How do you solve the system of equations #1/2x + 3y = 5/6# and #1/3x - 5y = 16/9#?

1 Answer
Jul 12, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Solve both equations for #x#:

  • Equation 1:

#1/2x + 3y = 5/6#

#color(red)(2)(1/2x + 3y) = color(red)(2) xx 5/6#

#(color(red)(2) xx 1/2x) + (color(red)(2) xx 3y) = cancel(color(red)(2)) xx 5/(color(red)(cancel(color(black)(6)))3)#

#color(red)(2)/2x + 6y = 5/3#

#1x + 6y = 5/3#

#x + 6y = 5/3#

#x + 6y - color(red)(6y) = 5/3 - color(red)(6y)#

#x + 0 = 5/3 - 6y#

#x = 5/3 - 6y#

  • Equation 2:

#1/3x - 5y = 16/9#

#color(red)(3)(1/3x - 5y) = color(red)(3) xx 16/9#

#(color(red)(3) xx 1/3x) - (color(red)(3) xx 5y) = cancel(color(red)(3)) xx 16/(color(red)(cancel(color(black)(9)))3)#

#color(red)(3)/3x - 15y = 16/3#

#1x - 15y = 16/3#

#x - 15y = 16/3#

#x - 15y + color(red)(15y) = 16/3 + color(red)(15y)#

#x - 0 = 16/3 + 15y#

#x = 16/3 + 15y#

Step 2) Because the left side of both equations are equal we can equate the right side of both equations and solve for #y#:

#5/3 - 6y = 16/3 + 15y#

#5/3 - color(blue)(16/3) - 6y + color(red)(6y) = 16/3 - color(blue)(16/3) + 15y + color(red)(6y)#

#(5 - color(blue)(16))/3 - 0 = 0 + (15 + color(red)(6))y#

#-11/3 = 21y#

#1/color(red)(21) xx -11/3 = (21y)/color(red)(21)#

#-(1 xx 11)/(color(red)(21) xx 3) = (color(red)(cancel(color(black)(21)))y)/cancel(color(red)(21))#

#-11/63 = y#

#y = -11/63#

Step 3) Substitute #-11/63# for #y# in the solution to either equation in Step 1 and calculate #x#:

#x = 16/3 + 15y# becomes:

#x = 16/3 + (15 xx -11/63)#

#x = (21/21 xx 16/3) + (15 xx -11)/63#

#x = 336/63 + (-165)/63#

#x = (336- 165)/63#

#x = 171/63#

The Solution Is:

#x = 171/63# and #y = -11/63#

Or

#(171/63, -11/63)#