How do you solve the system of equations #y=50+3x# and #y=100-2x#?

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Answer 1 · 2018-05-18T12:32:42

See a solution process below:

Step 1) Because the left side of both equations are the same we can equate the right sides of both equations and solve for #x#:

#50 + 3x = 100 - 2x#

#50 - color(red)(50) + 3x + color(blue)(2x) = 100 - color(red)(50) - 2x + color(blue)(2x)#

#0 + (3 + color(blue)(2))x = 50 - 0#

#5x = 50#

#(5x)/color(red)(5) = 50/color(red)(5)#

#(color(red)(cancel(color(black)(5)))x)/cancel(color(red)(5)) = 10#

#x = 10#

Step 2) Substitute #10# for #x# in either of the original equations and calculate #y#:

#y = 50 + 3x# becomes:

#y = 50 + (3 * 10)#

#y = 50 + 30#

#y = 80#

The Solution Is:

#x = 10# and #y = 80#

Or

#(10, 80)#