Sure! Let's solve this problem step-by-step.
Let's denote the numerator of the fraction as [tex]\( x \)[/tex] and the denominator as [tex]\( y \)[/tex].
We are given two conditions:
1. If 1 is added to both the numerator and the denominator, the result is [tex]\( \frac{4}{5} \)[/tex].
2. If 5 is subtracted from both the numerator and the denominator, the result is [tex]\( \frac{1}{2} \)[/tex].
Using the first condition, we can write the equation as:
[tex]\[
\frac{x + 1}{y + 1} = \frac{4}{5}
\][/tex]
Cross-multiplying to eliminate the fraction, we get:
[tex]\[
5(x + 1) = 4(y + 1)
\][/tex]
Expanding the equation:
[tex]\[
5x + 5 = 4y + 4
\][/tex]
Rearranging terms to isolate [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
5x - 4y = -1 \quad \text{(Equation 1)}
\][/tex]
Now, using the second condition, we can write the equation as:
[tex]\[
\frac{x - 5}{y - 5} = \frac{1}{2}
\][/tex]
Cross-multiplying to eliminate the fraction, we get:
[tex]\[
2(x - 5) = 1(y - 5)
\][/tex]
Simplifying the equation:
[tex]\[
2x - 10 = y - 5
\][/tex]
Rearranging terms to isolate [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
2x - y = 5 \quad \text{(Equation 2)}
\][/tex]
Now, we have a system of linear equations:
1. [tex]\( 5x - 4y = -1 \)[/tex]
2. [tex]\( 2x - y = 5 \)[/tex]
We can solve this system using the method of substitution or elimination. Here, we'll use elimination.
First, multiply Equation 2 by 4 to align the coefficients of [tex]\( y \)[/tex]:
[tex]\[
4(2x - y) = 4 \cdot 5
\][/tex]
[tex]\[
8x - 4y = 20 \quad \text{(Equation 3)}
\][/tex]
Now, subtract Equation 1 from Equation 3:
[tex]\[
(8x - 4y) - (5x - 4y) = 20 - (-1)
\][/tex]
Simplifying:
[tex]\[
3x = 21
\][/tex]
[tex]\[
x = 7
\][/tex]
Now, substitute [tex]\( x = 7 \)[/tex] back into Equation 2:
[tex]\[
2(7) - y = 5
\][/tex]
[tex]\[
14 - y = 5
\][/tex]
[tex]\[
y = 9
\][/tex]
So, the numerator [tex]\( x \)[/tex] is 7, and the denominator [tex]\( y \)[/tex] is 9.
Thus, the fraction is:
[tex]\[
\frac{7}{9}
\][/tex]
