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Solving a Third Cost and Revenue Application Using a System of Equations

A system of linear equations can be used to solve cost and revenue applications to find the break-even point. For example, if a manufacturer spends $120 to build each item, sells them for $170, and has fixed costs of $150,000, the cost function is C(x)=120x+150000C(x) = 120x + 150000 and the revenue function is R(x)=170xR(x) = 170x. Setting the costs equal to revenue to find the break-even point gives 120x+150000=170x120x + 150000 = 170x. Solving this equation yields 50x=15000050x = 150000, or x=3000x = 3000. Therefore, the break-even point is at 3,000 items, where both the cost and revenue are $510,000.

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Updated 2026-05-26

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