The problem set involves solving a system of linear algebraic equations by simplification, pairwise comparison and substitution or elimination.
This is a set of linear algebraic equations. In this system, the first step is to simplify each equation. The first equation becomes: 150c + 4g = 0; the second equation becomes 140c + 2.64g = 0; the third equation becomes 130 + 2.5g = 0.
Next, to solve this, we take two equations at a time and solve them as a pair, eliminating one variable in the process. For instance, you could subtract the second equation from the first to eliminate the 'c' variable: (150c + 4g) - (140c + 2.64g) = 10c + 1.36g = 0. You can then solve the third equation for its 'g' value, and substitute it into the first equation to find the value of 'c'. You'll have to do this for each set of equations till you get solutions for all variables.
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