How do you solve the system of equations #-3c + 7d = 47# and #- 6c - 5d = - 1#?

1 Answer
Mia
Nov 25, 2016

#c=-4" "and" " d=5#.

Explanation:

Solving this system is determined by the elimination method.
#" "#
#-3c+7d=47" " EQ1#
#" "#
#-6c-5d=-1" "EQ2#
#" "#
Method :
#" "#
First, Multiply one of the equations by an integer inorder to obtain
#" "#
opposite coefficients of the same variable in both equations.
#" "#
Here, in the given system we will multiply #" "EQ1" "# by #" "-2" "#.
#" "#
#EQ1xx-2" "# will be:
#" "#
#-3xx-2c+7xx(-2)d=-2xx47#
#" "#
#+6c-14d=-94#
#" "#
The system becomes :
#" "#
#+6c-14d=-94" "EQ1#
#" "#
#-6c-5d=-1" "EQ2#
#" "#
Second, Add both equation to obtain an equation with one unknown.
#" "#
#EQ1+EQ2#
#" "#
#cancel(+6c)cancel(-6c)-14d-5d=-94-1#
#" "#
#-19d=-95#
#" "#
#d=(-95)/(-19)#
#" "#
Therefore, #" "d=5#.
#" "#
Finally, substitute #" "d=5" "# in #" "EQ2" "# to find #" "c#.
#" "#
#-6c-5(5)=-1" "EQ2#
#" "#
#rArr-6c-25=-1#
#" "#
#rArr-6c = -1+25#
#" "#
#rArr-6c = +24#
#" "#
#rArrc = (+24)/-6#
#" "#
Therefore, #" "c=-4#
#" "#
Hence, #" "c=-4" "and" " d=5#.

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