Spencer H.

asked • 05/03/18

How to solve linear systems in Mrs Dupree’s class it’s very hard

6b+5x=40 5b+4x=33

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Arturo O. answered • 05/03/18

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Try substitution.  From the first equation, get
 
b = (40 - 5x)/6
 
Substitute this for b in the second equation.  You will be left with an equation in x only.  Solve it for x.  Then substitute the resulting value of x into either equation, and solve for b.  It is not hard at all.
 
 
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