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Solving a Second 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 $15 to build each item, sells them for $32, and has fixed costs of $25,500, the cost function is C(x)=15x+25500C(x) = 15x + 25500 and the revenue function is R(x)=32xR(x) = 32x. Setting the costs equal to revenue to find the break-even point gives 15x+25500=32x15x + 25500 = 32x. Solving this equation yields 17x=2550017x = 25500, or x=1500x = 1500. Therefore, the break-even point is at 1,500 items, where both the cost and revenue are $48,000.

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Updated 2026-05-26

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