How do you find the exact solutions to the system $y - x = 1$ $y-x=1$ and $x^{2} + y^{2} = 25$ $x^2+y^2=25$ ?

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Answer 1 · 2016-11-25T00:58:38

You substitute $y$ $y$ as function of $x$ $x$ in the equation of higher degree, using the equation of lower degree, then you solve the former.

$y - x = 1 \Rightarrow y = 1 + x$ $y-x=1 => y=1+x$

Then
$x^{2} + y^{2} = 25$ $x^2+y^2=25$

$x^{2} + {(1 + x)}^{2} = 25$ $x^2 +(1+x)^2 = 25$

$2 x^{2} + 2 x + 1 = 25$ $2x^2+2x+1=25$

$2 x^{2} + 2 x - 24 = 0$ $2x^2+2x-24 = 0$

$x = \frac{- 2 \pm \sqrt{4 + 192}}{4} = \frac{- 2 \pm 14}{4}$ $x= (-2 +-sqrt(4+192))/4=(-2+-14)/4$

$x_{1} = - 4$ $x_1=-4$
$x_{2} = 3$ $x_2=3$

and

$y_{1} = - 3$ $y_1=-3$
$y_{2} = 4$ $y_2=4$

How do you find the exact solutions to the system y−x=1y-x=1 and x2+y2=25x^2+y^2=25?