How do you solve this system of equations: #y= 14x ^ { 2} + 12x - 28 and y = 12x - 14#?

1 Answer
Dec 4, 2017

#(-1,-26)# and #(1,-2)#

Explanation:

#y=14x^2+12x-28# and #y=12x-14#

#14x^2+12x-28=12x-14#

#14x^2+12x-12x-28+14#

#14x^2-14=0#

#14(x^2-1)=0#

#(x^2-1)=0#

#x=+-1#

for x=1

#y=14(1)^2+12(1)-28#

#y=14+12-28 -> y=-2# #(1,-2)#

for x=-1

#y=12(-1)-14=-12-14=-26 -> y=-26#

#(-1,-26)# and #(1,-2)#