How do you find an ordered pair for 4y + 1 = -3x and 5x + 6y = -3?

1 Answer
Feb 12, 2017

Answer:

See the entire solution process below:

Explanation:

Step 1) Solve the first equation for #y#:

#4y + 1 = -3x#

#4y + 1 - color(red)(1) = -3x - color(red)(1)#

#4y + 0 = -3x - 1#

#4y = -3x - 1#

#(4y)/color(red)(4) = (-3x - 1)/color(red)(4)#

#(color(red)(cancel(color(Black)(4)))y)/cancel(color(red)(4)) = (-3x - 1)/4#

#y = (-3x - 1)/4#

Step 2) Substitute #(-3x - 1)/4# for #y# in the second equation and solve for #x#:

#5x + 6y = -3# becomes:

#5x + 6((-3x - 1)/4) = -3#

#5x - (18x)/4 - 6/4 = -3#

#color(red)(4)(5x - (18x)/4 - 6/4) = color(red)(4) xx -3#

#(color(red)(4) xx 5x) - (color(red)(4) xx (18x)/4) - (color(red)(4) xx 6/4) = -12#

#20x - (cancel(color(red)(4)) xx (18x)/color(red)(cancel(color(black)(4)))) - (cancel(color(red)(4)) xx 6/color(red)(cancel(color(black)(4)))) = -12#

#20x - 18x - 6 = -12#

#2x - 6 = -12#

#2x - 6 + color(red)(6) = -12 + color(red)(6)#

#2x - 0 = -6#

#2x = -6#

#(2x)/color(red)(2) = -6/color(red)(2)#

#(color(red)(cancel(color(black)(2)))x)/cancel(color(red)(2)) = -3#

#x = -3#

Step 3) Substitute #-3# for #x# in the solution to first equation at the end of Step 1 and calculate #y#:

#y = (-3x - 1)/4# becomes:

#y = ((-3 xx -3) - 1)/4#

#y = (9 - 1)/4#

#y = 8/4#

#y = 2#

The solution is: #x = -3# and #y = 2# or #(-3, 2)#