How do you solve for #x# in the system of equations #y = x^2 + 6x - 17# and #y = 3x - 7#?
1 Answer
Feb 2, 2016
x= 2 , x = - 5
Explanation:
Given 2 values for y that are equal , then
# x^2 + 6x - 17 = 3x - 7 # This is a quadratic equation , so collect terms to left side and
equate to zero.hence
#x^2 + 6x - 17- 3x + 7 = 0# gives:
#x^2 + 3x -10 = 0# to factorise require factors of -10 which sum to +3
these are +5 and -2
(x + 5 )(x - 2 ) =0
x+ 5 =0
#rArr x = -5 # or x - 2 = 0
# rArr x = 2 #
