How do you solve #6x+y=13# and #y-x= -8# using substitution?

Step 1) Solve the second equation for #y#:

#y - x = -8#

#y - x + color(red)(x) = -8 + color(red)(x)#

#y - 0 = -8 + x#

#y = -8 + x#

Step 2) Substitute #(-8 + x)# for #y# in the first equation and solve for #x#:

#6x + y = 13# becomes:

#6x + (-8 + x) = 13#

#6x - 8 + x = 13#

#6x - 8 + 1x = 13#

#6x + 1x - 8 = 13#

#(6 + 1)x - 8 = 13#

#7x - 8 = 13#

#7x - 8 + color(red)(8) = 13 + color(red)(8)#

#7x - 0 = 21#

#7x = 21#

#(7x)/color(red)(7) = 21/color(red)(7)#

#(color(red)(cancel(color(black)(7)))x)/cancel(color(red)(7)) = 3#

#x = 3#

Step 3) Substitute #3# for #x# in the solution to the second equation at the end of Step 1 and calculate #y#:

#y = -8 + x# becomes:

#y = -8 + 3#

#y = -5#

The Solution Is: #x = 3# and #y = -5# or #(3, -5)#

#6x+color(red)(y)=13to(1)#

#color(red)(y)-x=-8to(2)#

#"from " (2)color(white)(x)color(red)(y)=x-8to(3)#

#"substitute "y=x-8" in "(1)#

#rArr6x+x-8=13#

#rArr7x-8=13#

#"add 8 to both sides"#

#7xcancel(-8)cancel(+8)=13+8#

#rArr7x=21#

#"divide both sides by 7"#

#(cancel(7) x)/cancel(7)=21/7#

#rArrx=3#

#"substitute this value in "(3)" and evaluate y"#

#rArry=3-8=-5#

#color(blue)"As a check"#

#"substitute these values into "(1)#

#(6xx3)+(-5)=13larr" True"#

#rArr"point of intersection "=(3,-5)#
graph{(y+6x-13)(y-x+8)((x-3)^2+(y+5)^2-0.04)=0 [-16.02, 16.02, -8.01, 8.01]}