1.3 Solve for \( x \) and \( y \) simultaneously: \( y-1=2 x \) \( x^{2}+x y-3 x-y+2=0 \)
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To solve the equations simultaneously, let's start with the first equation: 1. \( y - 1 = 2x \) Rearranging gives us \( y = 2x + 1 \). Now, substitute \( y \) into the second equation: 2. \( x^2 + xy - 3x - y + 2 = 0 \) Replacing \( y \) with \( 2x + 1 \): \[ x^2 + x(2x + 1) - 3x - (2x + 1) + 2 = 0 \] This simplifies to: \[ x^2 + 2x^2 + x - 3x - 2x - 1 + 2 = 0 \] Combine like terms: \[ 3x^2 - 4x + 1 = 0 \] Now, we can solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \): Here, \( a = 3, b = -4, c = 1 \): \[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 3 \cdot 1}}{2 \cdot 3} \] \[ x = \frac{4 \pm \sqrt{16 - 12}}{6} \] \[ x = \frac{4 \pm \sqrt{4}}{6} \] \[ x = \frac{4 \pm 2}{6} \] Calculating the two possible values: 1. \( x = \frac{6}{6} = 1 \) 2. \( x = \frac{2}{6} = \frac{1}{3} \) Now, use these \( x \) values to find the corresponding \( y \) values: For \( x = 1 \): \[ y = 2(1) + 1 = 3 \] For \( x = \frac{1}{3} \): \[ y = 2\left(\frac{1}{3}\right) + 1 = \frac{2}{3} + 1 = \frac{5}{3} \] Thus, the solutions are: 1. \( (x, y) = (1, 3) \) 2. \( (x, y) = \left(\frac{1}{3}, \frac{5}{3}\right) \)
