Chapter 11: Problem 10
Solve the equation. Remember to check for extraneous solutions. $$\frac{4}{x(x+1)}=\frac{3}{x}$$
Chapter 11: Problem 10
Solve the equation. Remember to check for extraneous solutions. $$\frac{4}{x(x+1)}=\frac{3}{x}$$
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Get started for freeEvaluate the function for \(x=0,1,2,3,\) and 4. $$f(x)=x^{2}-1$$
Evaluate the expression. $$\left(-4^{-2}\right)^{-1}$$
Simplify the expression \(\frac{x}{x-1}-\frac{1}{2 x+1}\) (A) \(\frac{x-1}{(x-1)(2 x+1)}\) (B) \(-\frac{x}{x-1}\) (C) \(\frac{2 x^{2}+1}{(x-1)(2 x+1)}\) (D) \(\frac{2 x^{2}-1}{(x-1)(2 x+1)}\)
You can use \(x-3\) as the LCD when finding the sum \(\frac{5}{x-3}+\frac{2}{3-x}\) What number can you multiply the numerator and the denominator of the second fraction by to get an equivalent fraction with \(x-3\) as the new denominator?
When you add rational expressions, you may need to factor a trinomial to find the LCD. Study the sample below. Then simplify the expressions in Exercises 46–49. $$\text { Sample: } \frac{2 x}{x^{2}-1}+\frac{3}{x^{2}+x-2}=\frac{2 x}{(x+1)(x-1)}+\frac{3}{(x-1)(x+2)}$$ The LCD is \((x+1)(x-1)(x+2)\) Note: If you just used \(\left(x^{2}-1\right)\left(x^{2}+x-2\right)\) as the common denominator, the factor \((x-1)\) would be included twice. $$\frac{5 x-1}{2 x^{2}-7 x-15}-\frac{-3 x+4}{2 x^{2}+5 x+3}$$
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