Other ways to find $x$ in $\frac{x}{x+7}=\frac49$

Question

During solving a problem I got this equation and I want to solve for $x$:

$$\frac{x}{x+7}=\frac49$$ It is equivalent to solving $9x=4(x+7)$ hence $x=\frac{28}5$.

But is it possible to solve it other ways? I mean we have $x$ in both numerator and denominator of the fraction so how we can write it so that we have only one $x$ and solve it faster (I'm preparing for a timed exam) ?

Answer 1

Use componendo-dividendo.

Here,

$$\require{cancel}\frac{x}{x+7} = \frac{4}{9} \Rightarrow \frac{x}{(\cancel{x}+7)-\cancel{x}} = \frac{4}{9-4} \Rightarrow x = \frac{28}{5}$$

Answer 2

Invert the fractions so that you have:

$$\frac{x+7}{x} = \frac{9}{4} \implies \frac{7}{x} = \frac{5}{4} \implies x= \frac{4}{5} \cdot 7 = \frac{28}{5}.$$

Answer 3

This isn’t really different from the others, but it’s approaching it with a different “feel”.

If $\frac{x}{x+7}$ is $\frac49$, then $\frac{7}{x+7}$ must be the “other” $\frac59$. Clearly the two numerators are in the ratio $7:5$, and the denominators haven’t changed, so that ratio also applies to the original numerators. Thus, $x$ is just $4\times 7/5$, or $28/5$.