How do you solve #12x-5y=8# and #3x=5/4y+2# using substitution?

1 Answer
May 23, 2017

Answer:

See a solution process below:

Explanation:

Step 1) Solve the second equation for #x#:

#3x = 5/4y + 2#

#color(red)(1/3) xx 3x = color(red)(1/3)(5/4y + 2)#

#color(red)(1/color(black)(cancel(color(red)(3)))) xx color(red)(cancel(color(black)(3)))x = (color(red)(1/3) xx 5/4y) + (color(red)(1/3) xx 2)#

#x = 5/12y + 2/3#

Step 2) Substitute #5/12y + 2/3# for #x# in the first equation and solve for #y#:

#12x - 5y = 8# becomes:

#12(5/12y + 2/3) - 5y = 8#

#(12 xx 5/12y) + (12 xx 2/3) - 5y = 8#

#(color(red)(cancel(color(black)(12))) xx 5/color(red)(cancel(color(black)(12)))y) + (color(red)(cancel(color(black)(12)))4 xx 2/color(red)(cancel(color(black)(3)))) - 5y = 8#

#5y + 8 - 5y = 8#

#5y - 5y + 8 = 8#

#0 + 8 = 8#

#8 = 8#

Because #8# does in fact equal #8# these two equations are parallel and the same. They have an infinite number of points the same.