How do you solve #y=4x-9# and #y=x-3# using substitution?

Step 1) Because the first equation is already solve for #y#, substitute #4x - 9# for #y# in the second equation and solve for #x#:

#y = x - 3# becomes:

#4x - 9 = x - 3#

#4x - 9 + color(red)(9) - color(blue)(x) = x - 3 + color(red)(9) - color(blue)(x)#

#4x - color(blue)(1x) - 9 + color(red)(9) = x - color(blue)(x) - 3 + color(red)(9)#

#(4 - color(blue)(1))x - 0 = 0 + 6#

#3x = 6#

#(3x)/color(red)(3) = 6/color(red)(3)#

#(color(red)(cancel(color(black)(3)))x)/cancel(color(red)(3)) = 2#

#x = 2#

Step 2) Substitute #2# for #x# in the first equation and calculate #y#:

#y = 4x - 9# becomes:

#y = (4 xx 2) - 9#

#y = 8 - 9#

#x = 2#

The solution is: #x = 2# and #x = 2# or #(2, -1)#

Since both equations are #y = #, substitute the first equation for the second "y":

#4x - 9 = x - 3#

Now solve for #x#:
#4x -x - 9 = x -x - 3#

#3x - 9 = - 3#

#3x -9 +9 = -3 +9#

#3x = 6#

#3/3x = 6/3 = 2#

Now substitute #x# into either equation to find #y#:

#y = 2 - 3 = -1#

Check by substituting both putting both values into the first equation:
#-1 = 4(2) -9#
#-1 = 8-9#
#-1 = -1# TRUE

So #x = 2, y = -1#

Which is the point of intersection: #(2, -1)#

Labelling the equations.

#color(red)(y)=4x-9to(1)#

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

Since both equations have y as the subject we can equate the right sides.

#rArr4x-9=x-3#

subtract x from both sides.

#4x-x-9=cancel(x)cancel(-x)-3#

#rArr3x-9=-3#

add 9 to both sides.

#3xcancel(-9)cancel(+9)=-3+9#

#rArr3x=6#

divide both sides by 3

#(cancel(3) x)/cancel(3)=6/3#

#rArrx=2#

Substitute this value into either of the equations

#"Substitute " x=2" in " (2)#

#rArry=2-3=-1#

#color(blue)"As a check"#

#"Substitute " x=2" in "(1)#

#rArry=(4xx2)-9=8-9=-1to" true"#

#rArr(2.-1)" is the point of intersection"#
graph{(y-4x+9)(y-x+3)=0 [-10, 10, -5, 5]}