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Can someone explain substitution method when there are more than two parts?

 

ie) 9m - 2n - 42 = 0 and 5m - 2n - 26 = 0

Guest Apr 8, 2017

Best Answer 

 #1
avatar+6672 
+2

Can someone explain substitution method when there are more than two parts?
ie)

\( 9m - 2n - 42 = 0 \)

and

5m - 2n - 26 = 0

 

\(9m - 2n - 42 = 0 \\m= \color{red}\frac{2n+42}{9}\)     

 

\(5m - 2n - 26 = 0\\5({\color{red}\frac{2n+42}{9}})-2n-26=0\\10n+210-18n-234=0\\-8n=24\\{\color{blue}n=-3}\)         [ substitute  \(m=\color{red}\frac{2n+42}{9}\)

 

\(9m - 2n - 42 = 0 \\9m-2(-3)-42=0\\9m=42-6\\m= \frac{36}{9}\\\color{blue}m=4\)

 

laugh  !

asinus  Apr 8, 2017
edited by asinus  Apr 8, 2017
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1+0 Answers

 #1
avatar+6672 
+2
Best Answer

Can someone explain substitution method when there are more than two parts?
ie)

\( 9m - 2n - 42 = 0 \)

and

5m - 2n - 26 = 0

 

\(9m - 2n - 42 = 0 \\m= \color{red}\frac{2n+42}{9}\)     

 

\(5m - 2n - 26 = 0\\5({\color{red}\frac{2n+42}{9}})-2n-26=0\\10n+210-18n-234=0\\-8n=24\\{\color{blue}n=-3}\)         [ substitute  \(m=\color{red}\frac{2n+42}{9}\)

 

\(9m - 2n - 42 = 0 \\9m-2(-3)-42=0\\9m=42-6\\m= \frac{36}{9}\\\color{blue}m=4\)

 

laugh  !

asinus  Apr 8, 2017
edited by asinus  Apr 8, 2017

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