How do you solve #4x + y = 4# and #2x + 8y = 0# using substitution?

Given: #4x+y=4#

This implies: #y=4-4x#
So we can substitute #(4-4x)# for #y#
in the second given equation: #2x+8y=0#
to get:
#color(white)("XXX")2x+8(4-4x)=0#

Simplifying:
#color(white)("XXX")2x+32-32x=0#

#color(white)("XXX")30x=32#

#color(white)("XXX")x=16/15#

The we can substitute #16/15# for #x#
in the first given equation: #4x+y=4#
to get:
#color(white)("XXX")4*(16/15)+y=4#

#color(white)("XXX")y=4-64/15 = -4/15#

Solve by substitution and elimination

#color(blue)(4x+y=4#

#color(blue)(2x+8y=0#

We can eliminate #4x# from the first equation by #2x# in the second equation (whole equation) with #-2# to get #-4x#

#rarr-2(2x+8y=0)#

Use the distributive property

#color(brown)(a(b+c=x)=ab+ac=ax#

#rarr-4x-16y=0#

Now, add the above equation with the first equation to eliminate #4x#

#rarr(4x+y=4)+(-4x-16y=0)#

#rarr-15y=4#

Divide both sides by #-15#

#rarr(cancel(-15)y)/cancel(-15)=4/-15#

#color(green)(rArry=-4/15#

Now,substitute the value of #y# to the first equation

#rarr4x+(-4/15)=4#

#rarr4x-4/15=4#

Add #4/15# both sides

#rarr4x-4/15+4/15=4+4/15#

#rarr4x=64/15#

Divide both sides by #4#

#rarr(cancel4x)/cancel4=64/15-:4#

(I have written #64/15/4# as #64/15-:4#)

Take the reciprocal and multiply

#rarrx=cancel64^16/15*1/cancel4^1#

#color(green)(rArrx=16/15#

#color(blue)(ul bar |(x,y)=(16/15,-4/15)|#