Chapter 7: Problem 13
Solve each rational equation. $$\frac{6}{x+3}=\frac{4}{x-3}$$
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Chapter 7: Problem 13
Solve each rational equation. $$\frac{6}{x+3}=\frac{4}{x-3}$$
These are the key concepts you need to understand to accurately answer the question.
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denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{6}{x-5}+\frac{2}{5-x}$$
After adding \(\frac{3 x+1}{4}\) and \(\frac{x+2}{4},\) I simplified the sum by dividing the numerator and the denominator by 4 I use similar procedures to find each of the following sums: $$ \frac{3}{8}+\frac{1}{8} \text { and } \frac{x}{x^{2}-1}+\frac{1}{x^{2}-1} $$
Add or subtract as indicated. Simplify the result, if possible. $$\frac{x-1}{6}+\frac{x+2}{3}$$
Add or subtract as indicated. Simplify the result, if possible. $$\frac{3}{x-2}+\frac{4}{x+3}$$
denominators are opposites, or additive inverses. Add or subtract as indicated. Simplify the result, if possible. $$\frac{3-x}{x-7}-\frac{2 x-5}{7-x}$$
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