How do you use substitution to solve the following equations: #y=-6x-32, 2y=10x+46#?

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Answer 1 · 2017-05-18T13:40:56

See a solution process below:

Step 1) Because the first equation is already solved for #y# we can substitute #-6x - 32# for #y# in the second equation and solve for #x#:

#2y = 10x + 46# becomes:

#2(-6x - 32) = 10x + 46#

#(2 * -6x) - (2 * 32) = 10x + 46#

#-12x - 64 = 10x + 46#

#color(red)(12x) - 12x - 64 - color(red)(46) = color(red)(12x) + 10x + 46 - color(red)(46)#

#0 - 110 = (color(red)(12) + 10)x + 0#

#-110 = 22x#

#-110/color(red)(22) = (22x)/color(red)(22)#

#-5 = (color(red)(cancel(color(black)(22)))x)/cancel(color(red)(22))#

#-5 = x#

#x = -5#

Step 2) Substitute #-5# for #x# in the first equation and calculate #y#:

#y = -6x - 32# becomes:

#y = (-6 xx -5) - 32#

#y = 30 - 32#

#y = -2#

The solution is: #x = -5# and #y = -2# or #(-5, -2)#