**Step 1)** Solve each equation for a common term, in this problem it can be #40y#:

#3x - 10y = -25#

#color(red)(-4)(3x - 10y) = color(red)(-4) xx -25#

#(color(red)(-4) xx 3x) + (color(red)(-4) xx -10y) = 100#

#-12x + 40y = 100#

#-12x + color(red)(12x) + 40y = 100 + color(red)(12x)#

#0 + 40y = 100 + 12x#

#40y = 100 + 12x#

#4x + 40y = 20#

#4x - color(red)(4x) + 40y = 20 - color(red)(4x)#

#0 + 40y = 20 - 4x#

#40y = 20 - 4x#

**Step 2)** Because the left side of both equations are now equal we can equate the right side of the equations and solve for #x#:

#100 + 12x = 20 - 4x#

#100 - color(red)(100) + 12x + color(blue)(4x) = 20 - color(red)(100) - 4x + color(blue)(4x)#

#0 + (12 + color(blue)(4))x = -80 - 0#

#16x = -80#

#(16x)/color(red)(16) = -80/color(red)(16)#

#(color(red)(cancel(color(black)(16)))x)/cancel(color(red)(16)) = -5#

#x = -5#

**Step 3)** Substitute #-5# for #x# in the solution to either equation in Step 1 and solve for #y#:

#40y = 100 + 12x# becomes:

#40y = 100 + (12 * -5)#

#40y = 100 + (-60)#

#40y = 100 - 60#

#40y = 40#

#(40y)/color(red)(40) = 40/color(red)(40)#

#(color(red)(cancel(color(black)(40)))y)/cancel(color(red)(40)) = 1#

#y = 1#

**The Solution Is:**

#x = -5# and #y = 1#

Or

#(-5, 1)#