Question

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

Answer 1

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

Explanation:

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