How do you solve the system $y = 2 x - 3$ $y = 2x -3$ , $x^{2} + y^{2} = 2$ $x^2 + y^2 = 2$ ?

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Answer 1 · 2015-11-15T05:11:22

$(\frac{7}{5}, - \frac{1}{5}), (- 1 . - 5)$ $(7/5, -1/5), (-1. -5)$

Replace $y$ $y$ with its equivalent based from the linear equation and solve for $x$ $x$

$y = 2 x - 3$ $y = 2x - 3$

$x^{2} + y^{2} = 2$ $x^2 + y^2 = 2$

$\Rightarrow x^{2} + {(2 x - 3)}^{2} = 2$ $=> x^2 + (2x - 3)^2 = 2$

$\Rightarrow x^{2} + 4 x^{2} - 12 x + 9 = 2$ $=> x^2 + 4x^2 - 12x + 9 = 2$

$\Rightarrow 5 x^{2} - 12 x + 9 = 2$ $=> 5x^2 - 12x + 9 = 2$

$\Rightarrow 5 x^{2} - 12 x + 7 = 0$ $=> 5x^2 - 12x + 7 = 0$

$\Rightarrow (5 x - 7) (x - 1) = 0$ $=> (5x - 7)(x - 1) = 0$

$\Rightarrow x = \frac{7}{5}, x = - 1$ $=> x = 7/5, x = -1$

Now let's find the corresponding value of $y$ $y$

$x = \frac{7}{5}$ $x = 7/5$

$\Rightarrow y = 2 (\frac{7}{5}) - 3$ $=> y = 2(7/5) - 3$
$\Rightarrow y = \frac{14}{5} - 3$ $=> y = 14/5 -3$

$\Rightarrow y = \frac{14 - 15}{5}$ $=> y = (14 - 15)/5$

$\Rightarrow y = - \frac{1}{5}$ $=> y = -1/5$

$x = - 1$ $x = -1$

$\Rightarrow y = 2 (- 1) - 3$ $=> y = 2(-1) - 3$

$\Rightarrow y = - 2 - 3$ $=> y = -2 - 3$

$\Rightarrow y = - 5$ $=> y = -5$

How do you solve the system y=2x−3y = 2x -3, x2+y2=2x^2 + y^2 = 2?