How do you solve the system of equations #4x + 6y = 5# and #8x - 9y = 3#?

1 Answer
smendyka
Mar 20, 2018

See a solution process below:

Explanation:

Step 1) Solve each equation for #8x#:

  • Equation 1:

#4x + 6y = 5#

#4x + 6y - color(red)(6y) = 5 - color(red)(6y)#

#4x + 0 = 5 - 6y#

#4x = 5 - 6y#

#color(red)(2) xx 4x = color(red)(2)(5 - 6y)#

#8x = 10 - 12y#

  • Equation 2:

#8x - 9y = 3#

#8x - 9y + color(red)(9y) = 3 + color(red)(9y)#

#8x - 0 = 3 + 9y#

#8x = 3 + 9y#

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

#10 - 12y = 3 + 9y#

#10 - color(blue)(3) - 12y + color(red)(12y) = 3 - color(blue)(3) + 9y + color(red)(12y)#

#7 - 0 = 0 + (9 + color(red)(12))y#

#7 = 21y#

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

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

#1/3 = y#

#y = 1/3#

Step 3) Substitute #1/3# for #y# into either of the equations in Step 1 and calculate #x#:

#8x = 10 - 12y# becomes:

#8x = 10 - (12 xx 1/3)#

#8x = 10 - 12/3#

#8x = 10 - 4#

#8x = 6#

#(8x)/color(red)(8) = 6/color(red)(8)#

#(color(red)(cancel(color(black)(8)))x)/cancel(color(red)(8)) = 3/4#

#x = 3/4#

The Solution Is:

#x = 3/4# and #y = 1/3#

Or

#(3/4, 1/3)#

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