+2440 1)
\(\frac{\textcolor{red}{\frac{3}{x}+\frac{1}{4}}}{\textcolor{blue}{1+\frac{3}{x}}}\) is quite the complex fraction. First, let's just deal with the numerator
| \(\textcolor{red}{\frac{3}{x}+\frac{1}{4}}\) | First. let's change this into fractions with common denominators. The LCM is 4x. |
| \(\frac{12}{4x}+\frac{x}{4x}\) | Now, combine the fractions because we have formed a common denominator. |
| \(\frac{12+x}{4x}\) | |
| \(\textcolor{blue}{1+\frac{3}{x}}\) | Just like before, transform the fractions to create a common denominator and combine. |
| \(\frac{x}{x}+\frac{3}{x}\) | |
| \(\frac{x+3}{x}\) | |
Now, write it as a fraction; you'll see how much easier it is to work with!
| \(\frac{\textcolor{red}{\frac{3}{x}+\frac{1}{4}}}{\textcolor{blue}{1+\frac{3}{x}}}=\frac{\frac{12+x}{4x}}{\frac{x+3}{x}}\) | |
| \(\frac{\frac{12+x}{4x}}{\frac{x+3}{x}}\) | Multiply by \(\frac{x}{x+3}\), the reciprocal of the complex denominator, to eliminate this complex fraction. |
| \(\frac{x(12+x)}{4x(x+3)}\) | The x in the numerator and the x in the denominator cancel out here. |
| \(\frac{12+x}{4(x+3)}\) | Now, distribute the 4 into every term. |
| \(\frac{12+x}{4x+12}\) | |
2) \(\frac{c^2-4c+4}{12c^3+30c^2}\div\frac{c^2-4}{6c^4+15c^3}\)
Dividing by a fraction is the same as multiplying by its reciprocal.
\(\frac{c^2-4c+4}{12c^3+30c^2}*\frac{6c^4+15c^3}{c^2-4}\)
Now, let's factor the numerators and denominators completely and fully and see if any canceling can occur to simplify this.
| \(\frac{c^2-4c+4}{12c^3+30c^2}*\frac{6c^4+15c^3}{c^2-4}\) | Factor everything fully. |
| \(\frac{(c-2)^2}{6c^2(2c+5)}*\frac{3c^3(2c+5)}{(c+2)(c-2)}\) | I see a lot of canceling that will occur here, dont you? |
| \(\frac{c-2}{2}*\frac{c}{c+2}\) | Now, combine. |
| \(\frac{c(c-2)}{2(c+2)}\) | This answer corresponds to the third one listed in the multiple guess. |
+2440 1)
\(\frac{\textcolor{red}{\frac{3}{x}+\frac{1}{4}}}{\textcolor{blue}{1+\frac{3}{x}}}\) is quite the complex fraction. First, let's just deal with the numerator
| \(\textcolor{red}{\frac{3}{x}+\frac{1}{4}}\) | First. let's change this into fractions with common denominators. The LCM is 4x. |
| \(\frac{12}{4x}+\frac{x}{4x}\) | Now, combine the fractions because we have formed a common denominator. |
| \(\frac{12+x}{4x}\) | |
| \(\textcolor{blue}{1+\frac{3}{x}}\) | Just like before, transform the fractions to create a common denominator and combine. |
| \(\frac{x}{x}+\frac{3}{x}\) | |
| \(\frac{x+3}{x}\) | |
Now, write it as a fraction; you'll see how much easier it is to work with!
| \(\frac{\textcolor{red}{\frac{3}{x}+\frac{1}{4}}}{\textcolor{blue}{1+\frac{3}{x}}}=\frac{\frac{12+x}{4x}}{\frac{x+3}{x}}\) | |
| \(\frac{\frac{12+x}{4x}}{\frac{x+3}{x}}\) | Multiply by \(\frac{x}{x+3}\), the reciprocal of the complex denominator, to eliminate this complex fraction. |
| \(\frac{x(12+x)}{4x(x+3)}\) | The x in the numerator and the x in the denominator cancel out here. |
| \(\frac{12+x}{4(x+3)}\) | Now, distribute the 4 into every term. |
| \(\frac{12+x}{4x+12}\) | |
2) \(\frac{c^2-4c+4}{12c^3+30c^2}\div\frac{c^2-4}{6c^4+15c^3}\)
Dividing by a fraction is the same as multiplying by its reciprocal.
\(\frac{c^2-4c+4}{12c^3+30c^2}*\frac{6c^4+15c^3}{c^2-4}\)
Now, let's factor the numerators and denominators completely and fully and see if any canceling can occur to simplify this.
| \(\frac{c^2-4c+4}{12c^3+30c^2}*\frac{6c^4+15c^3}{c^2-4}\) | Factor everything fully. |
| \(\frac{(c-2)^2}{6c^2(2c+5)}*\frac{3c^3(2c+5)}{(c+2)(c-2)}\) | I see a lot of canceling that will occur here, dont you? |
| \(\frac{c-2}{2}*\frac{c}{c+2}\) | Now, combine. |
| \(\frac{c(c-2)}{2(c+2)}\) | This answer corresponds to the third one listed in the multiple guess. |
