\( 1.1 .4 \quad \sqrt{5-x}=x+1 \) Solve for \( x \) and \( y \) simultaneously if: \( x+4=2 y \) and \( y^{2}-x y+21=0 \)
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Let's solve the first equation \( \sqrt{5 - x} = x + 1 \) by squaring both sides: 1. Squaring both sides gives: \[ 5 - x = (x + 1)^2 \] Expanding the right side: \[ 5 - x = x^2 + 2x + 1 \] Rearranging terms: \[ x^2 + 3x - 4 = 0 \] 2. Now, we can factor it: \[ (x + 4)(x - 1) = 0 \] So, \( x = -4 \) or \( x = 1 \). 3. Next, we need to check both solutions in the original equation: - For \( x = 1 \): \[ \sqrt{5 - 1} = 1 + 1 \implies 2 = 2 \, \text{(valid)} \] - For \( x = -4 \): \[ \sqrt{5 - (-4)} = -4 + 1 \implies 3 = -3 \, \text{(invalid)} \] Thus, the valid solution for \( x \) is \( x = 1 \). Now using \( x = 1 \) in the second set of equations: 4. From the equation \( x + 4 = 2y \): \[ 1 + 4 = 2y \implies 5 = 2y \implies y = \frac{5}{2} \] 5. Now, let's verify if this \( y \) satisfies \( y^2 - xy + 21 = 0 \): \[ \left(\frac{5}{2}\right)^2 - 1 \cdot \frac{5}{2} + 21 = \frac{25}{4} - \frac{5}{2} + 21 \] Converting \( -\frac{5}{2} \) to fourths: \[ -\frac{5}{2} = -\frac{10}{4} \] Thus, \[ \frac{25}{4} - \frac{10}{4} + \frac{84}{4} = \frac{25 - 10 + 84}{4} = \frac{99}{4} \neq 0 \, \text{(not valid)} \] Since it does not satisfy this equation, we need to analyze our valid \( x \) to find \( y \): Solving \( y^2 - xy + 21 = 0 \) with \( x = 1 \): \[ y^2 - y + 21 = 0 \] Using the quadratic formula: \[ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 21}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 84}}{2} = \frac{1 \pm \sqrt{-83}}{2} \] Thus, \( x = 1 \) and \( y = \frac{1 \pm i\sqrt{83}}{2} \) (complex solutions). In summary, the solutions are: \[ (x,y) = \left(1, \frac{1 \pm i\sqrt{83}}{2}\right) \]
