How do you solve the system of equations #y=7/2x-5# and #y=-5#?
1 Answer
Jun 7, 2017
Explanation:
#color(red)(y)=7/2x-5to(1)#
#color(red)(y)=-5to(2)#
#"since both equations have y as the subject we can"#
#"equate the right sides"#
#rArr7/2x-5=-5#
#"add 5 to both sides"#
#7/2xcancel(-5)cancel(+5)=-5+5#
#rArr7/2x=0rArrx=0#
#"the point of intersection "=(0,-5)#
graph{(y-7/2x+5)(y-0.001x+5)=0 [-20, 20, -10, 10]}
