How do you solve the system $x^{2} + y^{2} - 4 x - 4 y = 26$ $x^2+y^2-4x-4y=26$ and $x^{2} + y^{2} - 4 x = 54$ $x^2+y^2-4x=54$ and $y = 3 x - 8$ $y=3x-8$ ?

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Answer 1 · 2016-09-08T09:26:55

$x = 5, y = 7$ $x = 5, y = 7$

$x^{2} + y^{2} - 4 x - 4 y = 26$ $x^2+y^2-4x-4y=26$
$x^{2} + y^{2} - 4 x = 54$ $x^2+y^2-4x=54$
$y = 3 x - 8$ $y=3x-8$

Subtracting the first from the second we get

Solving

${(4 y = 28), (y = 3 x - 8):}$

we obtain

$x = 5, y = 7$

Substituting those results in the first

$25+49-20-28=26$

Substituting now in the second

$25+49-20 = 54$

so the system has the solution $x = 5, y = 7$

How do you solve the system x^2+y^2-4x-4y=26x2+y2−4x−4y=26x^2+y^2-4x-4y=26 and x^2+y^2-4x=54x2+y2−4x=54x^2+y^2-4x=54 and y=3x-8y=3x−8y=3x-8?