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# Substitution method

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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

#1
+7641
+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  $$m=\color{red}\frac{2n+42}{9}$$

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

!

asinus  Apr 8, 2017
edited by asinus  Apr 8, 2017
#1
+7641
+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  $$m=\color{red}\frac{2n+42}{9}$$

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

!

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