How do you solve #2c - d = - 2# and #4c + d = 20#?

1 Answer
Nov 30, 2015

Answer:

#(c,d)=(3,8)#

Explanation:

Given:
#[1]color(white)("XXX")2c-d=-2#
#[2]color(white)("XXX")4c+d=20#

Add [1] and [2]
#{: (,color(white)("XXX"),2c,-d,=,-2), (+,,4c,+d,=, 20), (,,"-------","-----","----","-----"), ([3],,6c,,=,18) :}#

Divide [3] by #6#
#[4]color(white)("XXXXXX")c = 3#

Substitute #3# (from [4]) for #c# in [1]
#[5]color(white)("XXX")2*(3)-d = -2#
Simplify
#[6]color(white)("XXX")6-d = -2#
Subtract #6# from both sides
#[7]color(white)("XXX")-d = -8#
Multiply both sides by #(-1)#
#[8]color(white)("XXX")d=8#