Can someone explain substitution method when there are more than two parts?
ie) 9m - 2n - 42 = 0 and 5m - 2n - 26 = 0
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 \(m=\color{red}\frac{2n+42}{9}\)
\(9m - 2n - 42 = 0 \\9m-2(-3)-42=0\\9m=42-6\\m= \frac{36}{9}\\\color{blue}m=4\)
!
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 \(m=\color{red}\frac{2n+42}{9}\)
\(9m - 2n - 42 = 0 \\9m-2(-3)-42=0\\9m=42-6\\m= \frac{36}{9}\\\color{blue}m=4\)
!