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

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Answer 1 · 2015-11-15T05:11:22

$(7/5, -1/5), (-1. -5)$

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

$y = 2x - 3$

$x^2 + y^2 = 2$

$=> x^2 + (2x - 3)^2 = 2$

$=> x^2 + 4x^2 - 12x + 9 = 2$

$=> 5x^2 - 12x + 9 = 2$

$=> 5x^2 - 12x + 7 = 0$

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

$=> x = 7/5, x = -1$

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

$x = 7/5$

$=> y = 2(7/5) - 3$
$=> y = 14/5 -3$

$=> y = (14 - 15)/5$

$=> y = -1/5$

$x = -1$

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

$=> y = -2 - 3$

$=> y = -5$

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