How do you find the solution of the system of equations #-4x=y+1# and #-8x-5y=-19#?

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Answer 1 · 2018-06-17T01:01:29

We solve these kind of equation simultaneously...

#-4x=y+1#
or, #y=-1-4x#...................i

And,
The other equation is :
#-8x-5y=-19#......ii

Now put the value of #y# (which is #-1-4x#) in the second equation:
#-8x-5(-1-4x)=-19#

or, #-8x+5+20x=-19#

or, #12x=-24#

Thus #x=-2#

And now we have the values o #x# so let's put it in the first equation:
#y=-1-4x#
#y=-1+8#
Thus, #y=7#

Answer 2 · 2018-06-17T01:03:46

Substitution is the easiest way to complete this problem.

#-4x = y + 1" " " " (1)#
#-8x - 5y = -19 " " " " (2)#

#(1)#

Subtract the #1# on both sides:

#-4x - 1 = y#

Replace the #y# in the second equation with #-4x - 1#

(2)

#-8x - 5(-4x - 1) = -19#

Distribute the #-5# into the parentheses

#-8x + 20x + 5 = -19#

Add #-8x# and #20x#

#12x + 5 = -19#

Subtract the #5# on both sides

#12x = -24#

Divide by #12# on both sides

#x = -2#

To find the #y# value, substitute the #x# value into one of the original equations:

#(1)#

#-4(-2) = y + 1#

Multiply #-4# and #-2#

#8 = y + 1#

Subtract the #1# on both sides

#7 = y#

Answer:

#(-2 , 7)#