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3. Solve the following inequality and put your answer in interval notation. \[ \left|\frac{x+1}{2}-\frac{x-1}{3}\right|<1 \text {. } \] 4. Solve the equation \( \sqrt{x+2}+\sqrt{x-2}=\sqrt{4 x-2} \). S. .

Ask by Fuentes Watkins. in Malawi
Mar 09,2025

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Answer

**Problem 3** The solution to the inequality \( \left|\frac{x+1}{2}-\frac{x-1}{3}\right|<1 \) is: \[ (-11,\,1) \] **Problem 4** The solution to the equation \( \sqrt{x+2}+\sqrt{x-2}=\sqrt{4x-2} \) is: \[ x=\frac{5}{2} \]

Solution

**Problem 3** We start with the inequality \[ \left|\frac{x+1}{2}-\frac{x-1}{3}\right|<1. \] 1. **Combine the fractions:** Find a common denominator, which is \(6\): \[ \frac{x+1}{2}-\frac{x-1}{3}=\frac{3(x+1)-2(x-1)}{6}. \] 2. **Simplify the numerator:** Expand and simplify: \[ 3(x+1)-2(x-1)=3x+3-2x+2=x+5. \] So the expression becomes \[ \frac{x+5}{6}. \] 3. **Rewrite the inequality:** The inequality is now \[ \left|\frac{x+5}{6}\right|<1. \] 4. **Remove the absolute value:** This implies \[ \left|x+5\right|<6. \] which can be written as the compound inequality \[ -6 < x+5 < 6. \] 5. **Solve for \(x\):** Subtract \(5\) from all parts: \[ -6-5 < x < 6-5, \] \[ -11 < x < 1. \] 6. **Interval notation:** The solution in interval notation is \[ (-11,\,1). \] --- **Problem 4** We solve the equation \[ \sqrt{x+2}+\sqrt{x-2}=\sqrt{4x-2}. \] 1. **Determine the domain:** - For \(\sqrt{x+2}\) we require \(x+2\ge0\) i.e. \(x\ge-2\). - For \(\sqrt{x-2}\) we require \(x-2\ge0\) i.e. \(x\ge2\). - For \(\sqrt{4x-2}\) we require \(4x-2\ge0\) i.e. \(x\ge\frac{1}{2}\). The most restrictive condition is \(x\ge2\). Thus, the domain is \(x\ge2\). 2. **Square both sides:** Square the equation: \[ \Bigl(\sqrt{x+2}+\sqrt{x-2}\Bigr)^2 = (\sqrt{4x-2})^2. \] Expand the left-hand side: \[ x+2+x-2+2\sqrt{(x+2)(x-2)} = 4x-2. \] Combine like terms: \[ 2x+2\sqrt{x^2-4}=4x-2. \] 3. **Isolate the radical term:** Subtract \(2x\) from both sides: \[ 2\sqrt{x^2-4}=2x-2. \] Divide the equation by \(2\): \[ \sqrt{x^2-4}=x-1. \] 4. **Square again:** Square both sides to remove the square root: \[ x^2-4=(x-1)^2. \] Expand the right-hand side: \[ x^2-4=x^2-2x+1. \] 5. **Solve for \(x\):** Cancel \(x^2\) from both sides: \[ -4=-2x+1. \] Rearranging: \[ -2x = -5,\quad \text{so}\quad x=\frac{5}{2}. \] 6. **Verify the solution:** Substitute \(x=\frac{5}{2}\) into the original equation: \[ \sqrt{\frac{5}{2}+2}+\sqrt{\frac{5}{2}-2} = \sqrt{4\left(\frac{5}{2}\right)-2}. \] Compute each term: \[ \sqrt{\frac{9}{2}}+\sqrt{\frac{1}{2}} = \sqrt{10-2}, \] \[ \frac{3}{\sqrt{2}}+\frac{1}{\sqrt{2}} = \sqrt{8}, \] \[ \frac{4}{\sqrt{2}} = 2\sqrt{2}. \] Since \[ \frac{4}{\sqrt{2}} = 2\sqrt{2}, \] the solution is valid. The solution to the equation is \[ x=\frac{5}{2}. \]

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To solve the inequality \( \left|\frac{x+1}{2}-\frac{x-1}{3}\right|<1 \), start by simplifying the expression inside the absolute value. Find a common denominator (which is 6) and rewrite the terms: \[ \frac{x+1}{2} = \frac{3(x+1)}{6} = \frac{3x + 3}{6}, \quad \frac{x-1}{3} = \frac{2(x-1)}{6} = \frac{2x - 2}{6} \] This gives us: \[ \left| \frac{(3x + 3) - (2x - 2)}{6} \right| < 1 \implies \left| \frac{x + 5}{6} \right| < 1 \] Multiplying through by 6 yields: \[ -6 < x + 5 < 6 \] Subtracting 5: \[ -11 < x < 1 \] In interval notation, the solution is \( (-11, 1) \). --- For the equation \( \sqrt{x+2}+\sqrt{x-2}=\sqrt{4x-2} \), start by realizing that all expressions within the square roots must be non-negative, giving the condition \( x \geq 2 \). Next, square both sides to eliminate the square roots: \[ \left(\sqrt{x+2}+\sqrt{x-2}\right)^2 = (4x - 2) \] This expands to: \[ (x+2) + (x-2) + 2\sqrt{(x+2)(x-2)} = 4x - 2 \] Simplifying leads to: \[ 2\sqrt{(x+2)(x-2)} = 2x - 4 \] Dividing by 2 gives: \[ \sqrt{(x+2)(x-2)} = x - 2 \] Now, square both sides again: \[ (x+2)(x-2) = (x - 2)^2 \] Expanding both sides results in: \[ x^2 - 4 = x^2 - 4x + 4 \] Subtract \( x^2 \) from both sides: \[ -4 = -4x + 4 \] Rearranging gives us: \[ 4x = 8 \quad \Rightarrow \quad x = 2 \] Finally, check that \( x = 2 \) satisfies the original equation. Indeed, it does, so the solution is \( x = 2 \).

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