Question

The numerator of a certain fraction is five times the denominator. If nine is added to both the numerator and the denominator, the resulting fraction is equivalent to two. What was the original fraction (not written in lowest terms)?

Answer 1

Step-by-step explanation

To solve the question,

Let

The numerator = x

The denominator = y

So that the original equation will be

`(x)/(y)`

Next, we are told that the numerator is five times the denominator.

So that

`x=5y`

Again, we are told that If nine is added to both the numerator and the denominator, the resulting fraction is equivalent to two. so

`(x+9)/(y+9)=2`

Hence

we can substitute x =5y into the above

`\begin{gathered} (5y+9)/(y+9)=2 \\ \\ cross\text{ multiplying} \end{gathered}`
`\begin{gathered} 5y+9=2(y+9) \\ 5y+9=2y+18 \\ Taking\text{ like terms} \\ 5y-2y=18-9 \\ 3y=9 \\ \\ y=(9)/(3) \\ \\ y=3 \end{gathered}`

Thus, the denominator is 3

The numerator will be

`\begin{gathered} x=5y \\ x=5*3 \\ x=15 \end{gathered}`

The numerator is 15

Therefore, the fraction is