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

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