How do you solve the following system?: # -x+4y=8 , x=2y+1 #

Step 1) Because the second equation is already solved for #x# we can substitute #(2y + 1)# for #x# in the first equation and solve for #y#:

#-x + 4y = 8# becomes:

#-(2y + 1) + 4y = 8#

#-2y - 1 + 4y = 8#

#4y - 2y - 1 = 8#

#(4 - 2)y - 1 = 8#

#2y - 1 = 8#

#2y - 1 + color(red)(1) = 8 + color(red)(1)#

#2y - 0 = 9#

#2y = 9#

#(2y)/color(red)(2) = 9/color(red)(2)#

#(color(red)(cancel(color(black)(2)))y)/cancel(color(red)(2)) = 9/2#

#y = 9/2#

Step 2) Substitute #9/2# for #y# in the second equation and calculate #x#:

#x = 2y + 1# becomes:

#x = (2 * 9/2) + 1#

#x = (color(red)(cancel(color(black)(2))) * 9/color(red)(cancel(color(black)(2)))) + 1#

#x = 9 + 1#

#x = 10#

The Solution Is:

#x = 10# and #y = 9/2# or #(10, 9/2)#

#-color(red)(x)+4y=8to(1)#

#color(red)(x)=2y+1to(2)#

#"substitute "color(red)(x)=2y+1" into "(1)#

#rArr-(2y+1)+4y=8#

#rArr-2y-1+4y=8#

#rArr2y=9rArry=9/2#

#"substitute this value into "(2)#

#rArrx=(2xx9/2)+1=9+1=10#

#color(blue)"As a check"#

#"substitute these values into "(1)#

#-10+(4xx9/2)=-10+18=8larr" True"#

#rArr"point of intersection "=(10,9/2)#

Notice that the #x#-terms are ADDITIVE INVERSES. That means that they will add together to give #0#.

Change the second equation into the same form so we have:

#color(white)(xxxxxx)color(blue)(-x)+4y =8" ".......A#
#color(white)(xxxxxx)color(blue)(+x)-2y =1" ".........B#

#A+B:" "color(blue)(0x)+2y=9#

#color(white)(xxxxxxxxxxx)y =4.5#

Substitute this value for #y# into the original equation for #x#

#x =2y+1#

#x = 2(4.5)+1#

#x = 10#

Check in the other equation:#" "-x+4y =8#

#-(10)+4(4.5)#

#-10+18#

#=8" "larr# the answer is correct.