Question

How many real solutions does this system of equations have? x² + y² = 25 and x-3y=15​

Answer 1

$x - 3y = 15 \\ x = 15 + 3y$

$(15 + 3y) {}^{2} + y {}^{2} = 25$

$225 + 90y + 9y {}^{2} + y {}^{2} = 25$

$10y {}^{2} + 90y + 200 = 0$

$y {}^{2} + 9y + 20 = 0$

$(y + 4)(y + 5) = 0$

$y = - 4 \: \: \: \: \: \: y = - 5$

$for \: y= - 4 \\ x = 15 + 3( - 4) = 15 - 12 = 3$

$for \: y = - 5 \\ x = 15 + 3( - 5) = 15 - 15 = 0$