How do you solve #10x – 2y = 8 and 4x – 10y = -6#?

1 Answer
May 27, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve the first equation for #y#:

#10x - 2y = 8#

#(10x - 2y)/color(red)(-2) = 8/color(red)(-2)#

#(10x)/color(red)(-2) + (-2y)/color(red)(-2) = -4#

#-5x + (color(red)(cancel(color(black)(-2)))y)/cancel(color(red)(-2)) = -4#

#-5x + y = -4#

#color(red)(5x) - 5x + y = color(red)(5x) - 4#

#0 + y = 5x - 4#

#y = 5x - 4#

Step 3) Substitute #(5x - 4)# for #y# in the second equation and solve for #x#:

#4x - 10y = -6# becomes:

#4x - 10(5x - 4) = -6#

#4x - (10 xx 5x) + (-10 xx -4) = -6#

#4x - 50x + 40 = -6#

#(4 - 50)x + 40 = -6#

#-46x + 40 = -6#

#-46x + 40 - color(red)(40) = -6 - color(red)(40)#

#-46x + 0 = -46#

#-46x = -46#

#(-46x)/color(red)(-46) = (-46)/color(red)(-46)#

#(color(red)(cancel(color(black)(-46)))x)/cancel(color(red)(-46)) = 1#

#x = 1#

Step 3) Substitute #1# for #x# in the solution to the first equation at the end of Step 1 and calculate #y#:

#y = 5x - 4# becomes:

#y = (5 xx 1) - 4#

#y = 5 - 4#

#y = 1#

The solution is: #x = 1# and #y = 1# or #(1, 1)#