How do you solve #-3w+ z=4# and #-9w+ 5z= -1#?

1 Answer
Jun 20, 2018

Answer:

See a solution process below:

Explanation:

Step 1)Solve both equations for #-9w#:

  • Equation 1:

#-3w + z = 4#

#color(red)(3)(-3w + z) = color(red)(3) xx 4#

#(color(red)(3) xx -3w) + (color(red)(3) xx z) = 12#

#-9w + 3z = 12#

#-9w + 3z - color(red)(3z) = 12 - color(red)(3z)#

#-9w + 0 = 12 - 3z#

#-9w = 12 - 3z#

  • Equation 2:

#-9w + 5z = -1#

#-9w + 5z - color(red)(5z) = -1 - color(red)(5z)#

#-9w + 0 = -1 - 5z#

#-9w = -1 - 5z#

Step 2)* Now that the left side of each equation is equal we can equate the right side of each equation and solve for #z#

#12 - 3z = -1 - 5z#

#12 - color(red)(12) - 3z + color(blue)(5z) = -1 - color(red)(12) - 5z + color(blue)(5z)#

#0 + (-3 + color(blue)(5))z = -13 - 0#

#2z = -13#

#(2z)/color(red)(2) = -13/color(red)(2)#

#(color(red)(cancel(color(black)(2)))z)/cancel(color(red)(2)) = -13/2#

#z = -13/2#

Step 3) Substitute #-13/2# for #z# in the solution to either equation in Step 1 and solve for #w#:

#-9w = 12 - 3z# becomes:

#-9w = 12 - (3 xx -13/2)#

#-9w = 12 - (-39/2)#

#-9w = 12 + 39/2#

#-9w = (2/2 xx 12) + 39/2#

#-9w = 24/2 + 39/2#

#-9w = (24 + 39)/2#

#-9w = 63/2#

#1/color(red)(-9) xx -9w = 1/color(red)(-9) xx 63/2#

#1/cancel(color(red)(-9)) xx color(red)(cancel(color(black)(-9)))w = -63/18#

#w = -7/2#

The Solution Is:

#w = -7/2# and #z = -13/2#