How do you solve #0x – 8y = -16# and #-8x + 2y = 36 #?

We can use the elimination method to solve this system.

Equation 1: #0x - 8y = -16#
Equation 2: #-8x + 2y = 36#

In the elimination method, we multiply each equation by a suitable number so that the two equations have a like term. For this problem, we will be focusing on #y#.

First, we will multiply equation 1 by the coefficient of #y# in equation 2, and multiply equation 2 by the coefficient of #y# in equation 1.

This means will will be multiplying equation 1 by #color(red)2# and multiplying equation 2 by #color(blue)(-8)#.

Equation 1: #color(red)2 \times (0x color(blue)( - 8)y = -16)#
Equation 2: #color(blue)(-8) \times (-8x + color(red)2y = 36)#

New Equation 1: #0x - 16y = -32#
New Equation 2: #64x -16y = -288#

Now that both equations have a like term (#-16y#), we can subtract the second equation from the first equation to eliminate them.

#=> (0x - 16y = -32) - (64x -16y = -288)#

#=> -64x +0y = 256#

#=> -64x = 256#

Now all we have to do is divide both sides by -64 to isolate and solve for #x#:

#(-64x)/-64 = (256)/-64#

#color(magenta)(x = -4)#

We can then substitute #x# into either equation 1 or equation 2 to solve for #y#. I will be using equation 2.

Substitute #x# with #-4#:
#64(-4) -16y = -288#

#-256 -16y = -288#

Add #256# to both sides:
#-256 color(red)(+ 256) -16y = -288color(red)(+ 256)#
#-16y = -32#

Divide both sides by #-16# to isolate #y#.
#(-16y)/-16 = (-32)/-16#

#color(magenta)(y = 2)#

- - Alternate method: - -

As you may have noticed, equation 1 has a term #0x#. #color(red)(0x)# will equate to #color(red)0# no matter the value of #x#, so the equation can be converted into:

#-8y = -16#

We can then divide both sides by #-8# to solve for #y#:

#(-8y)/-8 = (-16)/-8 #

#color(magenta)(y = 2)#

Then substitute the value of #y# in to the second equation and solve for #x#:

#-8x + 2(2) = 36#

#-8x + 4 = 36#

Subtract 4 from both sides of the equation:

#-8x + 4 color(red)(-4)= 36 color(red)(-4)#

#-8x = 32#

Then divide both sides by -8 to solve for #x#:

#(-8x)/-8 = (32)/-8#

#color(magenta)(x = -4)#

The point #(-4,2)# is the point of intersection between the two lines.

graph{(-8y+16)(2y-8x-36)=0 [-12.66, 12.65, -6.33, 6.33]}

#0x# simply means #0#... so you can add as many variables as you want if there's a #0# behind them...

So... in the first equation
#=>0x-8y=-16#
#=>0-8y=-16#
#=>-8y=-16#
Cancel out the minuses
#=>8y=16#
#=>y=16/8#
#=>color(red)(y=2#
Put this value in the second equation

Second equation
#-8x+2y=36#
Put value
#-8x+2xx2=36#
Multiply
#-8x+4=36#
Transfer the value 4
#-8x=32#
Transfer #-8#
#x=32/(-8)#
#color(red)(x=-4#