How do you solve this system of equations: #-5x + y = - 4 and 2x - y = - 5#?

2 Answers
smendyka
Dec 7, 2017

See a solution process below:

Explanation:

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

#-5x + y = -4#

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

#0 + y = 5x - 4#

#y = 5x - 4#

Step 2) Substitute #(5x - 4)# for #y# in the second equation and solve for #x#:

#2x - y = -5# becomes:

#2x - (5x - 4) = -5#

#2x - 5x + 4 = -5#

#(2 - 5)x + 4 = -5#

#-3x + 4 = -5#

#-3x + 4 - color(red)(4) = -5 - color(red)(4)#

#-3x + 0 = -9#

#-3x = -9#

#(-3x)/color(red)(-3) = (-9)/color(red)(-3)#

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

#x = 3#

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

#y = 5x - 4# becomes:

#y = (5 xx 3) - 4#

#y = 15 - 4#

#y = 11#

The Solution Is: #x = 3# and #y = 11# or #(3, 11)#

Barney V.
Dec 7, 2017

#y=11, x=3#

Explanation:

#-5x+y=-4-------(1)#

#2x-y=-5--------(2)#

#:.(1)+(2)#

#:.-3x=-9#

#:.x=9/3#

#:.x=3#

substitute #x=3# in #(2)#

#:.2(3)-y=-5#

#6-y=-5#

#:.-y=-5-6#

#:.-y=-11#

#:.y=11#

~~~~~~~~~~~~~~

check:-

substitute #y=11# and #x=3# in (1)

#:.-5(3)+(11)=-4#

#:.-15+11=-4#

#:.-4=-4#

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