Let x = the numerator part of the original fraction. Since the ratio of the numerator to denominator is one to four, this means 4x is the denominator.
[tex] \frac{x}{4x} [/tex] is our original fraction.
Now we work with adding and subtracing to get to something that looks like one-third.
[tex] \frac{x+3}{4x-3} = \frac{x}{3x } [/tex]
We write the second half as x / 3x because the fraction can simplify to 1/3. The rest of the problem is solved by cross multiplying (essentially, clearing fractions and simplifying) and solving for x.
[tex] \frac{x+3}{4x-3} = \frac{x}{3x } [/tex]
(3x)(x + 3) = (4x -3)(x)
3x² + 9x = 4x² - 3x
Since it's a quadratic, let's set one side equal to zero and use the Zero Product Property (ZPP).
9x = x² - 3x
0 = x² - 12x
x is a common factor in x²-12x, so we can use the ZPP and factor
0 = x (x - 12)
So x = 0 or x =12.
Having zero as an answer won't work as we'd be dividing by zero. Let's check x = 12.
12/48 is the starting fraction, which simplifies to 1/4
12+3 / 48 -3 = 15 / 45 = 1/3.
Thus, 12/48 was the starting fraction.
Answer: The required original fraction is [tex]\dfrac{12}{48}.[/tex]
Step-by-step explanation: Given that the ratio of the numerator to the denominator of a certain fraction is one to four.
Also, if three is added to the numerator and subtracted from the denominator, the new fraction reduces to one-third.
We are to find the original fraction.
Let x and y represents the numerator and denominator of the original fraction.
Then, according to the given information, we have
[tex]x:y=1:4\\\\\Rightarrow \dfrac{x}{y}=\dfrac{1}{4}\\\\\Rightarrow y=4x~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
and
[tex]\dfrac{x+3}{y-3}=\dfrac{1}{3}\\\\\Rightarrow 3(x+3)=1(y-3)\\\\\Rightarrow 3x+9=y-3\\\\\Rightarrow 3x+9=4x-3~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~[\textup{using equation (i)}]\\\\\Rightarrow 4x-3x=9+3\\\\\Rightarrow x=12.[/tex]
From equation (i), we get
[tex]y=4\times12=48.[/tex]
Thus, the required original fraction is [tex]\dfrac{12}{48}.[/tex]
