How do you solve the system #y=2x# and #x+3y=-14# using substitution?

1 Answer
Apr 17, 2017

Answer:

#-2,-4)#

Explanation:

#"Label the equations"#

#color(white)(xxxx)color(red)(y)=2xto(1)#

#x+3color(red)(y)=-14to(2)#

#"Substitute " color(red)(y)=2x" in " (2)#

#rArrx+(3xx2x)=-14#

#rArr7x=-14#

#"divide both sides by 7"#

#(cancel(7) x)/cancel(7)=(-14)/7#

#rArrx=-2#

#"Substitute this value into " (1)#

#y=(2xx-2)=-4#

#color(blue)"As a check"#

#"substitute values for x and y in " (2)#

#-2+(3xx-4)=-2-12=-14to" true"#

graph{(y-2x)(y+1/3x+14/3)=0 [-10, 10, -5, 5]}
#rArr" point of intersection is " (-2,-4)#