# How do you solve \frac{2400}{100}=\frac{180}{x}?

Nov 13, 2017

See a couple of solution processed below:

#### Explanation:

Because both sides of the equation are pure fractions we can flip the fractions giving:

$\frac{100}{2400} = \frac{x}{180}$

Next, we can reduce the fraction on the left side of the equation:

$\frac{1}{24} = \frac{x}{180}$

Now, we can multiply each side of the equation by $\textcolor{red}{180}$ to solve for $x$ while keeping the equation balanced:

$\textcolor{red}{180} \times \frac{1}{24} = \textcolor{red}{180} \times \frac{x}{180}$

$\textcolor{red}{\left(12 \times 15\right)} \times \frac{1}{\left(12 \times 2\right)} = \cancel{\textcolor{red}{180}} \times \frac{x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{180}}}}$

$\textcolor{red}{\left(\textcolor{b l a c k}{\cancel{\textcolor{b l a c k}{12}}} \times 15\right)} \times \frac{1}{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{12}}} \times 2\right)} = x$

$\frac{15}{2} = x$

$x = \frac{15}{2}$

Another process would be to first reduce the fraction on the left side of the equation:

$\frac{24}{1} = \frac{180}{x}$

$24 = \frac{180}{x}$

Now, multiply each side of the equation by $\frac{\textcolor{red}{x}}{\textcolor{b l u e}{24}}$ to solve for $x$ while keeping the equation balanced:

$\frac{\textcolor{red}{x}}{\textcolor{b l u e}{24}} \times 24 = \frac{\textcolor{red}{x}}{\textcolor{b l u e}{24}} \times \frac{180}{x}$

$\frac{\textcolor{red}{x}}{\cancel{\textcolor{b l u e}{24}}} \times \textcolor{b l u e}{\cancel{\textcolor{b l a c k}{24}}} = \frac{\cancel{\textcolor{red}{x}}}{\textcolor{b l u e}{12 \times 2}} \times \frac{12 \times 15}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}$

$x = \frac{1}{\textcolor{b l u e}{\textcolor{b l a c k}{\cancel{\textcolor{b l u e}{12}}} \times 2}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{12}}} \times 15}{1}$

$x = \frac{15}{2}$