Question

Find the solution by elimination method if we add- I to the numerator and subtract I from the denominator a fraction reduces to 1. It becomes if we only add I to the denominator. What is the fraction.​

Answer 1

Step-by-step explanation:

Let the numerator = x,

denominator = y,

Required fraction = \frac{x}{y}---(1)

y

x

−−−(1)

According to the problem given,

if we add 1 to numerator, and subtract 1 from the denomination,the fraction reduce the 1,we get

\frac{x+1}{y-1}=1

y−1

x+1

=1

\implies x+1=y-1⟹x+1=y−1

\implies x = y-1-1⟹x=y−1−1

\implies x = y-2 ---(2)⟹x=y−2−−−(2)

And ,

If we add 1 to the denominator it becomes 1/2.

\frac{x}{y+1}=\frac{1}{2}

y+1

x

=

2

1

we add 1 to the denominator it becomes 1/2.

\frac{x}{y+1}=\frac{1}{2}

y+1

x

=

2

1

\implies x = \frac{y+1}{2}--(3)⟹x=

2

y+1

−−(3)

/* from (1) & (2), we get

y-2=\frac{y+1}{2}y−2=

2

y+1

\implies 2(y-2)=y+1⟹2(y−2)=y+1

\implies 2y-4=y+1⟹2y−4=y+1

\implies 2y-y=1+4⟹2y−y=1+4

\implies y = 5⟹y=5

Now, substitute y=5 in equation (2) ,we get

\implies x = 5-2=3⟹x=5−2=3