$$\frac{5c+38}{4c+40}-1 = \frac{1}{c+4}$$
Answer
$$ \begin{matrix}c_1 = -6 & c_2 = 8 \\[1 em] \end{matrix} $$
Explanation
$$ \begin{aligned} \frac{5c+38}{4c+40}-1 &= \frac{1}{c+4}&& \text{multiply ALL terms by } \color{blue}{ (4c+40)(c+4) }. \\[1 em](4c+40)(c+4)\frac{5c+38}{4c+40}-(4c+40)(c+4)\cdot1 &= (4c+40)(c+4)\cdot\frac{1}{c+4}&& \text{cancel out the denominators} \\[1 em]5c^2+58c+152-(4c^2+56c+160) &= 4c+40&& \text{simplify left side} \\[1 em]5c^2+58c+152-4c^2-56c-160 &= 4c+40&& \\[1 em]c^2+2c-8 &= 4c+40&& \text{move all terms to the left hand side } \\[1 em]c^2+2c-8-4c-40 &= 0&& \text{simplify left side} \\[1 em]c^2-2c-48 &= 0&& \\[1 em] \end{aligned} $$
$ x^{2}-2x-48 = 0 $ is a quadratic equation.
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