How do you solve the following system: #3x + 8y = -2, 2x – 4y = 8#?

Step 1) Solve each equation to #8y#

• Equation 1:

#3x + 8y = -2#

#3x - color(red)(3x) + 8y = -2 - color(red)(3x)#

#0 + 8y = -2 - 3x#

#8y = -2 - 3x#

• Equation 2:

#2x - 4y = 8#

#2x - color(red)(2x) - 4y = 8 - color(red)(2x)#

#0 - 4y = 8 - 2x#

#-4y = 8 - 2x#

#color(red)(-2) xx -4y = color(red)(-2)(8 - 2x)#

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

#8y = -16 - (-4x)#

#8y = -16 + 4x#

Step 2) Because the left side of both equations are the same we can equate the right side of each equation and solve for #x#:

#-2 - 3x = -16 + 4x#

#-2 + color(blue)(16) - 3x + color(red)(3x) = -16 + color(blue)(16) + 4x + color(red)(3x)#

#14 - 0 = 0 + (4 + color(red)(3))x#

#14 = 7x#

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

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

#2 = x#

#x = 2#

Step 3) Substitute #2# for #x# in the solution to either equation in Step 1 and solve for #y#:

#-4y = 8 - 2x# becomes:

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

#-4y = 8 - 4#

#-4y = 4#

#(-4y)/color(red)(-4) = 4/color(red)(-4)#

#(color(red)(cancel(color(black)(-4)))y)/cancel(color(red)(-4)) = -1#

#y = -1#

The Solution Is:

#x = 2# and #y = -1#

Or

#(2, -1)#