How do you solve the system #5x + 6y = 2 # and #-2x + 3y = 37 #?

1 Answer
Mar 18, 2018

Answer:

See a solution process below:

Explanation:

Step 1) Solve each equation for #6y#:

  • Equation 1:

#5x + 6y = 2#

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

#0 + 6y = 2 - 5x#

#6y = 2 - 5x#

  • Equation 2:

#-2x + 3y = 37#

#-2x + color(red)(2x) + 3y = 37 + color(red)(2x)#

#0 + 3y = 37 + 2x#

#3y = 37 + 2x#

#color(red)(2) xx 3y = color(red)(2)(37 + 2x)#

#6y = (color(red)(2) xx 37) + (color(red)(2) xx 2x)#

#6y = 74 + 4x#

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

#2 - 5x = 74 + 4x#

#2 - color(red)(2) - 5x - color(blue)(4x) = 74 - color(red)(2) + 4x - color(blue)(4x)#

#0 + (-5 - color(blue)(4))x = 72 + 0#

#-9x = 72#

#(-9x)/color(red)(-9) = 72/color(red)(-9)#

#(color(red)(cancel(color(black)(-9)))x)/cancel(color(red)(-9)) = -8#

#x = -8#

Step 3) Substitute #-8# into either of the equations in Step 1 and solve for #y#:

#3y = 37 + 2x# becomes:

#3y = 37 + (-8 xx 2)#

#3y = 37 - 16#

#3y = 21#

#(3y)/color(red)(3) = 21/color(red)(3)#

#(color(red)(cancel(color(black)(3)))y)/cancel(color(red)(3)) = 7#

#y = 7#

The Solution Is:

#x = -8# and #y = 7#

Or

#(-8, 7)#