Question

Solve the following system algerbraically: y=-3(x-2) squared+4 y=-6x+16

Answer 1

$y = - 3(x - 2) ^{2} + 4$
$y = - 6x + 16$
distribute the -3 and the square into the first equation
$y = - 3x ^{2} + 12x - 8$
$y = - 6x + 16$
multiply the second equation by 2
$y = - 3x ^{2} + 12x - 8$
$y = - 12x + 32$
add the equations together
$y = - 3x ^{2} + 24$
factor out -3
$y = - 3(x ^{2} - 8)$
factor inside parenthesis
$y = - 3(x + \sqrt{8} )( x - \sqrt{8})$
refine
$y = - 3(x + 2 \sqrt{2} )(x - 2 \sqrt{2} )$