How do you solve the system #y=-x+2# and #y=x-4#?

1 Answer
Mar 2, 2018

Answer:

#(x,y)to(3,-1)#

Explanation:

#y=-x+2to(1)#

#y=x-4to(2)#

#"adding "(1)" and "(2)" will eliminate term in x"#

#"adding term by term on both sides"#

#(y+y)=(-x+x)+(2-4)#

#rArr2y=-2#

#"divide both sides by 2"#

#(cancel(2) y)/cancel(2)=(-2)/2#

#rArry=-1#

#"substitute this value into either equation "(1)" or "(2)#

#"substituting in equation "(2)" gives"#

#-1=x-4rArrx=-1+4=3#

#"point of intersection "=(3,-1)#
graph{(y+x-2)(y-x+4)((x-3)^2+(y+1)^2-0.04)=0 [-10, 10, -5, 5]}