How do you solve #4x+5y=-23# and #x= -2-5y# using substitution?

Question

1432 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-05T19:35:49

#x=-7#
#y=1#

Given:
#" "4x+5y=-23# ...........................(1)
#" "x=-2-5y#..................................(2)

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Substitute for #x# in equation (1) using equation (2) giving:

#" "4(-2-5y)+5y=-23" "...................(1_a)#

Multiply out the bracket

#" "-8-20y+5y=-23#

#color(brown)(" "-8-15y=-23)#

Add #color(blue)(15y)# to both sides

#color(brown)(" -8-15ycolor(blue)(+15y)=-23color(blue)(+15y))#

But #color(brown)(-15y)color(blue)(+15y)color(green)(=0)#

#-8color(green)(+0)=-23+15y#

Add 23 to both sides

#-8+23=15y#

#15=15y#
Divide both sides by 15
#y=1#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~

Substitute #y=1# into equation (2)

#x=-2-5(1) = -7#