How do you solve the system #0.5x+4y=-1# and #x+2.5y=3.5# using substitution?

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Answer 1 · 2017-07-26T16:22:14

See a solution process below:

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

#0.5x + 4y = -1#

#0.5x + 4y - color(red)(4y) = -1 - color(red)(4y)#

#0.5x + 0 = -1 - 4y#

#color(red)(2) xx 0.5x = color(red)(2)(-1 - 4y)#

#1x = (color(red)(2) xx -1) - (color(red)(2) xx 4y)#

#x = -2 - 8y#

Step 2) Substitute #(-2 - 8y)# for #x# in the second equation and solve for #y#:

#x + 2.5y = 3.5# becomes:

#(-2 - 8y) + 2.5y = 3.5#

#-2 - 8y + 2.5y = 3.5#

#color(red)(2) - 2 - 8y + 2.5y = color(red)(2) + 3.5#

#0 - 8y + 2.5y = 5.5#

#-8y + 2.5y = 5.5#

#(-8 + 2.5)y = 5.5#

#-5.5y = 5.5#

#(-5.5y)/color(red)(-5.5) = 5.5/color(red)(-5.5)#

#(color(red)(cancel(color(black)(-5.5)))y)/cancel(color(red)(-5.5)) = -1#

#y = -1#

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

#x = -2 - 8y# becomes:

#x = -2 - (8 * -1)#

#x = -2 - (-8)#

#x = -2 + 8#

#x = 6#

The Solution Is: #x = 6# and #y = -1# or #(6, -1)#