-8}=\sqrt{x+2}\end{array} \)

Ask by Tyler Bright. in Zambia
Dec 11,2024

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(a) \( x = -2, 3 \) (b) \( x = \frac{1}{3}, 2 \) (c) \( x = -\frac{3}{2}, -1.5 \) (d) No real solution (e) \( x = 5 \) (f) \( x = 5 \)

Solution

Solve the equation \( |x-\frac{1}{2}|=\frac{5}{2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|x-\frac{1}{2}\right|=\frac{5}{2}\) - step1: Separate into possible cases: \(\begin{align}&x-\frac{1}{2}=\frac{5}{2}\\&x-\frac{1}{2}=-\frac{5}{2}\end{align}\) - step2: Solve the equation: \(\begin{align}&x=3\\&x=-2\end{align}\) - step3: Rewrite: \(x_{1}=-2,x_{2}=3\) Solve the equation \( |4x-3|=|1+2x| \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|4x-3\right|=\left|1+2x\right|\) - step1: Evaluate: \(\begin{align}&4x-3=1+2x\\&4x-3=-\left(1+2x\right)\end{align}\) - step2: Calculate: \(\begin{align}&x=2\\&x=\frac{1}{3}\end{align}\) - step3: Rewrite: \(x_{1}=\frac{1}{3},x_{2}=2\) Solve the equation \( |x+1|-|x+2|=0 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|x+1\right|-\left|x+2\right|=0\) - step1: Rewrite the expression: \(\left|x+1\right|=\left|x+2\right|\) - step2: Evaluate: \(\begin{align}&x+1=x+2\\&x+1=-\left(x+2\right)\end{align}\) - step3: Calculate: \(\begin{align}&x \in \varnothing \\&x=-\frac{3}{2}\end{align}\) - step4: Rearrange the terms: \(x=-\frac{3}{2}\) Solve the equation \( |x+1|=-3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\left|x+1\right|=-3\) - step1: The statement is false for any value of \(x:\) \(x \notin \mathbb{R}\) Solve the equation \( \sqrt{2x-1}=3 \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{2x-1}=3\) - step1: Find the domain: \(\sqrt{2x-1}=3,x\geq \frac{1}{2}\) - step2: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{2x-1}\right)^{2}=3^{2}\) - step3: Evaluate the power: \(2x-1=9\) - step4: Move the constant to the right side: \(2x=9+1\) - step5: Add the numbers: \(2x=10\) - step6: Divide both sides: \(\frac{2x}{2}=\frac{10}{2}\) - step7: Divide the numbers: \(x=5\) - step8: Check if the solution is in the defined range: \(x=5,x\geq \frac{1}{2}\) - step9: Find the intersection: \(x=5\) - step10: Check the solution: \(x=5\) Solve the equation \( \sqrt{3x-8}=\sqrt{x+2} \). Solve the equation by following steps: - step0: Solve for \(x\): \(\sqrt{3x-8}=\sqrt{x+2}\) - step1: Find the domain: \(\sqrt{3x-8}=\sqrt{x+2},x\geq \frac{8}{3}\) - step2: Raise both sides to the \(2\)-th power\(:\) \(\left(\sqrt{3x-8}\right)^{2}=\left(\sqrt{x+2}\right)^{2}\) - step3: Evaluate the power: \(3x-8=x+2\) - step4: Move the variable to the left side: \(3x-8-x=2\) - step5: Subtract the terms: \(2x-8=2\) - step6: Move the constant to the right side: \(2x=2+8\) - step7: Add the numbers: \(2x=10\) - step8: Divide both sides: \(\frac{2x}{2}=\frac{10}{2}\) - step9: Divide the numbers: \(x=5\) - step10: Check if the solution is in the defined range: \(x=5,x\geq \frac{8}{3}\) - step11: Find the intersection: \(x=5\) - step12: Check the solution: \(x=5\) The solutions to the given equations are as follows: (a) \( x = -2, 3 \) (b) \( x = \frac{1}{3}, 2 \) (c) \( x = -\frac{3}{2}, -1.5 \) (d) No real solution (e) \( x = 5 \) (f) \( x = 5 \)

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Let’s solve each of the equations one by one: **(a)** \(\left|x - \frac{1}{2}\right| = \frac{5}{2}\) This absolute value equation gives us two cases to consider: 1. \(x - \frac{1}{2} = \frac{5}{2}\) \(x = \frac{5}{2} + \frac{1}{2} = 3\) 2. \(x - \frac{1}{2} = -\frac{5}{2}\) \(x = -\frac{5}{2} + \frac{1}{2} = -2\) **Solutions:** \(x = 3\) or \(x = -2\) --- **(b)** \(|4x - 3| = |1 + 2x|\) We can set up four cases from the absolute values. **Case 1:** \(4x - 3 = 1 + 2x\) \(2x = 4 \Rightarrow x = 2\) **Case 2:** \(4x - 3 = - (1 + 2x)\) \(4x - 3 = -1 - 2x\) \(6x = 2 \Rightarrow x = \frac{1}{3}\) **Case 3:** \(- (4x - 3) = 1 + 2x\) \(-4x + 3 = 1 + 2x\) \(-6x = -2 \Rightarrow x = \frac{1}{3}\) (already found) **Case 4:** \(- (4x - 3) = - (1 + 2x)\) \(-4x + 3 = -1 - 2x\) \(-2 = 2x \Rightarrow x = -1\) **Solutions:** \(x = 2, x = \frac{1}{3}, x = -1\) --- **(c)** \(|x + 1| - |x + 2| = 0\) This simplifies to \(|x + 1| = |x + 2|\). There are two cases: 1. \(x + 1 = x + 2 \Rightarrow\) (not possible). 2. \(x + 1 = - (x + 2) \Rightarrow 2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\) **Solution:** \(x = -\frac{3}{2}\) --- **(d)** \(|x + 1| = -3\) An absolute value cannot equal a negative number, so there are no solutions here. **Solution:** No solution. --- **(e)** \(\sqrt{2x - 1} = 3\) Squaring both sides: \(2x - 1 = 9\) \(2x = 10 \Rightarrow x = 5\) **Solution:** \(x = 5\) --- **(f)** \(\sqrt{3x - 8} = \sqrt{x + 2}\) Squaring both sides: \(3x - 8 = x + 2\) \(2x = 10 \Rightarrow x = 5\) We must verify \(x = 5\) in the original equation: \(\sqrt{3(5) - 8} = \sqrt{5 + 2}\) \(\sqrt{15 - 8} = \sqrt{7}\) Both sides equal \(\sqrt{7}\). **Solution:** \(x = 5\) --- **Final Solutions Summary:** - (a) \(x = 3\) or \(x = -2\) - (b) \(x = 2, \frac{1}{3}, -1\) - (c) \(x = -\frac{3}{2}\) - (d) No solution - (e) \(x = 5\) - (f) \(x = 5\)

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