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)#